Question

How can an inner function call a parent function after it has expired?

setTimeout(main, 2000);
function main(){
    /* .... code */
    setTimeout(console.log("hello after 5 seconds"), 5000);
}

The intended action is to print hello after 5 seconds in 5 seconds (7 total); with the above code it prints it in 2 seconds.

Answer 1

You need to pass setTimeout function references. With setTimeout(console.log("hello after 5 seconds"), 5000);, you call console.log immediately. Any time you write () after a function name, you’re invoking it.

console.log returns undefined, which is what is passed to setTimeout. It just ignores the undefined value and does nothing. (And it doesn’t throw any errors.)

If you need to pass parameters to your callback function, there are a few different ways to go.

Anonymous function:

setTimeout(function() {
    console.log('...');
}, 5000);

Return a function:

function logger(msg) {
    return function() {
        console.log(msg);
    }
}

// now, whenever you need to do a setTimeout...
setTimeout(logger('...'), 5000);

This works because invoking logger simply returns a new anonymous function that closes over msg. The returned function is what is actually passed to setTimeout, and when the callback is fired, it has access to msg via the closure.