Passing a function as a parameter in javascript and not as the function's returned value

Question

Suppose I have a function roll() that randomly returns a value 1-6. Now, if I have another function called repeatFunction that takes a function as a parameter and a number n. The repeatFunction's purpose is to call whatever function it has as a parameter n times. However, if I were to pass roll() as a parameter, the repeatFunction would interpret it as whichever value 1-6 the roll() function returned, rather than a function. My code currently is as follows:

function repeatFunction(func, n){
    for(i = 0; i < n; i++){
        func;
    }
}
repeatFunction(roll(), 10);

How do I get it so repeatFunction interprets the func parameter as a function rather than a returned value so that I can call it again within the repeatFunction?

Aaditya Thakkar · Accepted Answer · 2018-02-03 21:54:06Z

3

pass a reference to the roll function and call it inside the repeat function as a callback. Like this,

function repeatFunction(func, n){
    for(i = 0; i < n; i++){
        func();
    }
}
repeatFunction(roll, 10);

answered Feb 3, 2018 at 21:54

Aaditya Thakkar

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Comments

Ele · Accepted Answer · 2018-02-03 21:58:34Z

1

You need to pass the function name rather than the returned execution.

You're executing the function roll, just pass roll.

repeatFunction(roll(), 10);
               ^

Look at this code snippet

The repeated function will execute the function fn recursively until i == n

function repeatFunction(fn, n, i){
    if (i === n) return;
    fn();
    
    repeatFunction(fn, 10, ++i);
}

var roll = function() {
  console.log('Called!');
};

repeatFunction(roll, 10, 0);

.as-console-wrapper {
  max-height: 100% !important
}

See? the function was called n times.

answered Feb 3, 2018 at 21:58

Ele

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