We Keep Coding

Multiply a string by itself in javascript [duplicate]

This question already has answers here:
Repeat a string in JavaScript a number of times
(24 answers)
Closed last year.
What is the best or most concise method for returning a string repeated an arbitrary amount of times?
The following is my best shot so far:
function repeat(s, n){
var a = [];
while(a.length < n){
a.push(s);
}
return a.join('');
}

Note to new readers: This answer is old and and not terribly practical - it's just "clever" because it uses Array stuff to get
String things done. When I wrote "less process" I definitely meant
"less code" because, as others have noted in subsequent answers, it
performs like a pig. So don't use it if speed matters to you.
I'd put this function onto the String object directly. Instead of creating an array, filling it, and joining it with an empty char, just create an array of the proper length, and join it with your desired string. Same result, less process!
String.prototype.repeat = function( num )
{
return new Array( num + 1 ).join( this );
}
alert( "string to repeat\n".repeat( 4 ) );

I've tested the performance of all the proposed approaches.
Here is the fastest variant I've got.
String.prototype.repeat = function(count) {
if (count < 1) return '';
var result = '', pattern = this.valueOf();
while (count > 1) {
if (count & 1) result += pattern;
count >>= 1, pattern += pattern;
}
return result + pattern;
};
Or as stand-alone function:
function repeat(pattern, count) {
if (count < 1) return '';
var result = '';
while (count > 1) {
if (count & 1) result += pattern;
count >>= 1, pattern += pattern;
}
return result + pattern;
}
It is based on wnrph's algorithm.
It is really fast. And the bigger the count, the faster it goes compared with the traditional new Array(count + 1).join(string) approach.
I've only changed 2 things:
replaced pattern = this with pattern = this.valueOf() (clears one obvious type conversion);
added if (count < 1) check from prototypejs to the top of function to exclude unnecessary actions in that case.
applied optimisation from Dennis answer (5-7% speed up)
UPD
Created a little performance-testing playground here for those who interested.
variable count ~ 0 .. 100:
constant count = 1024:
Use it and make it even faster if you can :)

