CODEWITHSUNDEEP
javascript
Mahdir
Mahdir

Reputation: 151

create a function that takes another function into itself

spread(someFunction, [1, true, "Foo", "bar"] )  // is the same as...
someFunction(1, true, "Foo", "bar")

I wrote this code but got some errors:

function spread(func, args) {
    func();
}
function func(args){
    console.log(args[0] + args[1]);
}
spread(func,[5,6])

TypeError: args is undefined, but I can't define it because it will be defined when call the function.

spread(func,[5,6])

And i can't use variable either. I can just this kind of function. (This is a test)

Upvotes: 0

Views: 130

Answers (2)

plalx
plalx

Reputation: 43718

There is only 2 ways to implement such function in JS. One which is generating a string that expresses the function call and evaluate it and the other using Function.prototype.apply.

spread(add, [1, 2]); //3

function add(num1, num2) { return num1 + num2; }

function spread(fn, args) {
    return fn.apply(null, args);
}

Upvotes: 2

flott
flott

Reputation: 231

You must pass the arguments to func and also return it if you want to keep the result in a variable. 

function spread(func, args) {
    return func(args);
}
function func(args){
    console.log(args[0] + args[1]);
    //do something more, maybe:
    return (args[0] + args[1])
}
var result = spread(func,[5,6])

Upvotes: 0

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