Tips for golfing in JavaScript

Question

What general tips do you have for golfing in JavaScript? I'm looking for ideas that can be applied to code golf problems in general that are at least somewhat specific to JavaScript (e.g. "remove comments" is not an answer).

Note: Also see Tips for Golfing in ECMAScript 6 and above

Answer 1

Splitting with numbers to save the quotemarks:

"alpha,bravo,charlie".split(",") // before
"alpha0bravo0charlie".split(0)   // after

Answer 2

Fancy For Loops

you can use the standard for loop in non-standard ways

for ( a; b; c )

is essentially equivalent to:

a;
while ( b )
{
  ...
  c;
}

so a good trick is to write your code with a while loop, and then split it into the a,b,c parts in a for loop.

A couple examples I've written:

for(x=y=n;!z;x--,y++)z=i(x)?x:i(y)?y:0
for(a=b=1;b<n;c=a+b,a=b,b=c);

Chain your setters

If you're initializing or resetting multiple values, chain the value to all the variables that need it:

a=b=1;

Implicit Casting

Don't check your types, just use them as they are. parseInt() costs 10 characters. If you need to cast out of a string, be creative:

a='30';
b='10';
c = a + b; //failure
c = parseInt(a) + parseInt(b) //too long

c = -(-a-b); //try these
c = ~~a+~~b;
c = +a+ +b;
c = a- -b;

Avoid Semicolons

JavaScript has automatic semi-colon insertion. Use it often and well.

One-liners

Save on brackets by shoving as much as possible into single lines, or parameters:

a( realParam1, realParam2, fizz='buzz' )

Increment/Decrement operators

a = a - 1;
foo(a);

and

foo(a);
a = a - 1;

can easily be rewritten as

foo(--a);

and

foo(a--);

respectively.

Use this or self instead of window in global context

self explanatory 2 character savings.

Use bracket notation for repeat property access

This is definitely a balancing act between property name length and number of accesses. Instead of calling a.longFunctionName() with dot notation twice, it's shorter to save the name and call the function via bracket notation:

a.longFunctionName(b)
a.longFunctionName(c)
//42

-vs-

a[f='longFunctionName'](b)
a[f](c)
//34

this is especially effective with functions like document.getElementById which can be reduced to d[e].

Note:

With bracket notation, the cost is 6 + name.length characters the first time. Each subsequent access has a cost of 3 characters.

For dot notation, all accesses cost name.length + 1 (+1 for the .) characters.

Use this method if 6 + name.length + (3 * (accesses - 1)) < accesses * (name.length + 1).

len = length of property name
i = minimum accesses to take advantage

len | i 
========
1   | ∞ 
2   | ∞ 
3   | 7 
4   | 4 
5   | 3 
6   | 3 
7   | 3 
8+  | 2 

The number of accesses can also span multiple objects. If you access .length 4 or more times on different arrays, you can use the same variable holding the string 'length'.

Answer 3

Use the comma operator to avoid braces (also applies to C):

if(i<10)m+=5,n-=3;

Instead of

if(i<10){m+=5;n-=3}

which is one character longer.

Answer 4

Shorter random number generation

If you need a random boolean (0 or 1):

new Date&1 // equivalent to Math.random()<0.5

If you need a random integer 0 <= n < 1337:

new Date%1337 // equivalent to Math.floor(Math.random()*1337))

This works because a Date is stored internally in JavaScript as the amount of milliseconds since an epoch, so the new Date is being coerced into 123somebignumber456 when you try to do integer math on it.

Of course, these "random" numbers really won't be as random, especially if you call them multiple times in quick succession, so keep that in mind.

