I have these two functions and I want to know which one is faster. I assume the first one, but what if I have hundreds of cases to evaluate?
function isSpecialKey(k) {
switch (k) {
case 9:
return true;
break;
case 16:
return true;
break;
case 17:
return true;
break;
case 18:
return true;
break;
case 20:
return true;
break;
case 37:
return true;
break;
case 38:
return true;
break;
case 39:
return true;
break;
case 40:
return true;
break;
default:
return false;
break;
}
}
function isSpecialKey(k) {
var arr = [9, 16, 17, 16, 8, 20, 37, 38, 39, 40]
for (i = 0; i < arr.length; i++) { if (k == arr[i]) { return true; } }
return false;
}
It is very unlikely to matter, not even with hundreds of cases. It might start mattering with thousands or tens of thousands but in that case, you probably shouldn't be using JavaScript anyway! At least not in a web browser.
Generally - the second way is the only way that makes sense from a maintenance perspective. I would absolutely take that.
However, for your specific case, there is a better solution.
Instead of doing this, just create a map:
var specialKeys = {
9: true,
16: true,
17: true,
...
40: true
};
Then you can just test it like so:
if(specialKeys[value]) {
...
}
How about?
function isSpecialKey(key) {
return [9, 16, 17, 16, 8, 20, 37, 38, 39, 40].indexOf(key) > -1;
}
Update. I've just remembered there are some browsers that don't support indexOf on arrays, but I forgot which of them, so be careful.
You can use fallthrough in the switch, which makes for a lot less code:
function isSpecialKey(k) {
switch (k) {
case 9:
case 16:
case 17:
case 18:
case 20:
case 37:
case 38:
case 39:
case 40:
return true;
}
return false;
}
Consider also:
function isSpecialKey(k) {
return (
k == 9 ||
k == 16 ||
k == 17 ||
k == 18 ||
k == 20 ||
k == 37 ||
k == 38 ||
k == 39 ||
k == 40
);
}
Use a map; faster yet. "Hundreds" of switches should never come close to passing a code review.
Related
Im new to coding and javascript, atm im doing som tests in school.
I have this function with two different switch cases with strings which i want to add together and return. But it only returns one of the strings. If i use switch(card.value) on the first and switch(card.suit) on the second it only returns the first one. But if i take it away on the frist one:
switch(value) and switch(card.suit) it return string from the lower switch-case. Why is that? And how do i get it to return A♥? Here is the code. Sorry for my messy description.
const prettyCard = function (card) {
let suit, value
switch (card.value) {
case 1:
return 'A';
break;
case 10:
return 'T';
break;
case 11:
return 'J';
break;
case 12:
return 'Q';
break;
case 13:
return 'K';
break;
case 2:
return '2';
break;
case 3:
return '3';
break;
case 4:
return '4';
break;
case 5:
return '5';
break;
case 6:
return '6';
break;
case 7:
return '7';
break;
case 8:
return '8';
break;
case 9:
return '9';
break;
}
switch (card.suit) {
case 'HEARTS':
return '♥';
break;
case 'SPADES':
return '♠';
break;
case 'DIAMONDS':
return '♦';
break;
case 'CLUBS':
return '♣';
break;
}
return value + suit
}
console.log(prettyCard({ suit: 'HEARTS', value: 1 }))
In addition to the fix suggested in comments, the code can shrink considerably by looking-up suits and values in an object, rather than a long switch.
const prettyCard = card => {
const values = { 1: 'A', 11: 'J', 12: 'Q', 13: 'K' }
const suits = { 'HEARTS': '♥', 'SPADES': '♠', 'DIAMONDS': '♦', 'CLUBS': '♣' };
const suit = suits[card.suit];
const value = card.value > 1 && card.value < 11 ? `${card.value}` : values[card.value];
return { suit, value };
}
console.log(prettyCard({ value: 12, suit: 'HEARTS' }))
function texas(val) {
var answer = "";
switch(val) {
case 1:
case 2:
case 3:
var answer = "low";
case 4:
case 5:
case 6:
var answer = "medium";
break;
} else if(val => 7) {
var answer = "Huge"
}
return answer;
}
it says error Declaration or statement expected. ts(1128) [13, 7]
and it poits at the else if statement
You can use the "default" keyword, but you should probably update your code in order to handle the cases in which the value of the parameter is not positive or not a number:
function texas(val) {
if (val <= 0 || isNan(val)) {
throw new InvalidOperationException("val should be a positive number");
}
switch(val) {
case 1:
case 2:
case 3:
return "low";
case 4:
case 5:
case 6:
return "medium";
default:
return "Huge"
}
}
It's >= and the elsehas to be deleted. The varfor answer is unnecesary, just declare it once with let. You forgot the break in case 3:.
