what happens behind the scenes when Javascript functions are called

Question

Can someone explain what happens in the below scenario:(this was an interview question)

function test(a){
    alert(a);
}

function test(a,b){
    alert(a,b);
}

function callTest(){
    test(a);
    test(a,b);
}

I was asked how the javascript calls the methods, and what happens behind the scenes in this scenario

What do you mean "behind the scenes"? Sounds like a pretty intense interview question. I would expect that maybe they would ask what the contents of the alert dialog would be... — Joe
– Joe, Commented Nov 10, 2014 at 3:37
alert only takes one argument, so the second will be ignored. — user663031
– user663031, Commented Nov 10, 2014 at 3:47

philz · Accepted Answer · 2014-11-10 03:34:22Z

3

If you run it in that order, test(a) will be replaced by test(a,b)

When callTest() is run, then test(a) will alert(a, undefined), then test(a,b) will alert(a,b)

answered Nov 10, 2014 at 3:34

philz

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1 Comment

Ramesh Over a year ago

Agreed - JS don't have the concept of method overloading.

sujin · Accepted Answer · 2014-11-10 03:39:37Z

1

function test(a){     // f1. never called.
    alert(a);
}

function test(a,b){   // f2. overwrite f1 function because name of two functions is same.
    alert(a,b);
}

function callTest(){
    test(1);     // called f2 with (1, undefined)
    test(1,2);   // called f2 with (1, 2)
}
callTest();

answered Nov 10, 2014 at 3:39

sujin

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1 Comment

Shruthi Over a year ago

Thank you, this was explained well.

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