Answer 5

You can use the object literal form of get/set to avoid using the keyword function.

var obj = {
  get f(){
    console.log("just accessing this variable runs this code");
    return "this is actually a function";
  },
  set f(v){
    console.log("you can do whatever you want in here, passed: " + v);
  }
};

1 && obj.f; // runs obj.[[get f]]
obj.f = Infinity; // runs obj.[[set f]](Infinity)

Answer 6

This one is lesser known and lesser used, but can be impressive if used in the right situation. Consider a function that takes no arguments and always returns a different number when called, and the returned number will be used in a calculation:

var a = [ 
    Math.random()*12|0,
    Math.random()*11|0,
    Math.random()*10|0,
    /* etc... */ 
];

You might normally shorten this function using a single-letter variable name:

var r=Math.random,a=[r()*12|0,r()*11|0,r()*10|0,r()*9|0,r()*8|0,r()*7|0,r()*6|0,r()*5|0];

A better way to reduce the length is by abusing valueOf, which gives you a saving of 2 characters per invocation. Useful if you call the function more than 5 times:

var r={valueOf:Math.random},a=[r*12|0,r*11|0,r*10|0,r*9|0r*8|0,r*7|0,r*6|0,r*5|0];

Answer 7

Taking advantage of short-circuit operators

Rather than long if statements or using ternary operators, you can make use of && and || to shorten your code. For instance:

var match = RegExp('[?&]' + name + '=([^&]*)').exec(window.location.search);

return match ? decodeURIComponent(match[1].replace(/\+/g, ' ')) : null;

can become

var match = RegExp('[?&]' + name + '=([^&]*)').exec(window.location.search);

return match && decodeURIComponent(match[1].replace(/\+/g, ' '));

The || operator is often used in this way for setting defaults:

evt = evt || window.event;

This is the same as writing

if (!evt)
    evt = window.event;

Creating repetitive strings using Array

If you want to initialize a long string of a particular character, you can do so by creating an array with a length of n+1, where n is the number of times you wish to repeat the character:

// Create a string with 30 spaces
str = "                              ";

// or
str = Array(31).join(" ");

The larger the string, the bigger the saving.

Parsing numbers

Use + and ~ operators instead of parseFloat() or parseInt() when coalescing a string type that is just a number to a number type:

var num = "12.6";
parseFloat(num) === +num;  // + is 10 characters shorter than parseFloat()

var num2 = "12"
parseInt(num2) === +num2;   // + is 8 characters shorter than parseInt()

var num3 = "12.6"
parseInt(num3) === ~~num3;  // ~~ is 7 characters shorter than parseInt()

var num4 = "12.6"
parseInt(num4) === num4|0;  // |0 is 7 characters shorter than parseInt()

Be wary though, other types can be coalesced with these operators (for instance, true would become 1) an empty string or a string containing just white space would become 0. This could be useful in certain circumstances, however.

Answer 8

Unicode shortcuts

If you use a hell of a built-in property at a big golfing challenge you can alias every property to a one character equivalent:

[Math,Number,S=String,Array].map(b=>
    Object.getOwnPropertyNames(b).map((p,i)=>
        b.prototype[S.fromCharCode(i+248)]=b[p]
    )
)

After executing the code above you can use it like this:
"foo".Č(/.*/,'bar') // replaces foo with bar

This costs 118 bytes, so it might not be useful in certain situations

It may be browser dependent and i'm not sure if it's shorter than with(Array){join(foo),...} or defining variables as used properties with(Array){j=join,m=map...} but still it is worth mentioning.

    Math        Number              String              Array

ø   toSource    prototype           prototype           prototype
ù   abs         NaN                 quote               join
ú   acos        POSITIVE_INFINITY   substring           reverse
û   asin        NEGATIVE_INFINITY   toLowerCase         sort
ü   atan        MAX_VALUE           toUpperCase         push
ý   atan2       MIN_VALUE           charAt              pop
þ   ceil        MAX_SAFE_INTEGER    charCodeAt          shift
ÿ   clz32       MIN_SAFE_INTEGER    contains            unshift
Ā   cos         EPSILON             indexOf             splice
ā   exp         isFinite            lastIndexOf         concat
Ă   floor       isInteger           startsWith          slice
ă   imul        isNaN               endsWith            filter
Ą   fround      toInteger           trim                isArray
ą   log         parseFloat          trimLeft            lastIndexOf
Ć   max         parseInt            trimRight           indexOf
ć   min         length              toLocaleLowerCase   forEach
Ĉ   pow         name                toLocaleUpperCase   map
ĉ   random      arguments           normalize           every
Ċ   round       caller              match               some
ċ   sin                             search              reduce
Č   sqrt                            replace             reduceRight
č   tan                             split   
Ď   log10                           substr  
ď   log2                            concat  
Đ   log1p                           slice   
đ   expm1                           fromCharCode    
Ē   cosh                            fromCodePoint   
ē   sinh                            localeCompare   
Ĕ   tanh                            length  
ĕ   acosh                           name    
Ė   asinh                           arguments   
ė   atanh                           caller  
Ę   hypot           
ę   trunc           
Ě   sign            
ě   cbrt            
Ĝ   E           
ĝ   LOG2E           
Ğ   LOG10E          
ğ   LN2         
Ġ   LN10            
ġ   PI          
Ģ   SQRT2           
ģ   SQRT1_2         

