I have a simple question here. Here is the code below
var num = 0;
var increment = function() {
return function() { num++; };
};
increment();
increment();
However, when I tried to run it, it errors with undefined is not a function. How come? Isn't increment clearly a function?
Also, when I wrote typeof increment, it returns undefined.
When increment() is called twice, it should modify num, and become 2.
I'm guessing you want to do something like this:
> var makeIncrement = function () { var num = 0; return function () { return num++ } }
undefined
> increment = makeIncrement()
function () { return num++ }
> increment()
0
> increment()
1
The function has to be called (makeIncrement()) in order to return the inner function, which you can then assign to a variable. Once it has been assigned, you can call it and it will work as you expect.
var num so the num variable isn't visible to the rest of the code. This means that you could have several increment functions, each with their own independent counter.var num = 0;
var increment = function() {
num++;
};
increment();
increment();
but what are you returning with return function()? You should just return num.
var num = 0;
var increment = function() {
num++
return num;
};
var something = 0;
something = increment();
something = increment();
alert(something);
As per your comment, if you want to return a function from your function, call the function returned by increment() :
var num = 0;
var increment = function() {
return function() { num++; };
};
increment()();
increment()();
Or, better suited :
var incrementer = increment();
incrementer(); //Call this each time you want to increment num
It's not a function ... exactly. Your function "increment" returns a function, not the incremented value. The correct way should be: var increment = function() { return num++; };
Phil
increment looks like a function to me. What it returns doesn't matter.Yes, the increment function does not increment the variable; it returns an inner function that would increment the variable when invoked.
As written, you would have to invoke the result of invoking the increment function:
increment()(); //results in num++
OR
var incrementFunction = increment(); //does not increment
incrementFunction(); //num++
If you want to assign to increment the inner function, then you need to execute the outer function:
var increment = (function() {
return function() { num++; };
}());
(This is called an immediately-invoked function expression, or IIFE for short.) A simpler approach, of course, is:
var increment = function() { num++; };
but perhaps you have other reasons for wanting to do it in a more complicated way.
var increment = function() { num++; }?increment()