This problem is a well-known / "classic" optimization issue for JavaScript, caused by the fact that JavaScript strings are "immutable" and addition by concatenation of even a single character to a string requires creation of, including memory allocation for and copying to, an entire new string.
Unfortunately, the accepted answer on this page is wrong, where "wrong" means by a performance factor of 3x for simple one-character strings, and 8x-97x for short strings repeated more times, to 300x for repeating sentences, and infinitely wrong when taking the limit of the ratios of complexity of the algorithms as n goes to infinity. Also, there is another answer on this page which is almost right (based on one of the many generations and variations of the correct solution circulating throughout the Internet in the past 13 years). However, this "almost right" solution misses a key point of the correct algorithm causing a 50% performance degradation.
JS Performance Results for the accepted answer, the top-performing other answer (based on a degraded version of the original algorithm in this answer), and this answer using my algorithm created 13 years ago
~ October 2000 I published an algorithm for this exact problem which was widely adapted, modified, then eventually poorly understood and forgotten. To remedy this issue, in August, 2008 I published an article http://www.webreference.com/programming/javascript/jkm3/3.html explaining the algorithm and using it as an example of simple of general-purpose JavaScript optimizations. By now, Web Reference has scrubbed my contact information and even my name from this article. And once again, the algorithm has been widely adapted, modified, then poorly understood and largely forgotten.
Original string repetition/multiplication JavaScript algorithm by
Joseph Myers, circa Y2K as a text multiplying function within Text.js;
published August, 2008 in this form by Web Reference:
http://www.webreference.com/programming/javascript/jkm3/3.html (The
article used the function as an example of JavaScript optimizations,
which is the only for the strange name "stringFill3.")
/*
* Usage: stringFill3("abc", 2) == "abcabc"
*/
function stringFill3(x, n) {
var s = '';
for (;;) {
if (n & 1) s += x;
n >>= 1;
if (n) x += x;
else break;
}
return s;
}
Within two months after publication of that article, this same question was posted to Stack Overflow and flew under my radar until now, when apparently the original algorithm for this problem has once again been forgotten. The best solution available on this Stack Overflow page is a modified version of my solution, possibly separated by several generations. Unfortunately, the modifications ruined the solution's optimality. In fact, by changing the structure of the loop from my original, the modified solution performs a completely unneeded extra step of exponential duplicating (thus joining the largest string used in the proper answer with itself an extra time and then discarding it).
Below ensues a discussion of some JavaScript optimizations related to all of the answers to this problem and for the benefit of all.
Technique: Avoid references to objects or object properties
To illustrate how this technique works, we use a real-life JavaScript function which creates strings of whatever length is needed. And as we'll see, more optimizations can be added!
A function like the one used here is to create padding to align columns of text, for formatting money, or for filling block data up to the boundary. A text generation function also allows variable length input for testing any other function that operates on text. This function is one of the important components of the JavaScript text processing module.
As we proceed, we will be covering two more of the most important optimization techniques while developing the original code into an optimized algorithm for creating strings. The final result is an industrial-strength, high-performance function that I've used everywhere--aligning item prices and totals in JavaScript order forms, data formatting and email / text message formatting and many other uses.
Original code for creating strings stringFill1()
function stringFill1(x, n) {
var s = '';
while (s.length < n) s += x;
return s;
}
/* Example of output: stringFill1('x', 3) == 'xxx' */
The syntax is here is clear. As you can see, we've used local function variables already, before going on to more optimizations.
Be aware that there's one innocent reference to an object property s.length in the code that hurts its performance. Even worse, the use of this object property reduces the simplicity of the program by making the assumption that the reader knows about the properties of JavaScript string objects.
The use of this object property destroys the generality of the computer program. The program assumes that x must be a string of length one. This limits the application of the stringFill1() function to anything except repetition of single characters. Even single characters cannot be used if they contain multiple bytes like the HTML entity .
The worst problem caused by this unnecessary use of an object property is that the function creates an infinite loop if tested on an empty input string x. To check generality, apply a program to the smallest possible amount of input. A program which crashes when asked to exceed the amount of available memory has an excuse. A program like this one which crashes when asked to produce nothing is unacceptable. Sometimes pretty code is poisonous code.
Simplicity may be an ambiguous goal of computer programming, but generally it's not. When a program lacks any reasonable level of generality, it's not valid to say, "The program is good enough as far as it goes." As you can see, using the string.length property prevents this program from working in a general setting, and in fact, the incorrect program is ready to cause a browser or system crash.
Is there a way to improve the performance of this JavaScript as well as take care of these two serious problems?
Of course. Just use integers.
Optimized code for creating strings stringFill2()
function stringFill2(x, n) {
var s = '';
while (n-- > 0) s += x;
return s;
}
Timing code to compare stringFill1() and stringFill2()
function testFill(functionToBeTested, outputSize) {
var i = 0, t0 = new Date();
do {
functionToBeTested('x', outputSize);
t = new Date() - t0;
i++;
} while (t < 2000);
return t/i/1000;
}
seconds1 = testFill(stringFill1, 100);
seconds2 = testFill(stringFill2, 100);
The success so far of stringFill2()
stringFill1() takes 47.297 microseconds (millionths of a second) to fill a 100-byte string, and stringFill2() takes 27.68 microseconds to do the same thing. That's almost a doubling in performance by avoiding a reference to an object property.
Technique: Avoid adding short strings to long strings
Our previous result looked good--very good, in fact. The improved function stringFill2() is much faster due to the use of our first two optimizations. Would you believe it if I told you that it can be improved to be many times faster than it is now?
Yes, we can accomplish that goal. Right now we need to explain how we avoid appending short strings to long strings.
The short-term behavior appears to be quite good, in comparison to our original function. Computer scientists like to analyze the "asymptotic behavior" of a function or computer program algorithm, which means to study its long-term behavior by testing it with larger inputs. Sometimes without doing further tests, one never becomes aware of ways that a computer program could be improved. To see what will happen, we're going to create a 200-byte string.
The problem that shows up with stringFill2()
Using our timing function, we find that the time increases to 62.54 microseconds for a 200-byte string, compared to 27.68 for a 100-byte string. It seems like the time should be doubled for doing twice as much work, but instead it's tripled or quadrupled. From programming experience, this result seems strange, because if anything, the function should be slightly faster since work is being done more efficiently (200 bytes per function call rather than 100 bytes per function call). This issue has to do with an insidious property of JavaScript strings: JavaScript strings are "immutable."
Immutable means that you cannot change a string once it's created. By adding on one byte at a time, we're not using up one more byte of effort. We're actually recreating the entire string plus one more byte.
In effect, to add one more byte to a 100-byte string, it takes 101 bytes worth of work. Let's briefly analyze the computational cost for creating a string of N bytes. The cost of adding the first byte is 1 unit of computational effort. The cost of adding the second byte isn't one unit but 2 units (copying the first byte to a new string object as well as adding the second byte). The third byte requires a cost of 3 units, etc.
