c++ - Failing the template function lookup -

consider following example.

#include <iostream> #include <boost/optional.hpp>  template < typename > int boo( const boost::optional< > &a );  template < typename > int foo( const &a ) {     return boo( ); }  template < typename > int boo( const boost::optional< > & ) {     return 3; }   int main() {     std::cout << "foo = " << foo( 3 ) << std::endl;     std::cout << "boo = " << boo( 3 ) << std::endl; } 

compiling using g++ 4.3.0 throws next compiling errors:

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

what should differently (if possible references c++ standard)? why happening , how fix it?

edit

the fix create correct type in foo:

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

but questions still stands: why not created automatically?

return boo( ); 

here type of a int, , there no function name boo accepts argument of type int. hence see error:

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

even if int can implicitly converted boost::optional<int>, compiler cannot deduce template argument boost::optional<t> calling site. it's 1 of non-deduced contexts explicitly need mention type as,

   return boo<a>(a); 

the standard says in $14.8.2.1,

if template-parameter not used in of function parameters of function template, or used in non-deduced context, its corresponding template-argument cannot deduced function call , template-argument must explicitly specified.