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JavaScript

Posted by Giovanni Dalla Rizza
Giovanni Dalla Rizza
10,637 Points

Function inside a function

function multiplier(factor) {
  return function(number) {
     return number*factor;
  };
}

var twice = multiplier(2);
console.log(twice(5));

// -> 10

Can't understand why the result printed to the console was 10. The argument of var twice shouldn't replace 'factor'? Thanks :)

Cheo R
Cheo R
37,150 Points 
function multiplier(factor) {
  return function(number) { <-- 3. The function, multiplier, returns an anonymous function. 
     return number*factor;  <-- 2. It is then passed along to the anonymous function and assigned to factor.
                            <-- 6. The anonymous function returns the number passed to it multiplied factor.
  };
}

var twice = multiplier(2);   <-- 1. Pass in a number to the multiplier function
                             <-- 4. Anonymous function is assigned to variable twice.

console.log(twice(5));       <-- 5. Calling function, twice, and passing in the number 2.

// -> 10                     <-- 7. So passing 2 to multiplier, and 5 to twice, returns 10.

1 Answer

Clayton Perszyk
MOD
Clayton Perszyk
Treehouse Moderator 49,057 Points
Clayton Perszyk
Clayton Perszyk
Treehouse Moderator 49,057 Points

Giovanni,

What you have created is a closure; even after the outer function has returned, the inner function retains a reference to any variables, including arguments/parameters, created in the context of the outer function. So, when you call multiplier(2), its returning a function that has access to the factor parameter, which has the value 2; when you call twice(5), 5 is being passed as the number argument (i.e. nothing is replacing factor, which is still 2); as such the call to twice(5) is returning the number, 5, multiplied by the factor, 2. What were you expecting as the value returned from twice(5), if not 10.?

Best,

Clayton