I have this two variables:
var a = 1,
b = 2;
My question is how to swap them? Only this variables, not any objects.
This question is related to
javascript
variables
swap
You can now finally do:
let a = 5;
let b = 10;
[a, b] = [b, a]; // ES6
console.log(a, b);
_x000D_
We can use the IIFE to swap two value without extra parameter
var a = 5, b =8;_x000D_
b = (function(a){ _x000D_
return a _x000D_
}(a, a=b));_x000D_
_x000D_
document.write("a: " + a+ " b: "+ b);
_x000D_
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
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
ES6 Destructuring:
Using an array: [a, b] = [b, a]; // my favorite
Using an object: {a, b} = {a:b, b:a}; // not bad neither
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).
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.
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
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
Swap using Bitwise
let a = 10;
let b = 20;
a ^= b;
y ^= a;
a ^= b;
Single line Swap "using Array"
[a, b] = [b, a]
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
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
ES6 (Firefox and Chrome already support it (Destructuring Assignment Array Matching)):
let a = 5, b = 6;_x000D_
[a, b] = [b, a];_x000D_
console.log(`${a} ${b}`);
_x000D_
ES6 array destructuring is used to swap two variables. See example
var [x,y]=[1,2];
[x,y]=[y,x];
Easier way possible with :
x === 1
and y === 2
;
But after destructuring, x
is y
, i.e. 2
, and y
is x
, i.e. 1
.
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;
Destructing assignment is the best way to solve your problem.
var a = 1;
var b = 2;
[a, b] = [b, a];
console.log("After swap a =", a, " and b =", b);
_x000D_
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
Single line swapping
a = a^b^(b^=(a^b));
(function(A, B){ b=A; a=B; })(parseInt(a), parseInt(b));
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;_x000D_
a = a ^ b;_x000D_
b = a ^ b;_x000D_
a = a ^ b;_x000D_
_x000D_
console.log('a is now:', a);_x000D_
console.log('b is now:', b);
_x000D_
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
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];
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
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
Because I hear this method runs slower:
b = [a, a = b][0];
If you plan on storing your vars in an object (or array), this function should work:
function swapVars(obj, var1, var2){
let temp = obj[var1];
obj[var1] = obj[var2];
obj[var2] = temp;
}
Usage:
let test = {a: 'test 1', b: 'test 2'};
console.log(test); //output: {a: 'test 1', b: 'test 2'}
swapVars(test, 'a', 'b');
console.log(test); //output: {a: 'test 2', b: 'test 1'}
I see kind of programming olympiad here. One more tricky one-line solution:
b = (function(){ a=b; return arguments[0]; })(a);
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);
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
Source: Stackoverflow.com