function texas(val) {
let answer = "";
switch(val) {
case 1:
case 2:
case 3:
answer = "low";
break;
case 4:
case 5:
case 6:
answer = "medium";
break;
}
if(val >= 7) {
answer = "Huge"
}
return answer;
}
console.log(texas(2));
console.log(texas(8));
You just need to return in the switch
function texas(val) {
var answer = "";
switch(val) {
case 1:
case 2:
case 3:
var answer = "low";
return answer;
case 4:
case 5:
case 6:
var answer = "medium";
return answer;
}
if(val => 7) {
var answer = "Huge"
}
return answer;
}
The syntax does not allow to put an else after a switch. else only makes sense in combination with an if statemen. But switch has a default: case which most closely matches your intention (hopefully):
function texas(val) {
var answer = "";
switch(val) {
case 1:
case 2:
case 3:
var answer = "low";
break;
case 4:
case 5:
case 6:
answer = "medium";
break;
default:
if(val >= 7) {
answer = "Huge"
}
// decide what should happen if val is 0, -1 or not even a number (e.g. texas('gotcha!')
break;
}
return answer;
}
Don't forget to put break in your cases, otherwise execution will "fall through" and execute the next cases. You would never end up with "low"
You can't use an if statement within a switch block.
You do have the default option tho -
function texas(val) {
var answer = "";
switch(val) {
case 1:
case 2:
case 3:
answer = "low";
case 4:
case 5:
case 6:
answer = "medium";
break;
default:
answer = val >= 7 ? "Huge" : "Invalid";
break;
}
return answer;
Note that if you have a minus / negative answer, it'll also fall into this clause, but you can the the value of answer with an inline ?: if statement...
You can't put the else after the switch block as people have stated above. switch statement is better for multi way branching and fixed data values. On the other side, if statement is better for boolean values. You can do something like this. It might not be the shortest line of codes, but just so you that there's another approach:
function texas(val) {
let answer = "";
switch (true) {
case (val == 1 || val == 2 || val == 3):
answer = "low";
break;
case (val == 4 || val == 5 || val == 6):
answer = "medium";
break;
case (val >= 7):
answer = "huge";
break;
}
return answer;
}
This question already has answers here:
Expression inside switch case statement
(8 answers)
Closed 3 years ago.
I use this code to enter a number and compare it using less and greater than within switch case how should I do to get the correct result, only default can works
var sum=prompt("enter sum:");
// sum=50;
switch(sum)
{
case sum=0:
alert("sucess");
break;
case sum>50:
alert("also sucess");
break;
case sum<50:
alert("failed");
default:
alert("there is errorrrr");
}
You can use switch (true):
switch (true) {
case sum === 0:
alert('success');
break;
case sum < 50:
alert('also success');
break;
case sum > 50:
alert('failed');
break;
default:
alert('there is an error.')
break;
}
Note that in your code, the first case is actually an assignment and modify sum to set it to 0.
It actually doesn't work, as you expect, the switch statement is compared against all cases, such as :
switch (something)
{
case 1: // something == 1 ?
// ....
}
Actually, what you have write was interpreted such as
var sum = 42;
switch(sum)
{
case sum < 50: // sum == sum < 50 ? -> 42 == 42 < 50 ? -> 42 == true ? false !
// ...
Instead, you can use a switch true statement.
// v------- convert the prompt to a numeric value
let sum = + prompt("enter sum:");
switch(true)
{
// VV----- notice the double equal
case sum == 0: // true == sum == 0 ?
alert("sucess");
break;
case sum > 50:
alert("also sucess");
break;
case sum < 50:
alert("failed");
break; // <---- You forgot a break; there
default:
alert("there is errorrrr");
break;
}
ur idea works fine
sum=20;
switch(true)
{
case 50:
alert("sucess");
break;
case (sum>50):
alert("also sucess");
break;
case sum<50:
alert("failed");
break;
default:
alert("there is errorrrr");
}
I have written a function where the values of the letters in a received string are shifted by 13 places.
My solution is highly inefficient and would need completely rewriting if the shift factor was changed.
