Templated Overloaded Member Function Question

[joseph cook]

I'm confused by this output. I would expect the more specific
overloaded function to be selected:

#include <iostream>
class Base
{
};
class Derived : public Base
{};

class Foo
{
public:
template<typena me U>

void bar(U* in, int a)
{
std::cout<<"Gen eric Call"<<std::end l;
}

void bar(Base* in, int a)
{
std::cout<<"A more appropriate choice"<<std::e ndl;
}
};

int main()
{
Derived *d = new Derived;

Foo foo;
foo.bar(d,1,1) // calling with derived

}

This prints "Generic Call". Is this correct? Why? I would have
expected the compiler to either pick the more specific overload, or
complain that the call was ambiguous.

What's going on ?
What's the fix ?

Thanks in advance,

[Victor Bazarov]

joseph cook wrote:

I'm confused by this output. I would expect the more specific
overloaded function to be selected:

#include <iostream>
class Base
{
};
class Derived : public Base
{};

class Foo
{
public:
template<typena me U>

void bar(U* in, int a)
{
std::cout<<"Gen eric Call"<<std::end l;
}

void bar(Base* in, int a)
{
std::cout<<"A more appropriate choice"<<std::e ndl;
}
};

int main()
{
Derived *d = new Derived;

Foo foo;
foo.bar(d,1,1) // calling with derived

Acually, it shouldn't compile - you gave three arguments when two
are expected for *either* of 'bar' members. I suppose it's a typo.

>
}

This prints "Generic Call". Is this correct? Why? I would have
expected the compiler to either pick the more specific overload, or
complain that the call was ambiguous.

Why would it be ambiguous? When deducing the type for the template
'bar', 'U' is 'Derived', and it's a better match than 'Base' since
it requires no conversions.

What's going on ?

It's called "deduction of template arguments".

What's the fix ?

The fix? To do what? Try casting 'd' to 'Base*'...

V
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