We Keep Coding

Javascript: How to recursively return a counter?

In the example below, I don't want to make a counter as a param. Rather, I just want to return '+ 1' each time so that what gets returned is the number of steps it takes. My issue lies with the base case. If I do return + 1, I get the correct number of steps plus one additional step so I tried just return but that delivers NaN. Is it even possible?
var numberOfSteps = function(num) {
if (num == 0) {
return;
} else {
if (num % 2 == 0) {
return 1 + numberOfSteps(num/2);
} else {
return 1 + numberOfSteps(num - 1);
}
}
};
edit : The goal is to track how many steps it takes to reduce a number to 0. If it's even, divide by 2 or else subtract by 1. Ultimately, I want to return the number of steps it takes for any given number to get reduced to 0 following those rules

I hope the point has gotten through in the long comment thread and other answers that return + 1 is equivalent to return (+1), that is, return the integer positive one. And since there are no steps to take once you've reached zero, +1 is the wrong answer. Similarly, a plain return is functionally equivalent to return undefined. But undefined is not a number, and you're going to run into problems if you later try to add 1 to it. So the solution from the comments or other answers to return the correct number of steps, which in this case 0, will fix your code.
I would like to point out another way to solve this, though:
const numberOfSteps = (n) =>
n <= 0
? 0
: 1 + numberOfSteps (n % 2 == 0 ? n / 2 : n - 1)
console .log (numberOfSteps (12))
There are superficial differences here from the other solutions, such as using an arrow function, using a conditional statement (ternary) rather than if-statements, and using <= 0 instead of < 0 to avoid possible infinite loops on negative numbers.
But the fundamental difference is that this code only has one recursive branch. I think this is a better match to the problem.
We can think of this as a function which answers "How many steps does it take to reach 0 from our input number if each step cuts even numbers in half and subtracts one from odd ones?" Well that logically leads to a base case (we're already at 0) so have to return 0, and a recursive one (we're at some positive integer) so have to add 1 to the total steps required from our next entry.
By doing this single recursive call and adding one to the result, we make it clearer what the recursion is doing.
If this is unclear, then this alternative might show what I mean:
const takeStep = (n) =>
n % 2 == 0 ? n / 2 : n - 1
const numberOfSteps = (n) =>
n <= 0
? 0
: 1 + numberOfSteps (takeStep (n))

Think you just need to return 0 when it's...zero.
var numberOfSteps = function(num) {
if (num == 0) {
return 0;
} else {
if (num % 2 == 0) {
return 1 + numberOfSteps(num/2);
} else {
return 1 + numberOfSteps(num - 1);
}
}
}
return + 1 maybe doesn't do what you think it does: it returns the number 1. + here means positive not negative, there is no addition or subtraction going on. It will also give you one too many steps.
return; by itself returns undefined, which when converted to a Number, translates to NaN, because, well, it's not a number.

Javascript Help - selfDividingNumbers Algorithm producing all 0's

Greetings Stack Overflow!
First off, this is my first question!
I am trying to solve the selfDividingNumbers algorithm and I ran into this interesting problem. This function is supposed to take a range of numbers to check if they are self dividing.
Self Dividing example:
128 is a self-dividing number because
128 % 1 == 0, 128 % 2 == 0, and 128 % 8 == 0.
My attempt with Javascript.
/*
selfDividingNumbers( 1, 22 );
*/
var selfDividingNumbers = function(left, right) {
var output = [];
while(left <= right){
// convert number into an array of strings, size 1
var leftString = left.toString().split();
// initialize digit iterator
var currentDigit = leftString[0];
for(var i = 0; i < leftString.length; i++){
currentDigit = parseInt(leftString[i])
console.log( left % currentDigit );
}
// increment lower bound
left++;
}
return output
};
When comparing the current lower bound to the current digit of the lower bound, left % currentDigit it always produces zero! I figure this is probably a type error but I am unsure of why and would love for someone to point out why!
Would also like to see any other ideas to avoid this problem!
I figured this was a good chance to get a better handle on Javascript considering I am clueless as to why my program is producing this output. Any help would be appreciated! :)
Thanks Stack Overflow!

