Why it is does not give error about ambiguity?

Question

4.75/5 (4 votes)

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Please see this code

C#

#include "stdafx.h"
#include "iostream"

void fun(int i)
{
    std::cout<<"Int Overload called";
}

void fun(int *i)
{
    std::cout<<"Pointer Overload called";
}

int main ()
{

    fun(0); \\gets called successfully 

  return 0;
}

Output : Int Overload called

My question is why call to fun does not leads to ambiguity, how compiler calls 'int' implementation and not pointer's implementation.

Posted 9-Apr-12 1:28am

Pranit Kothari

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2 solutions

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Pablo Aliskevicius · Accepted Answer · 2012-04-09T01:50:00

Solution 1

Nice question!

Found an answer at http://www.learncpp.com/cpp-tutorial/76-function-overloading/[^].
Basically, int is considered (by your compiler) a better match than int * for 0.
You may also like this: http://xania.org/200711/ambiguous-overloading[^]

Hope this helps,
Pablo.

Posted 9-Apr-12 1:50am

Pablo Aliskevicius

Comments

[no name] 9-Apr-12 7:57am

Thanks. 5!

Lakamraju Raghuram · Accepted Answer · 2012-04-09T01:51:00

Solution 2

Compiler calls int implementation because int* type is unsigned int not just int

Now try changing

C++

void fun(int i)
{
    std::cout<<"Int Overload called";
}

to

C++

void fun(unsigned int i)
{
    std::cout<<"Int Overload called";
}

The compiler cribs and says
'error C2668: 'fun' : ambiguous call to overloaded function'

Good question

Posted 9-Apr-12 1:51am

Lakamraju Raghuram

Comments

[no name] 9-Apr-12 7:57am

Hey nice explanation. Thanks.5!