C(N) = 1 + 2 + 3 + ... + N = N(N+1)/2 = O(N^2). The symbol O(N^2) is pronounced Big O of N squared, and it means that the computational cost in the long run is proportional to the square of the string length. To create 100 characters takes 10,000 units of work, and to create 200 characters takes 40,000 units of work.
This is why it took more than twice as long to create 200 characters than 100 characters. In fact, it should have taken four times as long. Our programming experience was correct in that the work is being done slightly more efficiently for longer strings, and hence it took only about three times as long. Once the overhead of the function call becomes negligible as to how long of a string we're creating, it will actually take four times as much time to create a string twice as long.
(Historical note: This analysis doesn't necessarily apply to strings in source code, such as html = 'abcd\n' + 'efgh\n' + ... + 'xyz.\n', since the JavaScript source code compiler can join the strings together before making them into a JavaScript string object. Just a few years ago, the KJS implementation of JavaScript would freeze or crash when loading long strings of source code joined by plus signs. Since the computational time was O(N^2) it wasn't difficult to make Web pages which overloaded the Konqueror Web browser or Safari, which used the KJS JavaScript engine core. I first came across this issue when I was developing a markup language and JavaScript markup language parser, and then I discovered what was causing the problem when I wrote my script for JavaScript Includes.)
Clearly this rapid degradation of performance is a huge problem. How can we deal with it, given that we cannot change JavaScript's way of handling strings as immutable objects? The solution is to use an algorithm which recreates the string as few times as possible.
To clarify, our goal is to avoid adding short strings to long strings, since in order to add the short string, the entire long string also must be duplicated.
How the algorithm works to avoid adding short strings to long strings
Here's a good way to reduce the number of times new string objects are created. Concatenate longer lengths of string together so that more than one byte at a time is added to the output.
For instance, to make a string of length N = 9:
x = 'x';
s = '';
s += x; /* Now s = 'x' */
x += x; /* Now x = 'xx' */
x += x; /* Now x = 'xxxx' */
x += x; /* Now x = 'xxxxxxxx' */
s += x; /* Now s = 'xxxxxxxxx' as desired */
Doing this required creating a string of length 1, creating a string of length 2, creating a string of length 4, creating a string of length 8, and finally, creating a string of length 9. How much cost have we saved?
Old cost C(9) = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 9 = 45.
New cost C(9) = 1 + 2 + 4 + 8 + 9 = 24.
Note that we had to add a string of length 1 to a string of length 0, then a string of length 1 to a string of length 1, then a string of length 2 to a string of length 2, then a string of length 4 to a string of length 4, then a string of length 8 to a string of length 1, in order to obtain a string of length 9. What we're doing can be summarized as avoiding adding short strings to long strings, or in other words, trying to concatenate strings together that are of equal or nearly equal length.
For the old computational cost we found a formula N(N+1)/2. Is there a formula for the new cost? Yes, but it's complicated. The important thing is that it is O(N), and so doubling the string length will approximately double the amount of work rather than quadrupling it.
The code that implements this new idea is nearly as complicated as the formula for the computational cost. When you read it, remember that >>= 1 means to shift right by 1 byte. So if n = 10011 is a binary number, then n >>= 1 results in the value n = 1001.
The other part of the code you might not recognize is the bitwise and operator, written &. The expression n & 1 evaluates true if the last binary digit of n is 1, and false if the last binary digit of n is 0.
New highly-efficient stringFill3() function
function stringFill3(x, n) {
var s = '';
for (;;) {
if (n & 1) s += x;
n >>= 1;
if (n) x += x;
else break;
}
return s;
}
It looks ugly to the untrained eye, but it's performance is nothing less than lovely.
Let's see just how well this function performs. After seeing the results, it's likely that you'll never forget the difference between an O(N^2) algorithm and an O(N) algorithm.
stringFill1() takes 88.7 microseconds (millionths of a second) to create a 200-byte string, stringFill2() takes 62.54, and stringFill3() takes only 4.608. What made this algorithm so much better? All of the functions took advantage of using local function variables, but taking advantage of the second and third optimization techniques added a twenty-fold improvement to performance of stringFill3().
Deeper analysis
What makes this particular function blow the competition out of the water?
As I've mentioned, the reason that both of these functions, stringFill1() and stringFill2(), run so slowly is that JavaScript strings are immutable. Memory cannot be reallocated to allow one more byte at a time to be appended to the string data stored by JavaScript. Every time one more byte is added to the end of the string, the entire string is regenerated from beginning to end.
Thus, in order to improve the script's performance, one must precompute longer length strings by concatenating two strings together ahead of time, and then recursively building up the desired string length.
For instance, to create a 16-letter byte string, first a two byte string would be precomputed. Then the two byte string would be reused to precompute a four-byte string. Then the four-byte string would be reused to precompute an eight byte string. Finally, two eight-byte strings would be reused to create the desired new string of 16 bytes. Altogether four new strings had to be created, one of length 2, one of length 4, one of length 8 and one of length 16. The total cost is 2 + 4 + 8 + 16 = 30.
In the long run this efficiency can be computed by adding in reverse order and using a geometric series starting with a first term a1 = N and having a common ratio of r = 1/2. The sum of a geometric series is given by a_1 / (1-r) = 2N.
This is more efficient than adding one character to create a new string of length 2, creating a new string of length 3, 4, 5, and so on, until 16. The previous algorithm used that process of adding a single byte at a time, and the total cost of it would be n (n + 1) / 2 = 16 (17) / 2 = 8 (17) = 136.
Obviously, 136 is a much greater number than 30, and so the previous algorithm takes much, much more time to build up a string.
To compare the two methods you can see how much faster the recursive algorithm (also called "divide and conquer") is on a string of length 123,457. On my FreeBSD computer this algorithm, implemented in the stringFill3() function, creates the string in 0.001058 seconds, while the original stringFill1() function creates the string in 0.0808 seconds. The new function is 76 times faster.
The difference in performance grows as the length of the string becomes larger. In the limit as larger and larger strings are created, the original function behaves roughly like C1 (constant) times N^2, and the new function behaves like C2 (constant) times N.
From our experiment we can determine the value of C1 to be C1 = 0.0808 / (123457)2 = .00000000000530126997, and the value of C2 to be C2 = 0.001058 / 123457 = .00000000856978543136. In 10 seconds, the new function could create a string containing 1,166,890,359 characters. In order to create this same string, the old function would need 7,218,384 seconds of time.
This is almost three months compared to ten seconds!
I'm only answering (several years late) because my original solution to this problem has been floating around the Internet for more than 10 years, and apparently is still poorly-understood by the few who do remember it. I thought that by writing an article about it here I would help:
Performance Optimizations for High Speed JavaScript / Page 3
Unfortunately, some of the other solutions presented here are still some of those that would take three months to produce the same amount of output that a proper solution creates in 10 seconds.
I want to take the time to reproduce part of the article here as a canonical answer on Stack Overflow.
Note that the best-performing algorithm here is clearly based on my algorithm and was probably inherited from someone else's 3rd or 4th generation adaptation. Unfortunately, the modifications resulted in reducing its performance. The variation of my solution presented here perhaps did not understand my confusing for (;;) expression which looks like the main infinite loop of a server written in C, and which was simply designed to allow a carefully-positioned break statement for loop control, the most compact way to avoid exponentially replicating the string one extra unnecessary time.

