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Here's my code:

var x = 3;
var y = x++;
y += 1;

Output: y = 4

I know the computer is right, but I'm not sure why the computer is right. y gets assigned the value 3 from x, then it increments it to 4 in line 2. So, the output from line 3 should be 5, correct?

From what I've read, y gets assigned the value of x before the increment happens, but when it does happen, why does the value of y not change?

SeriousBurn
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    Assignment of a value from one variable to another in JavaScript involves a *copy* of the value. In `y = x++;` no permanent relationship between `x` and `y` is established; it's a one-time copy. – Pointy Jan 10 '19 at 18:42
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    *"y gets assigned the value of x"* That's the crucial point. `x++` is equivalent to `x = x + 1`. So `y = x++` is the same as `y = x; x = x + 1;`. `x` is assigned a new value, but that has no effect on the value of `y`. – Felix Kling Jan 10 '19 at 18:45

3 Answers3

4

In your assignment y = x++; the value of y is first assigned to x and then the variable x gets incremented by 1. By performing this operation y becomes 3 and x is 4. Then after running y +=1 computer will calculate 3+1 = 4

If you're expecting y to be 5 you should do y = ++x;. By doing this x will first get incremented by 1 and then assigned to y so we will have y = 4 and x = 4 following the y += 1 (4+1=5)

rollingthedice
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  • Oh okay, I see. **y** gets assigned **x**, and then **x** gets incremented by **1** because that's the variable being targetted by the program in this instance. So **y** becomes **3**. Then the value of **x** gets incremented by **1** since they have no relation to each other. – SeriousBurn Jan 10 '19 at 18:57
  • Exactly, yes. :) – rollingthedice Jan 10 '19 at 18:59
3

There is a difference between pre-increment (++x) and post-increment (x++).

A pre-increment operator is used to increment the value of a variable before using it in a expression. In the pre-increment, value is first incremented and then used inside the expression. Let's say we have:

a = ++x;

Here, if the value of ‘x’ is 10 then value of ‘a’ will be 11 because the value of ‘x’ gets modified before using it in the expression. This is equivalent with:

x = x + 1;
a = x;

A post-increment operator is used to increment the value of variable after executing expression completely in which post increment is used. In the Post-Increment, value is first used in a expression and then incremented. Let's say we have:

a = x++;

Here, suppose the value of ‘x’ is 10 then value of variable ‘a’ will be 10 because old value of ‘x’ is used. This is equivalent with:

a = x;
x = x + 1;

You can read more on the interned about this (for example, here or here).

Cheers!

// Post-increment example
console.log("post-increment examples");
let x = 10;
a = x++;
console.log(x, a);

x = 10;
a = x;
x = x + 1;
console.log(x, a);

// Pre-increment example
console.log("pre-increment examples");
x = 10;
a = ++x;
console.log(x, a);

x = 10;
x = x + 1;
a = x;
console.log(x, a);
Adrian Pop
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1

x++ return the value and then add 1. See this: 

var x = 3;
var y = x++;// x return 3 and then add 1, y is 3 
y += 1;//3 + 1 = 4
console.log(y)
console.log(x)//x return 4
Emeeus
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