Hi I'm trying to run the following function:
function add (a, b) { return a + b; } var make_lazy = function (add, a, b) { return function () { add(a,b); } } Basically what I'm trying to do is to pass another function as an argument and its parameters to the make_lazy function - and then run the function that was passed in as the argument along with the other two parameters. I get undefined is not function as an error when I try to run the code.
You forgot the return statement in the anonymous function that you're returning from make_lazy:
var make_lazy = function (add, a, b) { return function () { return add(a,b) // <----- here } } I think you are trying for something like this.
function add (a, b) { return a + b; } var make_lazy = function (a, b) { return function () { return add(a,b); } } Then you can call var lazy = make_lazy(3,5); and later call lazy() to get 8
add if it's the only function you're ever going to pass.
Wrap you lazy function body inside a self calling function and return.
function add (a, b) { return a + b; } var make_lazy = function (add, a, b) { return (function () { add(a,b); })(); } Here's my point of view.
When you assign a function to make_lazy variable, after that you should make an invocation make_lazy() with the same params as they were in the definition of that function:
make_lazy(function expression, a, b); This portion:
function (add, a, b) just makes add a local variable, this is not the same as add(a,b) which is defined above.
To make the code work, try to invoke make_lazy as
make_lazy(add, 3, 4) 