A short question about resolving Overloaded Function Templates

[CoolPint]

In the code below, shouldn't the function call minimum(a,a); result in
compilation error? I read in Lippman's C++ Primer 3rd Edition on page
521, the call should be ambiguous.

But on g++ (versions 3.2.3 and 3.4.2) it compiles fine and the result
shows that minimum(a,a) is resolved using minimum(T,T)

I also tried it on http://www.comeaucomputing.com/tryitout/ and got
the
same result.

So what is the correct behaviour according to the standard?

I tried reading overloading resolution rules from a few sources
(including C++ Templates and online IBM C++ reference) but I still
cannot figure out.

So can anybody tell me what the correct behaviour is according to the
latest standard? If the correct behaviour is to choose minimum(T,T)
over minimum(T,U) in this case, what's the rule which determines this
behaviour? Has it anything to do with Partial Ordering rules?

Thank you very much in advance.
#include <iostream>
using std::cout;
using std::endl;

template <class T>
T minimum(T,T)
{ cout << "minimum(T,T)" << endl; }

template <class T, class U>
int minimum(T, U)
{ cout << "minimum(T,U)" << endl; }

int main()
{
int a = 10;
unsigned int b = 20;
minimum(a,b);
minimum(a,a); // shouldn't this call ambiguous?
}

[Charlie]

"CoolPint" <co******@yahoo.co.uk> wrote in message
news:15**************************@posting.google.c om...

In the code below, shouldn't the function call minimum(a,a); result in
compilation error? I read in Lippman's C++ Primer 3rd Edition on page
521, the call should be ambiguous.

But on g++ (versions 3.2.3 and 3.4.2) it compiles fine and the result
shows that minimum(a,a) is resolved using minimum(T,T)

I also tried it on http://www.comeaucomputing.com/tryitout/ and got
the
same result.

So what is the correct behaviour according to the standard?

I tried reading overloading resolution rules from a few sources
(including C++ Templates and online IBM C++ reference) but I still
cannot figure out.

So can anybody tell me what the correct behaviour is according to the
latest standard? If the correct behaviour is to choose minimum(T,T)
over minimum(T,U) in this case, what's the rule which determines this
behaviour? Has it anything to do with Partial Ordering rules?

Thank you very much in advance.
#include <iostream>
using std::cout;
using std::endl;

template <class T>
T minimum(T,T)
{ cout << "minimum(T,T)" << endl; }

template <class T, class U>
int minimum(T, U)
{ cout << "minimum(T,U)" << endl; }

int main()
{
int a = 10;
unsigned int b = 20;
minimum(a,b);
minimum(a,a); // shouldn't this call ambiguous?
}

Did you try 'double a'?

[Rob Williscroft]

CoolPint wrote in news:15**************************@posting.google.c om in
comp.lang.c++:

In the code below, shouldn't the function call minimum(a,a); result in
compilation error? I read in Lippman's C++ Primer 3rd Edition on page
521, the call should be ambiguous.
#include <iostream>
using std::cout;
using std::endl;

template <class T>
T minimum(T,T)
{ cout << "minimum(T,T)" << endl; }

template <class T, class U>
int minimum(T, U)
{ cout << "minimum(T,U)" << endl; }

int main()
{
int a = 10;
unsigned int b = 20;
minimum(a,b);
minimum(a,a); // shouldn't this call ambiguous?
}

No, but there is a rule that when two or more function templates are
selected they can be disambiguated by effectivly "selecting the most
specialized".

See 14.5.5.2 Partial ordering of function templates, for the details.

Rob.
--
http://www.victim-prime.dsl.pipex.com/

[CoolPint]

> No, but there is a rule that when two or more function templates are

selected they can be disambiguated by effectivly "selecting the most
specialized".

See 14.5.5.2 Partial ordering of function templates, for the details.

Rob.

Thank you for the reply. I found out why from GotW#49.

So it looks like the book (C++ Primer 3rd edition, page 521) has an
error unless I misinterprested the content.

I tried looking at the errata for the book, but I didn't find one
about it.

Thank you again.