1

In the below example. lname is callback function inside fname function. When executed, first output is Smith and the Billy. As per my understanding, callback function lname should be executed once fname function is finished executing. Why am wrong here? 

function fname(){ console.log("Billy");}
function lname(){ console.log("Smith");}
fname(lname());
vegabond
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    Because you're not passing `lname` as callback (function as argument), you're actually passing _the result of `lname`_ as argument... – IronGeek Mar 19 '19 at 03:09

4 Answers4

4

The arguments to a function are evaluated before it's called. So fname(lname()) results in lname being called, followed by fname. The result of lname is passed to fname as the first argument but it is not used by the called function.

jspcal
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3

In the below example. lname is callback function inside fname function.

No, it is not. You are invoking lname (because you added the parentheses), and passing the result of its invocation (which happens to be undefined, as is the case any time a function does not explicitly return a value) to fname. Just like Math.floor(Math.sqrt(10)) - there are no callbacks here. There is a big difference between fname(lname()) and fname(lname): the latter one passes a function itself into another function; and we'll call it a "callback" if it is, in fact, called back.

As per my understanding, callback function lname should be executed once fname function is finished executing.

Again, no. A callback is executed when it is invoked, no more, no less. For example:

function printsSomething() {
  console.log("Something");
}
function invokesCallbackAtEnd(callback) {
  console.log("invokesCallbackAtEnd doing something");
  callback();
}
function invokesCallbackAtStart(callback) {
  callback();
  console.log("invokesCallbackAtStart doing something");
}
function invokesCallbackAtStartAndEnd(callback) {
  callback();
  console.log("invokesCallbackAtStartAndEnd doing something");
  callback();
}
function invokesCallbackNever(callback) {
  console.log("invokesCallbackNever doing something");
}
invokesCallbackAtEnd(printsSomething); console.log("---");
invokesCallbackAtStart(printsSomething); console.log("---");
invokesCallbackAtStartAndEnd(printsSomething); console.log("---");
invokesCallbackNever(printsSomething); console.log("---");
Amadan
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  • If callback function is called back by outer most function, then what do it means when you say "A callback is executed when it is invoked, no more, no less." I understood from your example now, but still getting confused. – vegabond Mar 19 '19 at 08:38
  • I mean a callback is called "callback" because it is meant to be "called back" from the function you passed it to, where "call" means the same as "invoke". – Amadan Mar 19 '19 at 08:41
3

You aren't defining a callback function as such, to do so you will have to pass the reference to the function rather than the function call itself. i.e fname(lname); And you have to change your function fname to

function fname (cb) {
  console.log("Billy");
  cb && cb.apply(this);
}
varun agarwal
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3

You did not pass lname as a callback function. You just invoked lname function and passed its return value undefined.

And even if you passed a callback function, its execution point depends on code.

function fname(func) {
  func("Foo");
  console.log("Billy");
  func("Bar");
}

function lname() {
  console.log("Smith");
  return function (arg) { console.log(arg); }
}
fname(lname());
zmag
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