I've finally been curious enough to find out why javascript does its voodoo magic to learn why not all object references are created equal.
Given the example:
var a, b, c, d;
a = 100; b = a;
c = {}; d = c;
b = 10; d.e = 'f';
console.log(a, b); // outputs 100, 10
console.log(c, d); // outputs object => e = 'f', object => e = 'f'
If all variables in javascript are objects, then what makes the use case with c and d cast explicitly as an Object so different than defining a and b as Number? Or, why will c and d be linked to one another, and not a and b?
All variables in JavaScript are not objects. There are native types as well.
c and d are not linked to one another. They are pointing to the same object reference. If you were to reassign d to something else, it will not affect c.
var c = {};
var d = c;
d = { foo: "bar" };
c === d // false
However, if you were to modify the object being referenced by c or d, it will modify the same object since c and d are both referring to the same object as in your example.
It looks to me that the difference is with b, you're reassigning the variable to a new object/value, while with d, you're modifying the existing object.
The value of a that is assigned to b is a Number. The value assigned from c to d is a reference to an Object.
var a, b, c, d;
a = 100; // a has value 100, a number
b = a; // b has value 100, a number
c = {}; // c has value p, a reference to some object P
d = c; // d has value p, a reference to P
b = 10; // b has value 10, a number
d.e = 'f'; // P.e has value 'f', a string
Related
let a = 'alpha', b = 'beta';
[a,b] = [b,a];
This swaps the values of a and b as intended;
but when placed inside a function it doesn't work
let c = 'charlie', d = 'delta';
swapVar = (x,y) => [x,y] = [y,x]
swapVar(c,d);
What am I missing here?
When you do
let a = 'alpha', b = 'beta';
[a,b] = [b,a];
You're swapping the values of a and b.
When you do
let c = 'charlie', d = 'delta';
swapVar = (x,y) => {
// x and y are separate variables scoped within this block
[x,y] = [y,x]
console.log(x,y); // it is swapped alright but isn't reflected on c and d
c = x;
d = y;
// Now the value will have been reflected.
}
swapVar(c,d);
So, within the function the values are swapped but aren't reflected outside. You could modify the program like this:
swapVar = (x,y) => [y,x]
[c, d] = swapVar(c, d); // now you're reflecting the swapped values on the outside
To have the intended effect.
You are doing the swap in a scope where the variables aren't being "exported".
In the first example, you act on the actual variables a and b in the scope they are defined in.
However, in the second example, you are acting on the variables x and y which are the same value as c and d but aren't the actual c and d since they are primitives so the c and d outside the scope of the arrow function is not affected.
{
let a = 'alpha',
b = 'beta';
console.log("Test 1");
console.log(`before a: ${a} b: ${b}`);
[a, b] = [b, a];
console.log(`after a: ${a} b: ${b}`);
}
{
let c = 'charlie',
d = 'delta';
console.log("Test 2");
console.log(`before c: ${c} d: ${d}`);
swapVar = (x, y) => [x, y] = [y, x]
/*
function swapVarExpanded(x, y) {
const tmp = [y, x];
x = tmp[0];
y = tmp[1];
// Doesn't actually matter
// because x and y die at the next closing curly brace due to scope
}
*/
swapVar(c, d);
console.log(`after c: ${c} d: ${d}`);
}
function partialize(){
}
function calculation(a,b,c){
console.log(a*b/c);
return a*b/c;
}
var a = 10, b= 20, c= 5;
var partialize1 = partialize(calculation, a);
partialize1(b,c)
var partialize2 = partialize(calculation, a, b);
partialize2(c)
var partialize3 = partialize(calculation, a, b, c);
partialize3()
I need to write partialize function which give same output in all three condition.
I tried like that it work .but i used spread operator .can we do this without spread operator ?
function partialize(fn,...a) {
console.log(...a);
return function (...b) {
console.log(...b);
fn(...a,...b);
}
}
function calculation(a, b, c) {
console.log(a * b / c);
return a * b / c;
}
var a = 10, b = 20, c = 5;
var partialize1 = partialize(calculation, a);
partialize1(b, c)
var partialize2 = partialize(calculation, a, b);
partialize2(c)
var partialize3 = partialize(calculation, a, b, c);
partialize3()
can we do the same thing without spread operator ?