Calling split() isn't buying you anything. Remove it and you'll get the results you expect. You still have to write the code to populate output though.

The answer by #Joseph may fix your current code, but I think there is a potentially easier way to go about doing this. Consider the following script:
var start = 128;
var num = start;
var sd = true;
while (num > 0) {
var last = num % 10;
if (start % last != 0) {
sd = false;
break;
}
num = Math.floor(num / 10);
}
if (sd) {
print("Is self dividing");
}
else {
print("Is NOT self dividing");
}
Demo
To test each digit in the number for its ability to cleanly divide the original number, you can simply use a loop. In each iteration, check num % 10 to get the current digit, and then divide the number by ten. If we never see a digit which can not divide evenly, then the number is not self dividing, otherwise it is.

So the string split method takes the string and returns an array of string parts. The method expects a parameter, however, the dividing element. If no dividing element is provided, the method will return only one part, the string itself. In your case, what you probably intended was to split the string into individual characters, which would mean the divider would be the empty string:
var leftString = left.toString().split('');
Since you are already familiar with console.log, note that you could also use it to debug your program. If you are confused about the output of left % currentDigit, one thing you could try is logging the variables just before the call,
console.log(typeof left, left, typeof currentDigit, currentDigit)
which might give you ideas about where to look next.

check whether the number is nearly equal javascript

I want to know whether it is possible?
Let Suppose:
var a = 2592;
var b = 2584;
if(a nearly equal to b) {
// do something
}

Like so.
var diff = Math.abs( a - b );
if( diff > 50 ) {
console.log('diff greater than 50');
}
That would compare if the absolute difference is greater than 50 using Math.abs and simple comparison.

Here's the old school way to do it...
approxeq = function(v1, v2, epsilon) {
if (epsilon == null) {
epsilon = 0.001;
}
return Math.abs(v1 - v2) < epsilon;
};
so,
approxeq(5,5.000001)
is true, while
approxeq(5,5.1)
is false.
You can adjust pass in epsilons explicitly to suit your needs. One part in a thousand usually covers my javascript roundoff issues.

var ratio = 0;
if ( a > b) {
ratio = b / a;
}
else {
ratio = a / b;
}
if (ratio > 0.90) {
//do something
}

One line Es6 way version of The Software Barbarian:
const approxeq = (v1, v2, epsilon = 0.001) => Math.abs(v1 - v2) <= epsilon;
console.log(approxeq(3.33333, 3.33322)); // true
console.log(approxeq(2.3, 2.33322)); // false
console.log(approxeq(3, 4, 1)); // true
I changed it to include the number in the margin. So with an epsilon margin of 1 approxeq between 1 and 2 is true

Floating point comparison gets complicated in a hurry. It's not as simple as diff less than epsilon in a lot of cases.
Here's an article about the subject, though not javascript specific.
https://floating-point-gui.de/errors/comparison/
TLDR:
When one of the numbers being compared is very close to zero, subtracting the smaller from the larger can lose digits of precision, making the diff appear smaller than it is (or zero).
Very small numbers with different signs work in a weird way.
Dividing by zero will cause problems.
In the article is a function (java) which solves better for these cases:
public static boolean nearlyEqual(float a, float b, float epsilon) {
final float absA = Math.abs(a);
final float absB = Math.abs(b);
final float diff = Math.abs(a - b);
if (a == b) { // shortcut, handles infinities
return true;
} else if (a == 0 || b == 0 || (absA + absB < Float.MIN_NORMAL)) {
// a or b is zero or both are extremely close to it
// relative error is less meaningful here
return diff < (epsilon * Float.MIN_NORMAL);
} else { // use relative error
return diff / Math.min((absA + absB), Float.MAX_VALUE) < epsilon;
}
}
Before you complain: Yes, that's Java, so you'd have to rewrite it in Javascript. It's just to illustrate the algorithm and it's just copied from the article.
I'm still looking for a thorough solution to this problem, ideally with an NPM package so I don't have to figure this out again every time I need it.
Edit: I've found a package which implements the solution from the article linked above (which has the same link in their readme).
https://www.npmjs.com/package/#intocode-io/nearly-equal
This will be a less error-prone solution than others shown in other answers. There are several npm packages which implement the naive solutions which have error cases near zero as described above. Make sure you look at the source before you use them.

