JavaScript curry function

Question

I have implemented a curry function this way:

function curry (fn) {
    var slice = Array.prototype.slice,
        args = slice.apply(arguments,         


        

Answer 1

You're missing a return in your curry function.

I.e.

return function () {
    return fn.apply(null, args.concat(slice.apply(arguments)));
};

That seems to work :)

Answer 2

This is an example of curry function:

function curry(func) {
    var initial_args = [].slice.apply(arguments, [1]);
    var func_args_length = func.length;

    function curried(args) {
        if (args.length >= func_args_length) {
            return func.apply(null, args);
        }

        return function () {
            return curried(args.concat([].slice.apply(arguments)));
        };
    }

    return curried(initial_args);
}

This curry function creates a curried version of any function. You can partially apply parameters to the new function. For each partial application, it will return a function that takes the rest of the parameters. Once all parameters of the original function are applied, it will return a value.