Answer 9

Combine nested for loops:

// before:
for(i=5;i--;)for(j=5;j--;)dosomething(i,j)

// after:
for(i=25;i--;)dosomething(0|i/5,i%5)

Example with different values for i/j:

// before:
for(i=4;i--;)for(j=7;j--;)dosomething(i,j)

// after:
for(i=28;i--;)dosomething(0|i/7,i%7)

Answer 10

Sneak variable initialization into the prompt() call for getting user input

n=prompt(i=5);     // sets i=5 at the same time as getting user input

instead of using

n=prompt();i=5;

As a side-effect, it displays the input value in the prompt window while saving 1 character.

Answer 11

Use ^ instead of != or == when comparing to an integer

//x!=3?a:b
  x^3?a:b

//x==3?a:b
  x^3?b:a

Replace calls to built-in Math functions with shorter expressions

//Math.ceil(n)
  n%1?-~n:n

//Math.floor(n)
  ~~n
  0|n

//Math.abs(n)
  n<0?-n:n

//Math.round(n)
  n+.5|0

//Math.min(x,y)
  x<y?x:y

//Math.max(x,y)
  y<x?x:y

Answer 12

Array sum / product / quotient

ES5: 17 bytes

eval(a.join('+'))

ES6: 15 bytes

eval(a.join`+`)

Of course you can swap the + for anything you want, e.g., * for product or / for quotient.

Answer 13

Abuse literals

The recent sample: Check whether "c" is uppercase or lowercase, doesn't matter if not letter

"c"<{} // returns false, lower case
"C"<{} // returns true, upper case

Answer 14

Exception abusing

in case string/character literals are prohibited, you can use a try catch block:

try{something0}catch(e){str=e.message.split(0)[0]}

now str equals "something"

if more strings are needed you can chain it with a number (e.g. zeros)

try{something0foo0bar0}catch(e){arr=e.message.split(0)}

now arr equals ["something", "foo", "bar", " is not defined"]

Answer 15

If you need to check for NaN, don't use isNaN(x), but use x!=x, which is shorter and also works.

if(isNaN(x)){
if(x!=x){

Note that this only works if typeof(x) === "number"; if it's a string for example, isNaN("string") returns true, but "string" != "string" returns false. Thanks to Cyoce for pointing this out!

Answer 16

Converting a while loop into a for loop is often equivalent:

while(i--);
for(;i--;);

But the second form can have variable initialization combined:

i=10;while(i--);
for(i=10;i--;);

Notice the second form is one character shorter than the first form.

Answer 17

If you're initializing a variable to 1 in every iteration of a loop (for example, resetting a variable in an outer loop for an inner loop), like the following (from my answer to this question):

for(j=n-2;p=1,j++<=n;r|=p)for(i=1;++i<j;)p=j%i?p:0;
          ^^^^

Since the result of a condition like j++<=n is 1 whenever its true, you can just assign the condition directly to the variable (because when it becomes false, the loop will stop executing and will no longer matter):

for(j=n-2;p=j++<=n;r|=p)for(i=1;++i<j;)p=j%i?p:0;
          ^^^^^^^^

You can usually save 2 characters using this method. Regards to @ugoren for the idea in the comments to that answer.

---

For another example, I also applied this trick to my answer here with the expression w=r=++c<S.length in my outer for loop, saving a total of 4 characters.