Converting a double to an int in Javascript without rounding

In C# the following code returns 2:
double d = 2.9;
int i = (int)d;
Debug.WriteLine(i);
In Javascript, however, the only way of converting a "double" to an "int" that I'm aware of is by using Math.round/floor/toFixed etc. Is there a way of converting to an int in Javascript without rounding? I'm aware of the performance implications of Number() so I'd rather avoid converting it to a string if at all possible.

Use parseInt().
var num = 2.9
console.log(parseInt(num, 10)); // 2
You can also use |.
var num = 2.9
console.log(num | 0); // 2

I find the "parseInt" suggestions to be pretty curious, because "parseInt" operates on strings by design. That's why its name has the word "parse" in it.
A trick that avoids a function call entirely is
var truncated = ~~number;
The double application of the "~" unary operator will leave you with a truncated version of a double-precision value. However, the value is limited to 32 bit precision, as with all the other JavaScript operations that implicitly involve considering numbers to be integers (like array indexing and the bitwise operators).
edit — In an update quite a while later, another alternative to the ~~ trick is to bitwise-OR the value with zero:
var truncated = number|0;

Similar to C# casting to (int) with just using standard lib:
Math.trunc(1.6) // 1
Math.trunc(-1.6) // -1

Just use parseInt() and be sure to include the radix so you get predictable results:
parseInt(d, 10);

There is no such thing as an int in Javascript. All Numbers are actually doubles behind the scenes* so you can't rely on the type system to issue a rounding order for you as you can in C or C#.
You don't need to worry about precision issues (since doubles correctly represent any integer up to 2^53) but you really are stuck with using Math.floor (or other equivalent tricks) if you want to round to the nearest integer.
*Most JS engines use native ints when they can but all in all JS numbers must still have double semantics.

A trick to truncate that avoids a function call entirely is
var number = 2.9
var truncated = number - number % 1;
console.log(truncated); // 2
To round a floating-point number to the nearest integer, use the addition/subtraction trick. This works for numbers with absolute value < 2 ^ 51.
var number = 2.9
var rounded = number + 6755399441055744.0 - 6755399441055744.0; // (2^52 + 2^51)
console.log(rounded); // 3
Note:
Halfway values are rounded to the nearest even using "round half to even" as the tie-breaking rule. Thus, for example, +23.5 becomes +24, as does +24.5. This variant of the round-to-nearest mode is also called bankers' rounding.
The magic number 6755399441055744.0 is explained in the stackoverflow post "A fast method to round a double to a 32-bit int explained".
// Round to whole integers using arithmetic operators
let trunc = (v) => v - v % 1;
let ceil = (v) => trunc(v % 1 > 0 ? v + 1 : v);
let floor = (v) => trunc(v % 1 < 0 ? v - 1 : v);
let round = (v) => trunc(v < 0 ? v - 0.5 : v + 0.5);
let roundHalfEven = (v) => v + 6755399441055744.0 - 6755399441055744.0; // (2^52 + 2^51)
console.log("number floor ceil round trunc");
var array = [1.5, 1.4, 1.0, -1.0, -1.4, -1.5];
array.forEach(x => {
let f = x => (x).toString().padStart(6," ");
console.log(`${f(x)} ${f(floor(x))} ${f(ceil(x))} ${f(round(x))} ${f(trunc(x))}`);
});

As #Quentin and #Pointy pointed out in their comments, it's not a good idea to use parseInt() because it is designed to convert a string to an integer. When you pass a decimal number to it, it first converts the number to a string, then casts it to an integer. I suggest you use Math.trunc(), Math.floor(), ~~num, ~~v , num | 0, num << 0, or num >> 0 depending on your needs.
This performance test demonstrates the difference in parseInt() and Math.floor() performance.
Also, this post explains the difference between the proposed methods.