Good news! String.prototype.repeat is now a part of JavaScript.
"yo".repeat(2);
// returns: "yoyo"
The method is supported by all major browsers, except Internet Explorer. For an up to date list, see MDN: String.prototype.repeat > Browser compatibility.
MDN has a polyfill for browsers without support.

This one is pretty efficient
String.prototype.repeat = function(times){
var result="";
var pattern=this;
while (times > 0) {
if (times&1)
result+=pattern;
times>>=1;
pattern+=pattern;
}
return result;
};

String.prototype.repeat is now ES6 Standard.
'abc'.repeat(3); //abcabcabc

Expanding P.Bailey's solution:
String.prototype.repeat = function(num) {
return new Array(isNaN(num)? 1 : ++num).join(this);
}
This way you should be safe from unexpected argument types:
var foo = 'bar';
alert(foo.repeat(3)); // Will work, "barbarbar"
alert(foo.repeat('3')); // Same as above
alert(foo.repeat(true)); // Same as foo.repeat(1)
alert(foo.repeat(0)); // This and all the following return an empty
alert(foo.repeat(false)); // string while not causing an exception
alert(foo.repeat(null));
alert(foo.repeat(undefined));
alert(foo.repeat({})); // Object
alert(foo.repeat(function () {})); // Function
EDIT: Credits to jerone for his elegant ++num idea!

Use Array(N+1).join("string_to_repeat")

Here's a 5-7% improvement on disfated's answer.
Unroll the loop by stopping at count > 1 and perform an additional result += pattnern concat after the loop. This will avoid the loops final previously unused pattern += pattern without having to use an expensive if-check.
The final result would look like this:
String.prototype.repeat = function(count) {
if (count < 1) return '';
var result = '', pattern = this.valueOf();
while (count > 1) {
if (count & 1) result += pattern;
count >>= 1, pattern += pattern;
}
result += pattern;
return result;
};
And here's disfated's fiddle forked for the unrolled version: http://jsfiddle.net/wsdfg/

/**
#desc: repeat string
#param: n - times
#param: d - delimiter
*/
String.prototype.repeat = function (n, d) {
return --n ? this + (d || '') + this.repeat(n, d) : '' + this
};
this is how to repeat string several times using delimeter.

Tests of the various methods:
var repeatMethods = {
control: function (n,s) {
/* all of these lines are common to all methods */
if (n==0) return '';
if (n==1 || isNaN(n)) return s;
return '';
},
divideAndConquer: function (n, s) {
if (n==0) return '';
if (n==1 || isNaN(n)) return s;
with(Math) { return arguments.callee(floor(n/2), s)+arguments.callee(ceil(n/2), s); }
},
linearRecurse: function (n,s) {
if (n==0) return '';
if (n==1 || isNaN(n)) return s;
return s+arguments.callee(--n, s);
},
newArray: function (n, s) {
if (n==0) return '';
if (n==1 || isNaN(n)) return s;
return (new Array(isNaN(n) ? 1 : ++n)).join(s);
},
fillAndJoin: function (n, s) {
if (n==0) return '';
if (n==1 || isNaN(n)) return s;
var ret = [];
for (var i=0; i<n; i++)
ret.push(s);
return ret.join('');
},
concat: function (n,s) {
if (n==0) return '';
if (n==1 || isNaN(n)) return s;
var ret = '';
for (var i=0; i<n; i++)
ret+=s;
return ret;
},
artistoex: function (n,s) {
var result = '';
while (n>0) {
if (n&1) result+=s;
n>>=1, s+=s;
};
return result;
}
};
function testNum(len, dev) {
with(Math) { return round(len+1+dev*(random()-0.5)); }
}
function testString(len, dev) {
return (new Array(testNum(len, dev))).join(' ');
}
var testTime = 1000,
tests = {
biggie: { str: { len: 25, dev: 12 }, rep: {len: 200, dev: 50 } },
smalls: { str: { len: 5, dev: 5}, rep: { len: 5, dev: 5 } }
};
var testCount = 0;
var winnar = null;
var inflight = 0;
for (var methodName in repeatMethods) {
var method = repeatMethods[methodName];
for (var testName in tests) {
testCount++;
var test = tests[testName];
var testId = methodName+':'+testName;
var result = {
id: testId,
testParams: test
}
result.count=0;
(function (result) {
inflight++;
setTimeout(function () {
result.start = +new Date();
while ((new Date() - result.start) < testTime) {
method(testNum(test.rep.len, test.rep.dev), testString(test.str.len, test.str.dev));
result.count++;
}
result.end = +new Date();
result.rate = 1000*result.count/(result.end-result.start)
console.log(result);
if (winnar === null || winnar.rate < result.rate) winnar = result;
inflight--;
if (inflight==0) {
console.log('The winner: ');
console.log(winnar);
}
}, (100+testTime)*testCount);
}(result));
}
}

Here's the JSLint safe version
String.prototype.repeat = function (num) {
var a = [];
a.length = num << 0 + 1;
return a.join(this);
};

For all browsers
This is about as concise as it gets :
function repeat(s, n) { return new Array(n+1).join(s); }
If you also care about performance, this is a much better approach :
function repeat(s, n) { var a=[],i=0;for(;i<n;)a[i++]=s;return a.join(''); }
If you want to compare the performance of both options, see this Fiddle and this Fiddle for benchmark tests. During my own tests, the second option was about 2 times faster in Firefox and about 4 times faster in Chrome!
For moderns browsers only :
In modern browsers, you can now also do this :
function repeat(s,n) { return s.repeat(n) };
This option is not only shorter than both other options, but it's even faster than the second option.
Unfortunately, it doesn't work in any version of Internet explorer. The numbers in the table specify the first browser version that fully supports the method :

function repeat(pattern, count) {
for (var result = '';;) {
if (count & 1) {
result += pattern;
}
if (count >>= 1) {
pattern += pattern;
} else {
return result;
}
}
}
You can test it at JSFiddle. Benchmarked against the hacky Array.join and mine is, roughly speaking, 10 (Chrome) to 100 (Safari) to 200 (Firefox) times faster (depending on the browser).