Answer 18

If you can accept Spidermonkey (for now) specific scripts, you can use ECMAScript 6 arrow functions. Insteading of writing code like the following.

a.map(function(x){return x*2}) // function? return?

You can shorten it like this.

a.map(x=>x*2)

Answer 19

Rounding

I know that alternatives to Math.floor() have been posted, but what about the others?

Flooring:

Math.floor(x) //before
0|x           //after

Rounding:

Math.round(x) //before
0|x+.5        //after

Ceiling:

Math.ceil(x) //before
x%1?-~x:x    //after - credits to @Tomas Langkaas

Answer 20

Use a bitwise operation to round a number toward zero:

// do this
T=Math.random()*6+1|0

// or do this
T=~~(Math.random()*6+1)

(Source: Random dice tipping)

Operator precedence determines which will be shorter in your program.

Answer 21

Looping Tip I

You can save 1 character when looping by changing the i on the last time used:

//not so god
for(i=0;i<3;i++){
  alert(i);
}

//best
for(i=0;i<3;){
  alert(i++);
}

Note: works with -- too (but modify the loop accordingly to avoid infinite looping)

---

Looping Tip II

There are certain scenarios where you can save one character by playing with the incrementing operator and values:

for(i=0;i++<9;)
for(i=0;++i<10;)

Note: you need to pay attention when for example 0 to -1. and 9 to 10, 99 to 100, so play around until you find a way to save the character

Answer 22

Very simple one, even so, no one had mentioned it.

If you're using Math.min() or Math.max() you can save 6 chars by doing this:

Math.min(a,b)  // 13 chars
a<b?a:b        //  7 chars

Math.max(a,b)
a>b?a:b

Answer 23

Free commas!

Often you'll want to include a comma in a string, perhaps like so:

f=(x,y,z)=>x+","+y+z

By abusing the string representation of arrays, this can be shortened by two bytes:

f=(x,y,z)=>[x,y]+z

This particular instance only works if you have three variables you want to concatenate as shown. You can use the same trick with two, but you need to be careful. There are three variants you might try:

f=(x,y)=>[x,y]
f=(x,y)=>[x,]+y
f=(x,y)=>x+[,y]

The first one will return an actual array rather than a string, which defeats the purpose. The second one looks like it will work, but in fact most modern browsers will remove the trailing comma when parsing the array. The third one will work though, at the same byte count as the second.

---

To put this to good use, say you have a function which creates the range [0...n]:

f=x=>x?[...f(x-1),x]:[0]

If returning a string with a separator is allowed, you might do something like this, saving a few bytes:

f=x=>x?f(x-1)+" "+x:0

However, you can save another byte with an array literal:

f=x=>x?f(x-1)+[,x]:0

Note that depending on how you arrange the recursion, you may end up with a leading or trailing separator, so you'll need to make sure your output format is allowed by the challenge.

Example usage

Answer 24

Prefer .map over .reduce

Consider the following code for summing an array:

a.reduce(function(x,y){return x+y})

Pretty long, right? What if I told you that you could get rid of the return? Well, you can:

a.map(function(x){t+=x},t=0)    // 7 bytes saved

(Although an even shorter way is eval(a.join("+")).)

How about reducing by multiplication, where you have to specify the starting number anyway?

a.reduce(function(x,y){return x*y},1)  // Looooong
a.map(function(x){t*=x},t=1)    // An easy 9 bytes shorter

(Again, eval(a.join("*")) works as well.)

Here, let's try one that doesn't work with eval(a.join()):

a.reduce(function(x,y){return x+f(y)})
a.map(function(x){t+=f(x)},t=0)

Note that this doesn't work quite as well with ES6, although it's still a little shorter:

a.reduce((x,y)=>x+f(y))
a.map(x=>t+=f(x),t=0)

Note: in all of the .map versions, you will need to call t afterwards to get the actual value.

Answer 25

Converting an array of strings into numbers

Take the array ["32", "0x30", "0o10", "7.642", "1e3"]. The simple way to convert this to numbers would be .map(n=>+n) or .map(Number).

However, assuming you know everything is a valid number, you can simply use .map(eval), saving at least a byte.