You can save the initial arguments that were passed and return a function that can be called with the rest of the arguments, then calling the original function with apply:
function partialize(fn) {
const initialArguments = Array.from(arguments).slice(1);
return function() {
const finalArguments = Array.from(arguments);
fn.apply(null, initialArguments.concat(finalArguments));
}
}
function calculation(a, b, c) {
console.log(a * b / c);
return a * b / c;
}
var a = 10,
b = 20,
c = 5;
var partialize1 = partialize(calculation, a);
partialize1(b, c)
var partialize2 = partialize(calculation, a, b);
partialize2(c)
var partialize3 = partialize(calculation, a, b, c);
partialize3()
If your code is currently working as is but you'd like to change it to not use the spread operator, you can use the arguments object instead.
arguments object:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Functions/arguments
Also check out this stackoverflow question for some example code working with the arguments object if helpful. How can I convert the "arguments" object to an array in JavaScript?
user944513, to simulate an overload of methods in javascript, yo can use the arguments object, which comes by default in the functions. To explain you better, i've written a block of code. Hope this can help you:
function suma(a = 0){
let resultado = 0;
console.log(arguments.length);
for(let i = 0; i < arguments.length; i++)
{
if(typeof(arguments[i] == "number")){
resultado += arguments[i];
}
}
}
suma(1,2,3,4); // 10
suma(1,2) // 3
When arrays are assigned to another variable, the reference is passed as opposed to values. This is confirmed when you compare two arrays using == operator and it returns true
var a = [[1,2],[3,4],[5,6]];
var b = a; // b = [[1,2],[3,4],[5,6]]
var c = [].concat(a); // c = [[1,2],[3,4],[5,6]]
a == b; //true
a == c; //false
With the above inputs, when I modify the array b, it mutates array a, but not c.
b.push([7,8]); // b = [[1,2],[3,4],[5,6], [7,8]]
a; //a = [[1,2],[3,4],[5,6], [7,8]]
c; //c = [[1,2],[3,4],[5,6]]
But when I do the below, it mutates c.
b[0].push(5); // b = [[1,2,5],[3,4],[5,6], [7,8]]
a; //a = [[1,2,5],[3,4],[5,6], [7,8]]
c; //c = [[1,2,5],[3,4],[5,6]]
Why does this happen? This behavior happens with the usage of array methods that mutates the array.
.concat() does a shallow copy. So after the line:
var c = [].concat(a);
c and a reference different arrays, but c[0] and b[0] and a[0] all reference the same array.
Quoting from MDN:
concat copies object references into the new array. Both the original and new array refer to the same object. That is, if a referenced object is modified, the changes are visible to both the new and original arrays.
I have this two variables:
var a = 1,
b = 2;
My question is how to swap them? Only this variables, not any objects.
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>";
// swap values for variables "a" and "b"
b = [a, a = b][0];
output.innerHTML+="<p>Swapped: "+a+", "+b+"</p>";
<div id="output"></div>
ES6 (Firefox and Chrome already support it (Destructuring Assignment
Array Matching)):
let a = 5, b = 6;
[a, b] = [b, a];
console.log(`${a} ${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;
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
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
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
You can now finally do:
let a = 5;
let b = 10;
[a, b] = [b, a]; // ES6
console.log(a, b);
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).
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
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
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/
ES6 Destructuring:
Using an array: [a, b] = [b, a]; // my favorite
Using an object: {a, b} = {a:b, b:a}; // not bad neither
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.
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/
Single line swapping
a = a^b^(b^=(a^b));
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/
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);
first way,
var a = 5, b = 9;
a = a - b;
b = a + b;
a = b - a;
console.log(a, b);
second way
var a = 19, b = 22;
[a, b] = [b, a];
console.log(a, b);
simple and clear answer.
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'}
We can use the IIFE to swap two value without extra parameter
var a = 5, b =8;
b = (function(a){
return a
}(a, a=b));
document.write("a: " + a+ " b: "+ b);
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
Know more about JavaScript ES6 destructuring
Although the same answer is given previously, but here is a png to describe it.
Simplest form possible:
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];
var a = 1, b = 9;
a, b = b, a;
console.log(a,b)
1 9
Is possible that the assignments were carried out in left-to-right order? So, "a" would take the value of "b" and and "b" the value of "a".
Swapping values should be pretty simple:
var temp = b;
b = a;
a = temp;
EDIT: if it's all about integers, swapping can happen even without additional variable:
b = b - a;
a = a + b;
b = a - b;