function firstFunction(num, callback) {
  callback(num);
};

function secondFunction(num) {
  return num + 99;
};

console.log(firstFunction(56, secondFunction));

undefined

If I call console.log from within secondFunction, it returns the value.

Why not? What's the point of setting up callbacks if I can't get the value out of them to use later? I'm missing something.

function firstFunction(num, callback) {
  callback(num);
};

function secondFunction(num) {
  return num + 99;
};

console.log(firstFunction(56, secondFunction));

undefined

If I call console.log from within secondFunction, it returns the value.

Why not? What's the point of setting up callbacks if I can't get the value out of them to use later? I'm missing something.

• javascript
• callback

• 2 You forgot the chain the return value from callback. – Ja͢ck Commented May 25, 2013 at 2:50

2 Answers 2

In your function firstFunction, you do:

callback(num);

Which evaluates to

56 + 99;

Which is then

155;

But you never return the value! Without a return value, a function will simply evaluate to undefined.

Try doing this:

function firstFunction(num, callback) {
  return callback(num);
};

firstFunction does not return anything, plain and simple! That is why when you console.log the return value, it is undefined.

The code in question:

callback(num);

calls callback and then does nothing with the returned value. You want:

return callback(num);