Question

The comma operator evaluates each comma-separated operand (and returns the value of the last one).

(i = 1 + 2), (j = 3 + 4);

is functionally equivalent to

i = 1 + 2;
j = 3 + 4;

Also, as far as I know, a statement, such as var, is not considered an operator, but rather part of the operand. (See https://developer.mozilla.org/en/JavaScript/Reference/Operators/Operator_Precedence )

So if each operand is being separately evaluated, why, then, does

function foobar () {
    var i = 3, j = 7, z;
}

create 3 variables – i, j, and z – in foobar’s scope?

I know that’s what actually happens, but I’ve been wondering for a while why it is that this actually happens. It would seem that i should be in foobar’s scope, but that j and z should end up in the global scope.

Answer 1

The commas used in a var statement are not actually comma operators.

Similarly, the commas you use to separate arguments in a function call are not comma operators either.