What about this:
if (stringToSearch.IndexOfAny( ".,;:?!".ToCharArray() ) == -1) { ... }

I think that the easiest solution is using the bitwise not operator twice:
const myDouble = -66.7;
console.log(myDouble); //-66.7
const myInt = ~~myDouble;
console.log(myInt); //-66
const myInt = ~~-myDouble;
console.log(myInt); //66

Best way to prevent/handle divide by 0 in javascript

What is the best way to prevent divide by 0 in javascript that is accepting user inputs.
If there is no particular way to achieve this what would be the best way to handle such a situation so as to not prevent other scripts from executing?
Any insights are much appreciated.

There is no way to do that with the normal / and /= operators.
The best way to do what you want is with guards:
function notZero(n) {
n = +n; // Coerce to number.
if (!n) { // Matches +0, -0, NaN
throw new Error('Invalid dividend ' + n);
}
return n;
}
and then do division like
numerator / notZero(denominator)
Alternatively you can always guard the output
function dividend(numerator, denominator) {
var quotient = numerator / denominator;
if (quotient !== quotient) { throw new Error(numerator + " / " + denominator); }
return quotient;
}
but that loses the readability and expressiveness of /=.

Off the top of my head you could:
Check the user input to see if the denominator is zero (or evaluates to zero, depending on what your script actually does).
Check if the result of the action isFinite() and if not then handle appropriately.

what would be the best way to handle such a situation so as to not prevent other scripts from executing
Division by zero doesn't seem to prevent other scripts from execution in JavaScript:
var a = 20;
var b = 0;
var result = a/b;
console.log(result); // returns Infinity
If you want something different to happen in case of division by zero, you could use
function divideIfNotZero(numerator, denominator) {
if (denominator === 0 || isNaN(denominator)) {
return null;
}
else {
return numerator / denominator;
}
}

Hope this is useful
(denominator != 0 ? numerator/denominator : Infinity)
or whatever value you want to put at the end.
Greetings.

To prevent (unwanted) execution
Always verify critical user input and/or results
Use logic and/or callbacks you can prevent to execute
On HTML forms etc. you can use i.e. return false; as value to stop submission.

Why not just check if the denominator is zero?
if(x != 0) z = y / x;
You can also check if the result is Infinity:
3 / 0 == Infinity
Results in true;
(Only tested in chrome.)

A bit different than stopping execution, but the ternary operator is a pretty slick way to customize variable assignment.
var one = 1,
zero = 0,
customValue = 1;
var quotient = zero===0 ? customValue : one / zero;
This way, by setting the customVariable to the integer of your choice, you can expect a predictable result when division by zero occurs.

The best way is contextual. But here's the easiest:
function myFunction( input ){
input = 0 ? 0.0001 : input; // same as if( input == 0 ){ input = 0.0001; }
return 1 / input;
}
Basically if the input is zero, turn it into a very small number before using as a denominator. Works great for integers, since after your division you can round them back down.
A couple caveats prevent this from being universal:
It could cause false positives if your input accepts really small numbers
It won't trigger any error-handling code, if you need to do something special if zero is entered
So it's best for general-purpose, non-critical cases. For example, if you need to return the result of a complex calculation and don't care if the answer is accurate to N digits (determined by 0.0001 vs. 0.00000001, etc.); you just don't want it to break on a divide-by-zero.
As another answer suggested, you could also create a reusable global function.
function divisor( n ){ return ( n = 0 ? 0.0001 : n ); }
function myFunction( input ){ return 1 / divisor( input ); }
Possible improvements:
function divisor( n, orError ){
if( typeof n == 'undefined' || isNaN( n ) || !n ){
if( orError ){ throw new Error( 'Divide by zero.' ); }
return 0.000000000000001;
}else{ return 0 + n; }
}
This would take any value (null, number, string, object) and if invalid or zero, return the failsafe zero-like value. It would also coerce the output to a number just in case it was a string and you were doing something odd. All this would ensure that your divisor function always worked. Finally, for cases where you wanted to handle such errors yourself, you could set the second parameter to true and use a try/catch.

Set a cap on what the value for the numerator can be and set the numerator to that value when the denominator equals 0.

This is a faster approach yet is confusing
let divisor;
let dividend;
let result =(dividend/divisor) || 0
if the result for instance if you are calculating percentage is infinite you can give it 0 as value;

const progress = goal == 0 ? 0 : total/goal