Answer 26

Something worth noting is that you can use a string in place of zero in some instances to save a couple of bytes here and there in loops:

s='';for(i=0;i++<9;)s+=i
for(i=s='';i++<9;)s+=i
// s="123456789", i=10

Answer 27

How to compare a number with help of how numbers turn into booleans:

If you are going to check if something is equal to a positive number, you can subtract that amount and reverse what was inside the if and else blocks:

//simplified examples:
x==3?"y":"n"; <- 13 Chars
x-3?"n":"y"; <- 12 Chars

//expanded examples:
if(x==3){
    yes();
}else{
    no();
}

if(x-3){
    no();
}else{
    yes();
}

And in case you are wanting to compare with a negative number (*different than -1), you just simply need to add this number instead of subtracting.

*well, you can surely use x.indexOf(y) + 1, but in the special case of -1 you have the opportunity to use ~x.indexOf(y) instead.

Answer 28

In cases where you are using the ternary operator to chose between two numbers, and the conditional is either a boolean or number 1 or 0, you can do math operations instead:

(x ? num1 : num2) conclusions:

    1)if num1 equals num2, there ARE savings
    2)if num1 is (+1) or (-1) than num2, there ARE savings
    3)if either num1 or num2 equals to 0, there ARE savings
    4)it is MORE LIKELY to find greater savings on num1>num2 instead of num1<num2
    5)in method (*A) and (*B), savings are NOT GUARANTEED

    a)num1>num2
        i)(num1==(num2+1))
            ex1: (x?5:4) to (x+4)
            ex2: (x?8:7) to (x+7)
        ii)num2==0
            ex1: (x?3:0) to (x*3)
            ex2: (x?7:0) to (x*7)
        iii)
            (*A) or (*B) //one might be shorter

    b)num1<num2
        i)((num1+1)==num2)
            ex1: (x?4:5) to (5-x)
            ex2: (x?7:8) to (8-x)
        ii)num1==0
            ex1: (x?0:3) to (!x*3)
            ex2: (x?0:7) to (!x*7)
        iii)
            (*A) or (*B) //one might be shorter

    c)num1==num2
        i)
            ex1: (x?5:5) to (5)
            ex2: (x?-3:-3) to (-3)

    (*A) use ((x*(num1-num2))+num2)
        ex1: (x?8:4)   to ((x*4)+4)
        ex2: (x?4:8)   to ((x*-4)+8)

        ex3: (x?6:-4)  to ((x*10)-4)
        ex4: (x?-4:6)  to ((x*-10)+6)

        ex5: (x?4:-6)  to ((x*10)-6)
        ex6: (x?-6:4)  to ((x*-10)+4)

        ex7: (x?-5:-9) to ((x*4)-9)
        ex8: (x?-9:-5) to ((x*-4)-5)

    (*B) use ((!x*(num2-num1))+num1)
        ex1: (x?8:4)   to ((!x*-4)+8)
        ex2: (x?4:8)   to ((!x*4)+4)

        ex3: (x?6:-4)  to ((!x*-10)+6)
        ex4: (x?-4:6)  to ((!x*10)-4))

        ex5: (x?4:-6)  to ((!x*-10)+4)
        ex6: (x?-6:4)  to ((!x*10)-6)

        ex7: (x?-5:-9) to ((!x*-4)-5)
        ex8: (x?-9:-5) to ((!x*4)-9)

Note: In addition to this, you will need to remove the unnecessary 0-, +0, +- etc.

Note2: there is an isolated case where (x) !== (x?1:0), as x must be typeof === "number" for it to work. However, in the case of (-x) it works just fine.

Note3: In case you don't find savings, simply use the former (x?y:z)

Previously I thought method B couldn't ever beat A, however exceptions do exist:

(x?97:100) //original

(-3*x+100)
(3*!x+97)

I created a github project that makes the simplification for us (jsFiddle demo)

Answer 29

Instead of writing true you can use !0.

Answer 30

Transforming to a Boolean:

if(b){b=true}else{b=false}
b=b?true:false;
b=b?!0:!1;
b=!!b;

Note: This changes 0, "",false, null, undefined and NaN to false (everything else to true)