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unexpected call to a function in Javascript

Answer 1

+7 A:

Javascript doesn't support method overloading; It just overwrites the function so the last one will get called every time.

A way around it is just to make 1 function and put some conditions in there to check if the variables are set.

D4V360 2010-01-24 17:45:37

okay. thank you very much!

Reigel 2010-01-24 17:50:39

Answer 2

A:

In javascript the parameters for a function aren't required. You'd want to check to see if the parameter is undefined before using it.

If used this before in order to implement getter/setter type functionality in javascript similar to below:

var _pageSize;

function pageSize(value) {
    if (value) {
       _pageSize = value;
    }
    else {
       return _pageSize;
    }
}

You can then use the pageSize method to both get and set the page size like below:

if (pageSize() < 15) {
    pageSize(15)
}

Brian Hasden 2010-01-24 17:45:59

Answer 3

A:

Ah, Javascript doesn't do polymorphism like that. It does have an arguments collection that you can look in and then do the right thing given the inputs..

stuff like:

function calcAverage() 
{ 
   var sum = 0 
   for(var i=0; i<arguments.length; i++) 
      sum = sum + arguments[i] 
   var average = sum/arguments.length 
   return average 
} 
document.write("Average = " + calcAverage(400, 600, 83))

Bill Bingham 2010-01-24 17:46:02

Answer 4

+2 A:

You declare twice the function test. The 2nd one overwrite the first definition.

Mic 2010-01-24 17:46:39

+1 thank you for that..

Reigel 2010-01-24 17:51:23

Answer 5

+1 A:

In JavaScript overloaded functions are not allowed. so in this case the second function overrides the first one and therefore you always see this result.

A work around for this is to check the values of the parameters:

function test(one,two) { if(two) alert('expected: second function with two parameters'); else alert('surprise! first function with one parameter'); }

Mendy 2010-01-24 17:47:46

Answer 6

A:

When calling such function:

function test(one,two)

With only one parameter, the argument "two" will have special type: undefined. Any attempt to access the variable two will fail with "two is undefined" error, to avoid such error use the typeof command:

function test(one,two)
{
   if (typeof two == "undefined") {
      //give default value
      two = "default value here";
   }
   ...
}

Shadow Wizard 2010-10-25 15:26:53

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unexpected call to a function in Javascript

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