Templates specialization
c++, specialization, templates
Solution
Functions #3 and #4 are specializations of #1, not of #1 and #2 respectively.
This means that your compiler will choose between #1 and #2 first. When it has selected #1 to be the best fit for funcT(x1), it then selects the specialization, #3. For funcT(&x1), it chooses #2 as the best fit and finds no specializations.
By writing #4 as
template<> void funcT<>( int * arg )
it becomes a specialization of #2 and you'll get the expected result that #4 is called for funcT(&x1).
Another option would be to simply write
void funcT(int *arg)
since regular functions will always be chosen instead of templated versions if they match.
Problem
I have the following set of templates: ``` //1 template< typename T > void funcT( T arg ) { std::cout<<"1: template< typename T > void funcT( T arg )"; } //2 template< typename T > void funcT( T * arg ) { std::cout<<"2: template< typename T > void funcT( T * arg )"; } //3 template<> void funcT< int >( int arg ) { std::cout<<"3: template<> void funcT< int >( int arg )"; } //4 template<> void funcT< int * >( int * arg ) { std::cout<<"4: template<> void funcT< int *>( int * arg )"; } //... int x1 = 10; funcT( x1 ); funcT( &x1 ); ``` Can someone please explain why `funcT( x1 );` calls function #3 and `funcT( &x1 );` calls function #2 but not #4 as expected? I have already read this article http://www.gotw.ca/publications/mill17.htm which says that "overload resolution ignores specializations and operates on the base function templates only". But according to this logic `funcT( x1 );` should call function #1, not #3. I am confused.