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sciwander · Guest

#include <iostream>
//using namespace std;
//const_match.cpp
#include <iostream>
using namespace std;

void f(const int* & p) //#1
{
int a=2;
int* p1=&a;
p=p1;
cout<<"#1:"<<p<<endl;
}
void f(const void* p) //#2
{
int a=2;
int* p2=&a;
p=p2;
cout<<"#2:"<<p<<endl;
}

void foo(const int* p)
{
cout << "A" << endl;
}
void foo(int* const p)
{
cout << "B" << endl;
}

typedef int* int_ptr;
void goo(const int* p)
{
cout << "A" << endl;
}
void goo(const int_ptr p)
{
cout << "B" << endl;
}

template<class T>
void ff(const T val)
{
foo(val);
goo(val);
}

int main(int argc, char *argv[])
{
int* p=0;
f( p );
const int* const& r = p;
cout<<p<<" "<<r<<endl;
ff(p);
return 0;
}
in G++ 3.1 I got :#2 0x22ff54
0 0
B
B
but In VC2003 The answer is #1 : 0012FDE1
0012FDE1 0012FDE1
B
B
why an (int *) implicited converted to (const void*) not the (const
int*)?

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Guest

By the way, Visual Studio 2005 agrees with G++ 3.1. I believe that VS
2003 is incorrect. The reason why (int*) is converted to (const void*)
is that the first overload to f requires a conversion to (const int*&)
-- notice the reference -- instead of (const int*).

If I comment out the second overload, VS20005 produces and error during
compilation:

error C2664: 'f' : cannot convert parameter 1 from 'int *' to
'const int *&'

However, if you remove the reference so that the first overload is:

void f(const int* p) //#1

then, VS2005 will choose answer #1. And this makes sense. (int*) will
be converted to (const int*) before being converted to (const void*).

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kanze · Guest

sciwander wrote: