CODEWITHSUNDEEP
c++templatesoverloading
Drax
Drax

Reputation: 13278

Template function overload resolution

Consider this code:

#include <iostream>

void    func(int&)
{
  std::cout << "mutable" << std::endl;
}

void    func(const int&)
{
  std::cout << "const" << std::endl;
}

template<typename T>
void    tpl_func(T&)
{
  std::cout << "mutable_tpl" << std::endl;
}

template<typename T>
void    tpl_func(const T&)
{
  std::cout << "const_tpl" << std::endl;
}

class number
{
public:
  operator int&()
  {
    return nb_;
  }

  operator const int&()
  {
    return nb_;
  }

private:
  int   nb_ = 42;
};

int     main()
{
  number n;

  func(n);     // This produces: error: call to 'func' is ambiguous
  tpl_func(n); // This compiles fine
}

Tested with clang3.5

Questions:

Upvotes: 3

Views: 127

Answers (2)

Jarod42
Jarod42

Reputation: 217075

func(n); requires a conversion, both func are viable and overload is ambiguous.

tpl_func(n); has an exact match (template<typename T> void tpl_func(T&)).

Upvotes: 1

BeyelerStudios
BeyelerStudios

Reputation: 4283

Because in func(n) there's an implicit function call (the int-conversion operator) which is ambiguous (you could choose any of the two) and in tpl_func(n) you're NOT int-converting, i.e. the template deduces to tpl_func<number>(number &) since n is lvalue.

Upvotes: 3

Related Questions