Explain, please, these ++ and -- operations

Why this code ou开发者_开发问答tputs 3, not 2?

var i = 1; 
i = ++i + --i; 
console.log(i);

I expected:

++i // i == 2
--i // i == 1
i = 1 + 1 // i == 2

Where I made mistake?

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The changes occur in this order:

1. Increment i (to 2)
2. Take i for the left hand side of the addition (2)
3. Decrement i (to 1)
4. Take i for the right hand side of the addition (1)
5. Perform the addition and assign to i (3)

… and seeing you attempt to do this gives me some insight in to why JSLint doesn't like ++ and --.

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Look at it this way

x = (something)
x = (++i) + (something)
x = (2) + (something)
x = (2) + (--i)
x = (2) + (1)

The terms are evaluated from left to right, once the first ++i is evaluated it won't be re-evaluated when you change its value with --i.

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Your second line is adding 2 + 1.

In order, the interpreter would execute:

++i  // i == 2
+
--i  // i == 1
i = 2 + 1

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++i equals 2, `--i' equals 1. 2 + 1 = 3.

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You're a little off on your order of operations. Here's how it goes:

1. i is incremented by 1 (++i) resulting in a value of 2. This is stored in i.
2. That value of two is then added to the value of (--i) which is 1. 2 + 1 = 3

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Because when you use ++i the value of i is incremented and then returned. However, if you use i++, the value of i is returned and then incremented. Reference

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++$a   Increments $a by one, then returns $a.
$a++   Returns $a, then increments $a by one.
--$a   Decrements $a by one, then returns $a.
$a--   Returns $a, then decrements $a by one.

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Because you're expecting this code to work as if this is a reference object and the values aren't collected until the unary operations are complete. But in most languages an expression is evaluated first, so i returns the value of i, not i itself.

If you had ++(--i) then you'd be right.

In short, don't do this.

The result of that operation isn't defined the same in every language/compiler/interpreter. So while it results in 3 in JavaScript, it may result in 2 elsewhere.