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Could you be a bit more specific as to what you find confusing? Is it: - That you can use the name of a variable, h, as if it was a function? That's because Ja

Could you be a bit more specific as to what you find confusing? Is it: - That you can use the name of a variable, h, as if it was a function? That's because Javascript has first class functions [1] - the language is defined to support passing them around as variables and calling them like that. - That you can use h at all even though it's neither a local variable nor a parameter of the anonymous function? That's because functions in javascript aren't simply procedures in the traditional sense - i.e. description (function signature) + code - they are also closures [2]. If you declare one function inside another, it can capture (have a reference to) variables and parameters of the outer one. You are also guaranteed that local variables and parameters will not get cleaned up while a closure still exists that holds a reference to them. [1] : http://en.wikipedia.org/wiki/First_class_function http://en.wikipedia.org/wiki/First_class_function [2] : Can't vouch for any particular article, try googling Javascript closures

lelandbatey 13y ago
OOOOOOh, ok I see how this is working a bit, I think. I am very inexperienced with Javascript, and I think I only just "got" what was being done here, so I'll try to explain my thought processes as a newbie: So, in Javascript, variables can be functions. Which means you can pass in a function as a variable to another function. And the part that says h(h(y)), basically says that "h" has to be a function. Which means you pass in a function, and then it get's applied to itself in the way specified within that function, "g". Another odd part is the function you pass in: g(function(x) { return x * x; })(3); because you are passing in a function, but I'd assumed that if you can only pass in one variable, and that variable has to be a function, then it seemed like you wouldn't be able to pass in an initial value for the function you want to apply. But I guess in Javascript you can pass in a value for an anonymous function defined inside a function call by using this syntax: g(function(x) { whatever it is that happens in this function; })(some_value_to_be_passed_into_the_previously_defined_anonymous_function);
- minikomi 13y ago
  Just to help in case it isn't obvious - g also returns a function - which could be assigned to a variable. In this case it's called straight away but you could do it like var powerOfFour = g(function(x) {return x*x;}); powerOfFour(3) === 81; //true
- Twisol 13y ago
  `g(f)` returns a function, which is then immediately called with another value. You're passing `3` to the result of `g`, not to the function you passed to `g`.
  - lelandbatey 13y ago
    Ah, I see, that makes sense. Thanks for the explanation.

lelandbatey 13y ago

OOOOOOh, ok I see how this is working a bit, I think. I am very inexperienced with Javascript, and I think I only just "got" what was being done here, so I'll try to explain my thought processes as a newbie: So, in Javascript, variables can be functions. Which means you can pass in a function as a variable to another function. And the part that says h(h(y)), basically says that "h" has to be a function. Which means you pass in a function, and then it get's applied to itself in the way specified within that function, "g". Another odd part is the function you pass in: g(function(x) { return x * x; })(3); because you are passing in a function, but I'd assumed that if you can only pass in one variable, and that variable has to be a function, then it seemed like you wouldn't be able to pass in an initial value for the function you want to apply. But I guess in Javascript you can pass in a value for an anonymous function defined inside a function call by using this syntax: g(function(x) { whatever it is that happens in this function; })(some_value_to_be_passed_into_the_previously_defined_anonymous_function);

minikomi 13y ago
Just to help in case it isn't obvious - g also returns a function - which could be assigned to a variable. In this case it's called straight away but you could do it like var powerOfFour = g(function(x) {return x*x;}); powerOfFour(3) === 81; //true
Twisol 13y ago
`g(f)` returns a function, which is then immediately called with another value. You're passing `3` to the result of `g`, not to the function you passed to `g`.
- lelandbatey 13y ago
  Ah, I see, that makes sense. Thanks for the explanation.