Here is my solution:
function rot13(str) {
var charArray = str.split("");
var myArray = [];
for (var i = 0; i < charArray.length; i++) {
switch (charArray[i]) {
case "A":
myArray.push("N");
break;
case "B":
myArray.push("O");
break;
case "C":
myArray.push("P");
break;
case "D":
myArray.push("Q");
break;
case "E":
myArray.push("R");
break;
case "F":
myArray.push("S");
break;
case "G":
myArray.push("T");
break;
case "H":
myArray.push("U");
break;
case "I":
myArray.push("V");
break;
case "J":
myArray.push("W");
break;
case "K":
myArray.push("X");
break;
case "L":
myArray.push("Y");
break;
case "M":
myArray.push("Z");
break;
case "N":
myArray.push("A");
break;
case "O":
myArray.push("B");
break;
case "P":
myArray.push("C");
break;
case "Q":
myArray.push("D");
break;
case "R":
myArray.push("E");
break;
case "S":
myArray.push("F");
break;
case "T":
myArray.push("G");
break;
case "U":
myArray.push("H");
break;
case "V":
myArray.push("I");
break;
case "W":
myArray.push("J");
break;
case "X":
myArray.push("K");
break;
case "Y":
myArray.push("L");
break;
case "Z":
myArray.push("M");
break;
case " ":
myArray.push(" ");
break;
case ",":
myArray.push(",");
break;
case "!":
myArray.push("!");
break;
case "?":
myArray.push("?");
break;
case ".":
myArray.push(".");
break;
}
}
console.log(myArray.join(""));
}
rot13("GUR DHVPX OEBJA QBT WHZCRQ BIRE GUR YNML SBK.");
Can you show me a more efficient less cumbersome solution?
Here's one possible implementation, with the ability to pass in any (positive) rotation value and a table of other replacements. Written in ES6.
function rotn(str, rotation = 13, map = {}) {
const table = {}; // New table, to avoid mutating the parameter passed in
// Establish mappings for the characters passed in initially
for (var key in map) {
table[map[key]] = key;
table[key] = map[key];
}
// Then build the rotation map.
// 65 and 97 are the character codes for A and a, respectively.
for (var i = 0; i < 26; i++) {
table[String.fromCharCode(65 + i)] = String.fromCharCode(65 + (i + rotation) % 26);
table[String.fromCharCode(97 + i)] = String.fromCharCode(97 + (i + rotation) % 26);
}
return str.split('').map((c) => table[c] || c).join('');
}
console.log(rotn("Gur dhvpx oebja Qbt whzcrq bire gur ynml Sbk.", 13, {'.': '!'}));
Here is an example using the reduce function:
function rot13(str) {
chars = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
return str.split("").reduce(function(a, b) {
if (chars.indexOf(b) == -1) {
return a + b;
}
return a + chars[(chars.indexOf(b)+13) % chars.length]
}, "");
}
console.log(rot13("GUR DHVPX OEBJA QBT WHZCRQ BIRE GUR YNML SBK."));
function ROT13 (str){
var result = "";
for (var i = 0; i < str.length; i++) {
var charCode = str.charCodeAt(i);
if (charCode >= 65 && charCode <= 90) {
result += String.fromCharCode(((charCode - 65 + 13) % 26) + 65);
} else if (charCode >= 97 && charCode <= 122) {
result += String.fromCharCode(((charCode - 97 + 13) % 26) + 97);
} else {
result += str[i];
}
}
return result;
}
Here's a solution that uses:
replace() with a /[a-z]/ regex so that only alphabetic characters will be modified
indexOf() to convert that alphabetic character to an index
then uses that index to lookup the corresponding cipher character
function rot13(txt) {
return txt.replace(/[a-z]/gi, c =>
"NOPQRSTUVWXYZABCDEFGHIJKLMnopqrstuvwxyzabcdefghijklm"
[ "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz".indexOf(c) ] );
}
console.log(rot13("GUR DHVPX OEBJA QBT WHZCRQ BIRE GUR YNML SBK."));
I have a problem with the 'case' statement in the 'switch' statement in java script.
