How to swap two variables in JavaScript

问题:

I have this two variables:

var a = 1,
    b = 2;

My question is how to swap them? Only this variables, not any objects.

回答1:

Here's a one-liner to swap the values of two variables.
Given variables a and b:

b = [a, a = b][0];

Demonstration below:

var a=1,
    b=2,
    output=document.getElementById('output');

output.innerHTML="<p>Original: "+a+", "+b+"</p>";

b = [a, a = b][0];

output.innerHTML+="<p>Swapped: "+a+", "+b+"</p>";

<div id="output"></div>

回答2:

ES6 (Firefox and Chrome already support it (Destructuring Assignment Array Matching)):

let a = 5, b = 6;
[a, b] = [b, a];
console.log(`${a} ${b}`);

回答3:

You can do this:

var a = 1,
    b = 2,
    tmp;
tmp = a;
a = b;
b = tmp;

For readability and maintainability, this can't be beat (at least in JavaScript). Anybody maintaining the code (including you six months from now) will know exactly what's going on.

Since these are integers, you can also use any number of clever tricks¹ to swap without using a third variable. For instance you can use the bitwise xor operator:

let a = 1, b = 2;
a = a ^ b;
b = a ^ b;
a = a ^ b;
    
console.log('a is now:', a);
console.log('b is now:', b);

This is called the XOR swap algorithm. Its theory of operation is described in this Wikipedia article.

¹"The competent programmer is fully aware of the limited size of his own skull. He therefore approaches his task with full humility, and avoids clever tricks like the plague." — Edsger W. Dijkstra

回答4:

Don't use the code below. It is not the recommended way to swap the values of two variables (simply use a temporary variable for that). It just shows a JavaScript trick.

This solution uses no temporary variables, no arrays, only one addition, and it's fast. In fact, it is sometimes faster than a temporary variable on several platforms.
It works for all numbers, never overflows, and handles edge-cases such as Infinity and NaN.

a = b + (b=a, 0)

It works in two steps:

• (b=a, 0) sets b to the old value of a and yields 0
• a = b + 0 sets a to the old value of b

回答5:

Since ES6, you can also swap variables more elegantly:

var a = 1,
    b = 2;

[a, b] = [b, a];

console.log('a:', a, 'b:', b); // a: 2 b: 1

回答6:

Here's a one-liner, assuming a and b exist already and have values needing to be swapped:

var c=a, a=b, b=c;

As @Kay mentioned, this actually performs better than the array way (almost 2x as fast).

回答7:

ES6+ method: Since ES6, you can swap variables more elegantly. You can use destructuring assignment array matching. It’s simply. var a=10 b=20;

[a, b] = [b, a]

console.log(a,b) // 20 10

回答8:

Use a third variable like this:

var a = 1,
    b = 2,
    c = a;

a = b; // must be first or a and b end up being both 1
b = c;

---

DEMO - Using a third variable

---

回答9:

You can now finally do:

var a = 5;
var b = 10;

[a, b] = [b, a]; // ES6

console.log(a, b);

回答10:

You could use a temporary swap variable or XOR.

a = a ^ b
b = a ^ b
a = a ^ b

This is just a basic logical concept and works in every language that supports XOR operation.

edit: see the Comments. Forgot to tell that this works for sure only with integer. Assumed the integer variables from question's thread

回答11:

ES6 Destructuring:

Using an array: [a, b] = [b, a]; // my favorite

Using an object: {a, b} = {a:b, b:a}; // not bad neither

回答12:

As your question was precious "Only this variables, not any objects. ", the answer will be also precious:

var a = 1, b = 2

a=a+b;
b=a-b;
a=a-b;

it's a trick

And as Rodrigo Assis said, it "can be shorter "

 b=a+(a=b)-b;

Demo: http://jsfiddle.net/abdennour/2jJQ2/

回答13:

You can use ES6 destructuring assignment like so:

let a = 10;
let b = 20;

[a, b] = [b, a]; 

console.log(a, b); // a = 20, b = 10

回答14:

How could we miss these classic oneliners

var a = 1, b = 2
a = ({a:b, _:(b=a)}).a;

And

var a = 1, b = 2
a = (_=b,b=a,_);

The last one exposes global variable '_' but that should not matter as typical javascript convention is to use it as 'dont care' variable.

回答15:

Single line swapping

a = a^b^(b^=(a^b));

回答16:

var a = 5;
var b = 10;

b = [a, a = b][0];
//or
b = [a, a = b];
b = b[0];

//or
b = [a, b];
a = b[1];
b = b[0];


alert("a=" + a + ',' + "b=" + b);

remove or comment the 2 //or's and run with the one set of code

http://jsfiddle.net/USdv8/57/

回答17:

In ES6 now there is destructuring assignment and you can do:

let a = 1;
let b = 2;
[b, a] = [a, b]  // a = 2, b = 1

回答18:

I see kind of programming olympiad here. One more tricky one-line solution:

b = (function(){ a=b; return arguments[0]; })(a);

Fiddle: http://jsfiddle.net/cherniv/4q226/

回答19:

We are able to swap var like this :

var val1 =  117,
    val2 = 327;

val2 = val1-val2; 
console.log(val2);
val1 = val1-val2;
console.log(val1);
val2 = val1+val2;
console.log(val2);

回答20:

let a = 2, b = 4;
[b, a] = [a, b];

a more verbose approach would be

let a = 2, b = 4;
a = [a, b];
b = a[0];
a = a[1];

回答21:

It's very simple, use the ES6 array destructuring syntax which is [y, x] = [x, y] for more information consult this link https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Destructuring_assignment

回答22:

Till ES5, to swap two numbers, you have to create a temp variable and then swap it. But in ES6, its very easy to swap two numbers using array destructuring. See example.

let x,y;
[x,y]=[2,3];
console.log(x,y);      // return 2,3

[x,y]=[y,x];
console.log(x,y);      // return 3,2

回答23:

(function(A, B){ b=A; a=B; })(parseInt(a), parseInt(b));