Just another repeat function:
function repeat(s, n) {
var str = '';
for (var i = 0; i < n; i++) {
str += s;
}
return str;
}

There are many ways in the ES-Next ways
1. ES2015/ES6 has been realized this repeat() method!
/**
* str: String
* count: Number
*/
const str = `hello repeat!\n`, count = 3;
let resultString = str.repeat(count);
console.log(`resultString = \n${resultString}`);
/*
resultString =
hello repeat!
hello repeat!
hello repeat!
*/
({ toString: () => 'abc', repeat: String.prototype.repeat }).repeat(2);
// 'abcabc' (repeat() is a generic method)
// Examples
'abc'.repeat(0); // ''
'abc'.repeat(1); // 'abc'
'abc'.repeat(2); // 'abcabc'
'abc'.repeat(3.5); // 'abcabcabc' (count will be converted to integer)
// 'abc'.repeat(1/0); // RangeError
// 'abc'.repeat(-1); // RangeError
2. ES2017/ES8 new add String.prototype.padStart()
const str = 'abc ';
const times = 3;
const newStr = str.padStart(str.length * times, str.toUpperCase());
console.log(`newStr =`, newStr);
// "newStr =" "ABC ABC abc "
3. ES2017/ES8 new add String.prototype.padEnd()
const str = 'abc ';
const times = 3;
const newStr = str.padEnd(str.length * times, str.toUpperCase());
console.log(`newStr =`, newStr);
// "newStr =" "abc ABC ABC "
refs
http://www.ecma-international.org/ecma-262/6.0/#sec-string.prototype.repeat
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/repeat
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padEnd

function repeat(s, n) { var r=""; for (var a=0;a<n;a++) r+=s; return r;}

This may be the smallest recursive one:-
String.prototype.repeat = function(n,s) {
s = s || ""
if(n>0) {
s += this
s = this.repeat(--n,s)
}
return s}

Fiddle: http://jsfiddle.net/3Y9v2/
function repeat(s, n){
return ((new Array(n+1)).join(s));
}
alert(repeat('R', 10));

Simple recursive concatenation
I just wanted to give it a bash, and made this:
function ditto( s, r, c ) {
return c-- ? ditto( s, r += s, c ) : r;
}
ditto( "foo", "", 128 );
I can't say I gave it much thought, and it probably shows :-)
This is arguably better
String.prototype.ditto = function( c ) {
return --c ? this + this.ditto( c ) : this;
};
"foo".ditto( 128 );
And it's a lot like an answer already posted - I know this.
But why be recursive at all?
And how about a little default behaviour too?
String.prototype.ditto = function() {
var c = Number( arguments[ 0 ] ) || 2,
r = this.valueOf();
while ( --c ) {
r += this;
}
return r;
}
"foo".ditto();
Because, although the non recursive method will handle arbitrarily large repeats without hitting call stack limits, it's a lot slower.
Why did I bother adding more methods that aren't half as clever as those already posted?
Partly for my own amusement, and partly to point out in the simplest way I know that there are many ways to skin a cat, and depending on the situation, it's quite possible that the apparently best method isn't ideal.
A relatively fast and sophisticated method may effectively crash and burn under certain circumstances, whilst a slower, simpler method may get the job done - eventually.
Some methods may be little more than exploits, and as such prone to being fixed out of existence, and other methods may work beautifully in all conditions, but are so constructed that one simply has no idea how it works.
"So what if I dunno how it works?!"
Seriously?
JavaScript suffers from one of its greatest strengths; it's highly tolerant of bad behaviour, and so flexible it'll bend over backwards to return results, when it might have been better for everyone if it'd snapped!
"With great power, comes great responsibility" ;-)
But more seriously and importantly, although general questions like this do lead to awesomeness in the form of clever answers that if nothing else, expand one's knowledge and horizons, in the end, the task at hand - the practical script that uses the resulting method - may require a little less, or a little more clever than is suggested.
These "perfect" algorithms are fun and all, but "one size fits all" will rarely if ever be better than tailor made.
This sermon was brought to you courtesy of a lack of sleep and a passing interest.
Go forth and code!

Firstly, the OP's questions seems to be about conciseness - which I understand to mean "simple and easy to read", while most answers seem to be about efficiency - which is obviously not the same thing and also I think that unless you implement some very specific large data manipulating algorithms, shouldn't worry you when you come to implement basic data manipulation Javascript functions. Conciseness is much more important.
Secondly, as André Laszlo noted, String.repeat is part of ECMAScript 6 and already available in several popular implementations - so the most concise implementation of String.repeat is not to implement it ;-)
Lastly, if you need to support hosts that don't offer the ECMAScript 6 implementation, MDN's polyfill mentioned by André Laszlo is anything but concise.
So, without further ado - here is my concise polyfill:
String.prototype.repeat = String.prototype.repeat || function(n){
return n<=1 ? this : this.concat(this.repeat(n-1));
}
Yes, this is a recursion. I like recursions - they are simple and if done correctly are easy to understand. Regarding efficiency, if the language supports it they can be very efficient if written correctly.
From my tests, this method is ~60% faster than the Array.join approach. Although it obviously comes nowhere close disfated's implementation, it is much simpler than both.
My test setup is node v0.10, using "Strict mode" (I think it enables some sort of TCO), calling repeat(1000) on a 10 character string a million times.

