Re: Function overloading

From:
Victor Bazarov <v.Abazarov@comAcast.net>
Newsgroups:
comp.lang.c++
Date:
Wed, 17 Sep 2008 18:01:09 -0400
Message-ID:
<garun4$3p8$1@news.datemas.de>
Frank Neuhaus wrote:

I am trying to overload a function for several types and I have
encountered some problems.


You basically have overloaded your function for an *infinite* set of
types since you mixed in a template. Not *several*, but *way too many*.

Below is the program I am considering:

#include <iostream>
#include <string>

struct Base {};

class Derived : public Base {};

class SomeClass {};

void f(int t)
{
    std::cout << "int" << std::endl;
}

void f(const Base* t)
{
    std::cout << "base" << std::endl;
}

template<typename T>
void f(const T& t)
{
    std::cout << "cref" << std::endl;
};

template<>
void f(const std::string& t)
{
    std::cout << "string" << std::endl;
};

int main()
{
    f(5); // expecting "int"

    Base b;
    f(&b); // expecting "base"
    Derived d;
    f(&d); // expecting "base"

    SomeClass c;
    f(c); // expecting "cref"

    std::string test="test";
    f(test); // expecting "string"
};

I want the call f(&b) and f(&d) to result in the output "base". What I
am getting is the result "cref" for both calls. The compiler appears
to prefer making T=Base*, and T=Derived* respectively in the template
function. What do I need to change in order to make this result in the
desired output? What exactly are the internal rules of the compiler
for function overloading - i.e in which cases does it prefer which of
the overloads?


The rules are complex. There is no way I can explain them all in a
single newsgroup posting (I'm just too lazy). Find a book and study.
Or get the Standard and read chapter 13. However, in your case I can
tell you this... Since you added a template, it is considered in the
resolution. The template argument for that function template is deduced
from the argument you supplied. Try printing out the type you get:

template<typename T>
void f(const T& t)
{
     std::cout << "f<T>, T == " << typeid(T).name() << std::endl;
};

What happens here is that the match is closer and the rank of the
conversion (if any) is higher than the conversion of 'Derived*' to
'Base*' and then to 'Base const*', I'm guessing. The determination of
the rank is done according to some table in the chapter 13, and you can
probably find it in a decent book that talks about overload resolution.

V
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