Following code is given:
var a = [ ], i = 0, j = 0;
for (i = 0; i < 5; i += 1) {
(function(c) {
a.push(function () {
console.log(c); });
})(i);
};
for (j = 0; j < 5; j += 1) { a[j](); }
Why does i always get bigger by 1 instead of staying 5? Hasn't the foor loop already been passed, so the i parameter given to the anonymous function should be 5?
If you referenced i from the inner closure then yes, you would see the result being 5 in all cases. However, you pass i by value to the outer function, which is accepted as parameter c. The value of c is then fixed to whatever i was at the moment you created the inner closure.
Consider changing the log statement:
console.log("c:" + c + " i:" + i);
You should see c going from 0 to 4 (inclusive) and i being 5 in all cases.
c variable, each of which is initialized to the value that i held when the closure was created.chhowie's answer is absolutely right (and I upvoted it), but I wanted to show you one more thing to help understand it. Your inner function works similarly to a more explicit function call like this:
var a = [ ], i = 0, j = 0;
function pushFunc(array, c) {
array.push(function () {
console.log(c);
});
}
for (i = 0; i < 5; i += 1) {
pushFunc(array, i);
}
for (j = 0; j < 5; j += 1) { a[j](); }
Which should also help you understand how c comes from the function argument, not from the for loop. Your inner function is doing exactly the same thing as this, just without an externally declared named function.
i.