Question

Consider following example.

#include <iostream>
#include <boost/optional.hpp>

template < typename A >
int boo( const boost::optional< A > &a );

template < typename A >
int foo( const A &a )
{
    return boo( a );
}

template < typename A >
int boo( const boost::optional< A > & )
{
    return 3;
}


int main()
{
    std::cout << "foo = " << foo( 3 ) << std::endl;
    std::cout << "boo = " << boo( 3 ) << std::endl;
}

Compiling this using g++ 4.3.0 throws the next compiling errors:

dfg.cpp: In function ‘int main()’:
dfg.cpp:25: error: no matching function for call to ‘boo(int)’
dfg.cpp: In function ‘int foo(const A&) [with A = int]’:
dfg.cpp:24:   instantiated from here
dfg.cpp:12: error: no matching function for call to ‘boo(const int&)’

What should I do differently (if possible with references from the C++ standard)?
Why is it happening and how do I fix it?

EDIT

The fix is to create the correct type in foo:

template < typename A >
int foo( const A &a )
{
    const boost::optional< A > optA( a );
    return boo( optA );
}

But the questions still stands: why is it not created automatically?

Answer 1

return boo( a );

Here type of a is int, and there is no function with name boo which accepts argument of type int. Hence you see this error:

dfg.cpp:25: error: no matching
function for call to ‘boo(int)’

Even if int can be implicitly converted into boost::optional<int>, the compiler cannot deduce the template argument for boost::optional<T> from the calling site. It’s one of the non-deduced contexts where you explicitly need to mention the type as,

   return boo<A>(a);

The Standard says in $14.8.2.1,

if a template-parameter is not used in
any of the function parameters of a
function template, or is used only in
a non-deduced context, its
corresponding template-argument cannot
be deduced from a function call and
the template-argument must be
explicitly specified.