Function Overloading concept

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In OverLoading concept,Why they are not consider return value and why they are consider only parameters in method? For ex: public int Add(int a,int b){...} public String Add(int a,int b){...} . how compiler will find which function called?

Posted 11-May-11 5:37am

shikha_14

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OriginalGriff · Answer 1 · 2011-05-11T05:47:00

That is exactly why the return type is not used to discriminate between overlaods - you will get a compiler error if you try.

Think about it - assume you have two overloaded methods:

public HSObject Add(int a, int b) { ... }
public HSSlide Add(int a, int b) { ... }

It looks pretty easy to decide which of these should be called, doesn't it?

HSObject obj = Add(1, 2);
HSSlide slide = Add(1, 2);

But what if you did this:

Add(1,2);

Perfectly legal, you are just ignoring the result. But which result are you ignoring?
And if HSSlide inherits from HSObject, which method should I call? I can put either result into an HSObject variable.

The rule is simple: parameters only. Eliminates all possible confusion, for the compiler, and for us as developers!

Sergey Alexandrovich Kryukov · Answer 2 · 2011-05-11T05:52:00

It has nothing to do with OOP!

Your example will not compile:

C#

int Add(int a, int b) { return a + b; }
string Add(int a, int b) { return (a + b).ToString(); }

Error message:

Error	1	Type 'ContraOrCovariance.Program' already defines a member called 'Add' with the same parameter types…

All correct.

Overloading works if the compiler has a way to determine statically which function should be compiled in what case. It can only be finally determined by the calling code. Your case is special: even before looking at the calling code the compiler denies to work with so close signatures: parameter list should be different for two or more methods of the same name.

Consider this:

C#

class Base { /*...*/ }
class Derived : Base { /*...*/ }

//...

static void Work(Derived i1, Base i2) { /*...*/ }
static void Work(Base i1, Derived i2) { /*...*/ }

Derived d = new Derived();
Derived b = new Derived();

//...

Work((Base)d, b);
Work(d, (Base)b);

//will not compile: The call is ambiguous between the following methods... [two Work methods]
Work(d, b);

This is the case when ambiguous situation is only detected by looking at the calling code.

—SA

Yusuf · Answer 3 · 2011-05-11T05:52:00

Solution 3

The example you provided is not valid. In order for function overloading to be valid, it needs to have different signature. If the signatures are the same but the return type is different ( as in your case ) then it is not valid. See an example here[^]

Posted 11-May-11 5:52am

Yusuf

Function Overloading concept

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