Javascript array with callback function

Question

I tried to have better understanding of JavaScript. Here is a piece of code that I read from JavaScript function closures.

var funcs = [];
// create a bunch of functions
for (var i = 0; i < 3; i++) {
 funcs.push(function() {
 console.log(i);
 })
}
// call them
for (var j = 0; j < 3; j++) {
 funcs[j]();
}

The array funcs has a push callback function. I don't why in the J loop, funcs[j]() will call this function to print the i in the console.
I have tried to understand this sequencey by adding some console messages:

var funcs = [];
console.log("start");
for (var i = 0; i < 3; i++) {
 console.log("i:" + i);
 funcs.push(function(){
 console.log(i);
 })
}
console.log("J loop");
for (var j=0; j<3; j++) {
 console.log("j:" + j);
 funcs[j]();
}

As expected, there is 3 for all three functions.
My question is: How does funcs[j]() calls the funcs.push(...) function? I understant the funcs[j] is reference the j element of the funcs array. But why having parentheses () will call the push(...) function?

• Pointy: It doesn't call the .push() function; it calls the function that was passed to .push(), the little function with console.log(i); inside.
• Shaohao: @Pointy can you explain a little bit more? That's what I am confused about. Why does it call the function that was passed to .push()? It is a syntax or something else?
• Darlyn: .push add element at the end of the array. In first loop you pushed a function to the array. In second loop , you are invoking the function with () [] indicating index.
• T.J. Crowder: Possibly separately from what confusing you, the code above also has the issue that the value of i may surprise you: stackoverflow.com/questions/750486/…

Answer 1

function() {console.log(i);} is an expression which evaluates to a value that is function that logs i.

funcs.push is a function that adds a value to an array.

Putting () after a function will call that function.

funcs.push(some_value) calls the push function and passes some_value as the value to put in the array.

funcs.push(function() {console.log(i);}) adds the function to the array.

The value of funcs[0] becomes that function.

Putting () after a function will call that function.

funcs[0]() calls the function that is the first value in the array.

• Shaohao: Thank you for the step by step explanation! It's very helpful!
• Shaohao: So putting () after a function will always invoke whatever function in the push()?
• Quentin: Putting () after a function will invoke that function. Putting it after something which represents the function which was pushed into the array will invoke that function.

Answer 2

First, the 'i' variable are global, and ending the loop, i=3 Then, the functions inside funcs, use the variable "i", then, all functions print "3" in console.

Maybe you wanted to do this:

var funcs = [];
console.log("start");
for (var i = 0; i < 3; i++) {
 console.log("i:" + i);
 funcs.push(function(i){
 console.log(i);
 })
}
console.log("J loop");
for (var j=0; j<3; j++) {
 console.log("j:" + j);
 funcs[j](j);
}

• Shaohao: Thanks for the answer, but I don't have a problem to understand why it prints 3 three times. I just have a problem about why () will invoke the console.log(i) function. And @Quentin did a great job to explain it.