If you think all those prototype definitions, array creations, and join operations are overkill, just use a single line code where you need it. String S repeating N times:
for (var i = 0, result = ''; i < N; i++) result += S;

Use Lodash for Javascript utility functionality, like repeating strings.
Lodash provides nice performance and ECMAScript compatibility.
I highly recommend it for UI development and it works well server side, too.
Here's how to repeat the string "yo" 2 times using Lodash:
> _.repeat('yo', 2)
"yoyo"

Recursive solution using divide and conquer:
function repeat(n, s) {
if (n==0) return '';
if (n==1 || isNaN(n)) return s;
with(Math) { return repeat(floor(n/2), s)+repeat(ceil(n/2), s); }
}

I came here randomly and never had a reason to repeat a char in javascript before.
I was impressed by artistoex's way of doing it and disfated's results. I noticed that the last string concat was unnecessary, as Dennis also pointed out.
I noticed a few more things when playing with the sampling disfated put together.
The results varied a fair amount often favoring the last run and similar algorithms would often jockey for position. One of the things I changed was instead of using the JSLitmus generated count as the seed for the calls; as count was generated different for the various methods, I put in an index. This made the thing much more reliable. I then looked at ensuring that varying sized strings were passed to the functions. This prevented some of the variations I saw, where some algorithms did better at the single chars or smaller strings. However the top 3 methods all did well regardless of the string size.
Forked test set
http://jsfiddle.net/schmide/fCqp3/134/
// repeated string
var string = '0123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789';
// count paremeter is changed on every test iteration, limit it's maximum value here
var maxCount = 200;
var n = 0;
$.each(tests, function (name) {
var fn = tests[name];
JSLitmus.test(++n + '. ' + name, function (count) {
var index = 0;
while (count--) {
fn.call(string.slice(0, index % string.length), index % maxCount);
index++;
}
});
if (fn.call('>', 10).length !== 10) $('body').prepend('<h1>Error in "' + name + '"</h1>');
});
JSLitmus.runAll();
I then included Dennis' fix and decided to see if I could find a way to eek out a bit more.
Since javascript can't really optimize things, the best way to improve performance is to manually avoid things. If I took the first 4 trivial results out of the loop, I could avoid 2-4 string stores and write the final store directly to the result.
// final: growing pattern + prototypejs check (count < 1)
'final avoid': function (count) {
if (!count) return '';
if (count == 1) return this.valueOf();
var pattern = this.valueOf();
if (count == 2) return pattern + pattern;
if (count == 3) return pattern + pattern + pattern;
var result;
if (count & 1) result = pattern;
else result = '';
count >>= 1;
do {
pattern += pattern;
if (count & 1) result += pattern;
count >>= 1;
} while (count > 1);
return result + pattern + pattern;
}
This resulted in a 1-2% improvement on average over Dennis' fix. However, different runs and different browsers would show a fair enough variance that this extra code probably isn't worth the effort over the 2 previous algorithms.
A chart
Edit: I did this mostly under chrome. Firefox and IE will often favor Dennis by a couple %.

Simple method:
String.prototype.repeat = function(num) {
num = parseInt(num);
if (num < 0) return '';
return new Array(num + 1).join(this);
}

People overcomplicate this to a ridiculous extent or waste performance. Arrays? Recursion? You've got to be kidding me.
function repeat (string, times) {
var result = ''
while (times-- > 0) result += string
return result
}
Edit. I ran some simple tests to compare with the bitwise version posted by artistoex / disfated and a bunch of other people. The latter was only marginally faster, but orders of magnitude more memory-efficient. For 1000000 repeats of the word 'blah', the Node process went up to 46 megabytes with the simple concatenation algorithm (above), but only 5.5 megabytes with the logarithmic algorithm. The latter is definitely the way to go. Reposting it for the sake of clarity:
function repeat (string, times) {
var result = ''
while (times > 0) {
if (times & 1) result += string
times >>= 1
string += string
}
return result
}

Concatenating strings based on an number.
function concatStr(str, num) {
var arr = [];
//Construct an array
for (var i = 0; i < num; i++)
arr[i] = str;
//Join all elements
str = arr.join('');
return str;
}
console.log(concatStr("abc", 3));
Hope that helps!

With ES8 you could also use padStart or padEnd for this. eg.
var str = 'cat';
var num = 23;
var size = str.length * num;
"".padStart(size, str) // outputs: 'catcatcatcatcatcatcatcatcatcatcatcatcatcatcatcatcatcatcatcatcatcatcat'

To repeat a string in a specified number of times, we can use the built-in repeat() method in JavaScript.
Here is an example that repeats the following string for 4 times:
const name = "king";
const repeat = name.repeat(4);
console.log(repeat);
Output:
"kingkingkingking"
or we can create our own verison of repeat() function like this:
function repeat(str, n) {
if (!str || !n) {
return;
}
let final = "";
while (n) {
final += s;
n--;
}
return final;
}
console.log(repeat("king", 3))
(originally posted at https://reactgo.com/javascript-repeat-string/)

How can I get all combinations of a String with the given replacements in JavaScript?

