I often run into compiling issues when writing a child class using one of its parent's template method. For example, I wrote this, but I don't know why it compiles:
#include <iostream>
#include <type_traits>
#include <algorithm>
/******************** ********** ********************/
template <class T, unsigned N>
struct A
{
T data[N];
template <class U>
inline auto operator= ( const U& other ) -> decltype(*this)
{ this->assign(other); return *this; }
template <class U>
void assign( const U& other )
{ assign_impl( other, std::is_arithmetic<U>() ); }
template <class U>
void assign_impl( const U& val, std::true_type const )
{ std::fill( data, data+N, static_cast<T>(val) ); }
};
// ------------------------------------------------------------------------
template <class T, unsigned N>
struct B
: public A<T,N>
{
// Needed in order to compile
using A<T,N>::operator=;
void show()
{
for (unsigned i=0; i<N; ++i)
std::cout<< this->data[i] << " ";
std::cout<< std::endl;
}
};
// ------------------------------------------------------------------------
int main()
{
B<double,5> b;
b = -5.1;
b.show();
b.assign(3.14159);
b.show();
}
Including the statement using A<T,N>::operator=;
is necessary if I want to use this operator with instances of B<T,N>
, but I never specified that the method assign
should be visible. Is it visible because operator=
uses it?
The method assign
is visible in your class B
because you are using struct
which has public default encapsulation and public
inheritance.
As for operator=
:
(13.5.3 Assignment) An assignment operator shall be implemented by a non-static member function with exactly one parameter. Because a copy assignment operator operator= is implicitly declared for a class if not declared by the user, a base class assignment operator is always hidden by the copy assignment operator of the derived class.