JavaScript "arguments" -passing a function with other parameters

Function multiply below is passed a callback function "addOne". Function multiply returns [3,5,7].

Since the callback function addOne is one of the arguments of function multiply, and since the arguments of function mu开发者_如何学运维ltiply are all multiplied by 2, why doesnt the callback function itself (i.e. addOne) get multiplied by 2? In other words, instead of function multiply returning [3,5,7], I would have expected it to return [3,5,7, NAN] since the function is not a number?

Does JavaScript interpreter just somehow know not to multiply it by 2?

function addOne(a) {
return a + 1;
}

function multiply(a,b,c, callback) {
    var i, ar = [];
    for (i = 0; i < 3; i++) {
        ar[i] = callback(arguments[i] * 2);
    }
    return ar;
}



myarr = multiply(1,2,3, addOne);
myarr;

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Because your loop's condition is <3 (hehe) which means it won't subscript the callback (the last element).

You should consider making the callback the first argument always, and splitting the arguments like so...

var argumentsArray = Array.prototype.slice.call(arguments),
    callback = argumentsArray.shift();

jsFiddle.

Then, callback has your callback which you can call with call(), apply() or plain (), and argumentsArray has the remainder of your arguments as a proper array.

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This line for (i = 0; i < 3; i++) { is protecting you.

You stop before it hits the callback argument

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Because you are running the the loop for the first 3 args only. i < 3 runs for i=0, i=1,i=2