I would like to return all possible combinations for a string while maintaining the proper order of everything and avoiding duplicates. The reason for this? I'd like to make answers for some Japanese quizzes more flexible by allowing a mix of kana and kanji. As such, I require all possible combinations for comparison against the user's answer.
This is the current syntax of the function: (located here)
Genki.getAlts('{月曜日}と{水曜日}と{金曜日}に{日本語}のクラスがあります', 'げつようび|すいようび|きんようび|にほんご');
The text within curly braces is the text that will be replaced by the alternative text in the second argument, I'll refer to these simply as replacements. HOWEVER, the alternate text should ONLY replace the same index. That is:
月曜日 can only be replaced with げつようび
水曜日 can only be replaced with すいようび
and so on...
To give a simple example of what I'd like to achieve. Say I have the following:
Genki.getAlts('...{A}...{B}...', '1|2', true);
I'd like it to return all combinations, such as below.
'...1...{B}...'
'...1...2...'
'...{A}...2...'
'...{A}...{B}...'
The current implementation works well with 2-7 given replacements, but when given more than 8, the total combo coverage begins to drop. The total amount of combinations can be calculated using this formula: Math.pow(2, 8), which would return "256" combinations for 8 replacements, but currently getAlts() is only returning 234 combos, which means we're missing 22, only giving us 91% combo coverage.
So that is where I'm currently stuck. You can review the current code via the links below. (and yes, it's rather hackish) Being self-taught I tried my best to get as many combos as possible, but I'm afraid that my skill with mathematics isn't that good. I'm sure there's a much simpler way of going about this and I'm just overthinking it.
code: Genki.getAlts()
test page: lesson-4/workbook-6 || page source (the console will show all current combinations)
As an example of the current algorithm's failure, open your console, and you should see a warning for the last problem, saying something along the lines of:
234/256 (91.40625% combo coverage for 8 replacements; 22 missing combos
Code for this problem:
Genki.getAlts('{1:私}はきのう{学校}で{1:写真}を{1:撮}りました。{2:私}は{家}でも{2:写真}を{2:撮}りました。', 'わたし|がっこう|しゃしん|と|わたし|いえ|しゃしん|と', true);
and a much simpler one with 10 replacements for performing test cases in the console:
Genki.getAlts('{A}{B}{C}{D}{E}{F}{G}{H}{I}{J}', '1|2|3|4|5|6|7|8|9|10', true)
Is there any possible and simplistic way of returning all the combinations for a string regardless of how many replacements are specified? While I do know how many combinations there are, using Math.pow(2, n), I'm unsure of how to properly get them all.
I am open to hearing about existing algorithms or frameworks for achieving this.
PS: as things are, the algorithm works fine for 2-7 replacements, with very few problems ever reaching or going above this threshold. However, when they do, there's a chance that the user's answer will erroneously be marked wrong and I'd like to avoid this. The simplest solution would obviously be to avoid ever breaking 7, but that's not always possible, and furthermore, the current way I'm achieving this isn't optimal, so I would like to optimize it as well.

You can solve this problem using binary math. Here's an approach that generates the array of strings:
function getAlts(str, alt) {
var subs = alt.split('|');
var length = subs.length;
var permutations = Math.pow(2, length);
var results = [];
for (var i = 0; i < permutations; ++i) {
var bitIndex = 0;
var result = str.replace(/\{(.*?)\}/g, function (match, p1) {
var subIndex = bitIndex++;
var bit = length - 1 - subIndex;
return ((1 << bit) & i) ? subs[subIndex] : p1;
});
results.push(result);
}
return results;
}
console.log(getAlts('...{A}...{B}...', '1|2'));
Or if you're able to use ES6 (ECMAScript 2015), you can write a generator function to use less memory:
function* getAlts(str, alt) {
var subs = alt.split('|');
var length = subs.length;
var permutations = Math.pow(2, length);
for (var i = 0; i < permutations; ++i) {
var bitIndex = 0;
var result = str.replace(/\{(.*?)\}/g, function (match, p1) {
var subIndex = bitIndex++;
var bit = length - 1 - subIndex;
return ((1 << bit) & i) ? subs[subIndex] : p1;
});
yield result;
}
}
var results = getAlts('{A}{B}{C}{D}{E}{F}{G}{H}{I}', '1|2|3|4|5|6|7|8|9');
var total = 0;
for (var result of results) {
console.log(result);
total++;
}
console.log('total:', total);

Javascript variable what is less efficient

I am learning Javascript currently.I was wondering if there is any difference between:
var factor=0.1;
var limit=10;
var x;
var y;
x= limit*factor;
y= limit*factor;
//or
var limit=10;
var x;
var y;
x=limit *0.1;
y=limit*0.1;
Does it make any difference (when looking at performance for example)? If so, why it is different? The second example looks less promising to me, because I keep thinking that I am declaring the variable 0.1 twice. Thanks for your help in advance.

There is a very small difference. When you use factor in the two multiplications, the JavaScript engine has to go look up the value of factor in the current lexical environment object each time — in theory, anyway; optimization may well be able to get rid of that, if and when the code is chosen for optimization by the JavaScript engine.
But regardless: Worry about performance problems when you have a performance problem to worry about. Instead, concentrate on readability and maintainability. If x and y are meant to be multiplied by the same value, put that value in something (a var, or perhaps a const with ES2015+), and use it in both places.

I would suggest you go ahead with the first example, but with a modification. Variables are meant to hold dynamic data, it is better to hold 0.1 in a variable, so you can change it over time if required.
// have a function so that you don't repeat the code
function getFactor(factor, limit) {
return limit * factor;
}
//declare the variables and set the required default values
var factor = 0.1,
limit = 10,
x, y;
//assign getFactor to x and y variables
x = y = getFactor(factor, limit);

How to compare a number string and it's reverse to find the next highest palindrome?

I am trying to build a function that takes a price with 2 decimal points and finds the next highest palindrome. I know there are several other ways to approach this, but I am curious why my method is not working. I am new to JS so this might be a simple one. Any advice would be great.
I broke it into smaller chunks with explanations of what I want it to do below:
var ask = prompt("Enter a price");
var reverseIt = function (x) {
x = (parseFloat(x) * 100).toString();
for (var i = (x.length - 1); i >= 0; i--) {
x.substr(i, 1);
}
return
};
The reverseIt function takes an argument removes the decimal (* 100) and reverses the number.
var removeDec = function (j) {
return (parseFloat(j) * 100).toString();
}
The removeDec function takes an argument, removes the decimal point (* 100), and converts it back to a string. Is this redundant for comparing two "number" strings? Should I use the Number() and String() functions instead?
var findDrome = function (i) {
for (var i; removeDec(i) != reverseIt(i); i += (1 / 100)) {
if ((removeDec(i) + 1).toString() == reverseIt(i)) {
document.write(i + (1 / 100));
}
} return
};
findDrome(ask);
The findDrome function takes the ask prompt at the start as an argument. If the number without a decimal doesn't match the reverse without a decimal, then increments it by 0.01. Right before the loop ends, I wanted it to check if the number prior +1 (since it is * 100) is equal to the reverse and if so write the next number.
It wasn't working, so I tried adding parseFloat and toString to specify stricter/more explicit conversions. I also used the loose equality operators, but it's still not working.
My questions: Is this a conversion or syntactical problem or can you not compare the output of 2 functions? Should I instead compare 2 variables and if so how do I assign the for loop in the reverseIt function to a variable?