My question is how to write more than one number in the 'case' statement and save all the work on writing multiple of commands for each number , ill try to explain myself better. i want to write in the case statement the
number 10-14 (10,11,12,13,14).
how can i write it?
thanks for helping and sorry for my bad english.
name = prompt("What's your name?")
switch (name)
{
case "Ori":
document.write("<h1>" + "Hello there Ori" + "<br>")
break;
case "Daniel":
document.write("<h1>" + "Hi, Daniel." + "<br>")
break;
case "Noa":
document.write("<h1>" + "Noa!" + "<br>")
break;
case "Tal":
document.write("<h1>" + "Hey, Tal!" + "<br>")
break;
default:
document.write("<h1>" + name + "<br>")
}
age = prompt ("What's your age?")
switch (age)
{
case "10":
document.write("you are too little" + name)
break;
case "14":
document.write("So , you are in junior high school" + name)
break;
case "18":
document.write("You are a grown man" + name)
break;
default:
document.write("That's cool" + name)
break;
}
Check out this answer Switch on ranges of integers in JavaScript
In summary you can do this
var x = this.dealer;
switch (true) {
case (x < 5):
alert("less than five");
break;
case (x > 4 && x < 9):
alert("between 5 and 8");
break;
case (x > 8 && x < 12):
alert("between 9 and 11");
break;
default:
alert("none");
break;
}
but that sort of defeats the purpose of a switch statement, because you could just chain if-else statments. Or you can do this:
switch(this.dealer) {
case 1:
case 2:
case 3:
case 4:
// Do something.
break;
case 5:
case 6:
case 7:
case 8:
// Do something.
break;
default:
break;
}
use this, if you dont provide break then control will fall down, In this way you can match for group of numbers in switch.
case 10:
case 11:
case 12:
case 14:
case 15: document.write("i am less than or equal to 15");break;
Say you wanted to switch on a number 10-14 (10,11,12,13,14) you can chain the cases together:
switch(number) {
case 10:
case 11:
case 12:
case 13:
case 14:
alert("I'm between 10 and 14");
break;
default:
alert("I'm not between 10 and 14");
break;
}
You can simply omit the break; statement.
switch (2) {
case 1: case 2: case 3:
console.log('1 or 2 or 3');
break;
default:
console.log('others');
break;
}
I wanted to play with the concept a bit, I do not recommend this approach, however you could also rely on a function that will create a control flow function for you. This simply allows some syntaxic sugar.
var createCaseFlow = (function () {
var rangeRx = /^(\d)-(\d)$/;
function buildCases(item) {
var match = item.match(rangeRx),
n1, n2, cases;
if (match) {
n1 = parseInt(match[1], 10);
n2 = parseInt(match[2], 10);
cases = [];
for (; n1 <= n2; n1++) {
cases.push("case '" + n1 + "':");
}
return cases.join('');
}
return "case '" + item + "':";
}
return function (cases, defaultFn) {
var fnStrings = ['switch (value.toString()) { '],
k;
for (k in cases) {
if (cases.hasOwnProperty(k)) {
fnStrings.push(k.split(',').map(buildCases).join('') + "return this[0]['" + k + "'](); break;");
}
}
defaultFn && fnStrings.push('default: return this[1](); break;');
return new Function('value', fnStrings.join('') + '}').bind(arguments);
};
})();
var executeFlow = createCaseFlow({
'2-9': function () {
return '2 to 9';
},
'10,21,24': function () {
return '10,21,24';
}
},
function () {
return 'default case';
}
);
console.log(executeFlow(5)); //2 to 9
console.log(executeFlow(10)); //10,21,24
console.log(executeFlow(13)); //default case
You have already gotten a few answers on how to make this work. However, I want to point out a few more things. First off, personally, I wouldn't use a switch/case statement for this as there are so many similar cases – a classic if/elseif/else chain feels more appropriate here.
Depending on the use-case you could also extract a function and then use your switch/case (with more meaningful names and values, of course):
function getCategory (number) {
if(number > 20) {
return 3;
}
if(number > 15) {
return 2;
}
if(number > 8) {
return 1;
}
return 0;
}
switch( getCategory( someNumber ) ) {
case 0:
// someNumber is less than or equal to 8
break;
case 1:
// someNumber is any of 9, 10, 11, 12, 13, 14, 15
break;
// ...
}
If the intervals are equally spaced, you could also do something like this:
switch( Math.floor( someNumber / 5 ) ) {
case 0:
// someNumber is any one of 0, 1, 2, 3, 4
break;
case 1:
// someNumber is any one of 5, 6, 7, 8, 9
break;
// ...
}
Also, it should be noted that some people consider switch/case statements with fall-throughs (= leaving out the break; statement for come cases) bad practice, though others feel it's perfectly fine.