Can somebody please explain why the below code returns undefined 2 times ?

    var test = function (theArr) {
        alert(theArr);
    };

    test.call(6);               //Undefined

    var theArgs = new Array();
    theArgs[0] = 6;

    test.apply(theArgs)         //Undefined

Can somebody please explain why the below code returns undefined 2 times ?

    var test = function (theArr) {
        alert(theArr);
    };

    test.call(6);               //Undefined

    var theArgs = new Array();
    theArgs[0] = 6;

    test.apply(theArgs)         //Undefined

• javascript
• function
• parameters
• call
• undefined

Share Follow edited Oct 15, 2013 at 12:38 Stephane Rolland 40k3838 gold badges126126 silver badges172172 bronze badges asked Oct 13, 2011 at 10:43 HerbalMartHerbalMart 1,73933 gold badges2727 silver badges5050 bronze badges 2

• Why do you need to use the call method? – Jamie Dixon Commented Oct 13, 2011 at 10:46
• You mean that it shows "Undefined" in the alert dialog, right? Because there are not return values anywhere. – Till Helge Commented Oct 13, 2011 at 10:46

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1 Answer 1

Sorted by: Reset to default 8

The syntax for the JavaScript call method:

fun.call(object, arg1, arg2, ...)

The syntax for the JavaScript apply method:

fun.apply(object, [argsArray])

The main difference is that call() accepts an argument list, while apply() accepts a single array of arguments.

So if you want to call a function which prints something and pass an object scope for it to execute in, you can do:

function printSomething() {
    console.log(this);
}

printSomething.apply(new SomeObject(),[]); // empty arguments array
// OR
printSomething.call(new SomeObject()); // no arguments