Your program has a number of issues. First, your reverseIt function never returns a reversed value. The variable x is passed in but it's never updated in the for loop - x.substr() creates a new string instance but it's never assigned back to x so its value never changes. As it is, your for loop in findDrome goes infinite since reverseIt returns undefined.
Another - possible - problem is that you're incrementing a floating-point number by 1/100 but floating point values have no exact representation. I don't know if this is actually affecting your code (since it currently never returns a proper value) but it's something you may have to worry about. This would likely affect parseFloat (which may return a slighly different floating-point value than the string it parses).
Using toFixed() would truncate the number to 2 decimal digits. You could then turn the number to a string and remove the decimal dot character, rather than converting the number back and forth between string and number.
You may want to read up on floating-point arithmetic (if you're not already familiar with it).
As a last comment, you should never, ever rely on Javascript terminating your statements - you should always use ; to terminate a statement like in other proper C-style languages. Leaving out ;-s (even if Javascript lets you get away with it) is considered very poor practice.

I figured it out thanks to the help above! Here is how the fixed program works:
var ask = prompt("Enter a price to find the next palindromic price");
var removeDec = function (j) {
return parseInt(j * 100);
};
var c = removeDec(ask);
This prompts a price and multiplies it by 100 to remove the decimal point and avoid floating point arithmetic. The parseInt removes any decimals smaller than the hundredths place.
var reverseIt = function (x) {
var a = ""
x = x.toString();
for (var i = (x.length - 1); i >= 0; i--) {
a = (a + String(x.substr(i, 1)));
}
return Number(a);
};
var b = reverseIt(c);
The reverseIt function takes an argument, converts it to string and adds each character in reverse to an empty string (a=""). Var a is then returned as a number. The empty string is important for storing the reverse number and is a big reason why my code wasn't working before.
var e = Math.pow(10, (String(c).length - 1));
Var e was added to take into account varying place values to left side of the decimal. Later this helps check if a number is equal to its reverse by adding a 1 to both sides of the number. Var e counts the length of var c (entered value with decimal removed) and finds the appropriate power of 10 to add later. So if you entered 14.40 * 100 then later it will check if 1440 + 1 is equal to 0441 + 1000.. or 10^3. This test is important later in order to exit the loop. This is where my code was failing before because I didn't take adding a number to the reverse into account and I was trying to add decimals which aren't as predictable.
if (c == b) {
document.write("$" + (c / 100) + "... This price is already palindrome!")
} else {
for (c; c !== b; c++) {
b = reverseIt(c);
if ((c + 1) == (b + e)) {
document.write("The next palindromic price is $" + ((Number(c) + 1) / 100));
break;
}
}
}
Here, If the original number and it's reverse are not equal then a loop begins that increments the value by 1 until the entered number + 1 is equal to the reversed number + e. So effectively the loop finds the number right before the loop ends, writes it and then breaks out of the loop. This palindrome finder seems to work smoothly with values big and small, no matter where you put the decimal point. Glad I got it working... it was a great learning experience figuring it out!

creating a simple one-way hash

Are there any standard hash functions/methods that maps an arbitrary 9 digit integer into another (unique) 9 digit integer, such that it is somewhat difficult to map back (without using brute force).
Hashes should not collide, so every output 1 ≤ y < 10^9 needs to be mapped from one and only one input value in 1 ≤ x < 10^9.

The problem you describe is really what Format-Preserving Encryption aims to solve.
One standard is currently being worked out by NIST: the new FFX mode of encryption for block ciphers.
It may be more complex than what you expected though. I cannot find any implementation in Javascript, but some examples exist in other languages: here (Python) or here (C++).

You are requiring a non-colliding hash function with only about 30 bits. That's going to be a tall order for any hash function. Actually, what you need is not a Pseudo Random Function such as a hash but a Pseudo Random Permutation.
You could use an encryption function for this, but you would obviously need to keep the key secret. Furthermore, encryption functions normally bits as input and output, and 10^9 is not likely to use an exact number of bits. So if you are going for such an option you may have to use format preserving encryption.
You may also use any other function that is a PRP within the group 0..10^9-1 (after decrementing the value with 1), but if an attacker finds out what parameters you are using then it becomes really simple to revert back to the original. An example would be a multiplication with a number that is relatively prime with 10^9-1, modulo 10^9-1.

This is what i can come up with:
var used = {};
var hash = function (num) {
num = md5(num);
if (used[num] !== undefined) {
return used[num];
} else {
var newNum;
do {
newNum = Math.floor(Math.random() * 1000000000) + 1;
} while (contains(newNum))
used[num] = newNum;
return newNum;
}
};
var contains = function (num) {
for (var i in used) {
if (used[i] === num) {
return true;
}
}
return false;
};
var md5 = function (num) {
//method that return an md5 (or any other) hash
};
I should note however that it will run into problems when you try to hash a lot of different numbers because the do..while will produce random numbers and compare them with already generated numbers. If you have already generated a lot of numbers it will get more and more unlikely to find the remaining ones.