#include <type_traits>
class A{};
template< typename T > struct is_allowed
{
enum { value = false };
};
template<> struct is_allowed<A>
{
enum { value = true };
};
template< typename T >
void f( std::enable_if_t< is_allowed<T>::value , T >* Parent )
{
}
int main()
{
A a;
f( &a );
return 0;
}
There's a simple function template. Why can't we say that function f doesn't make type inference for class A? The development environment is VC2017.
Error message: error C2783: 'voidf(enable_if<_Test,T>:type *)': could not delete template argument for 'T'
c++ template
http://en.cppreference.com/w/cpp/language/template_argument_deduction#Non-deduced_contexts
contains the following information:
In the following cases, the types, templates, and non-type values that are used to compose P do not participate in template argument deduction, but instead use the template arguments that were either deduced elsewhere or explicitly specified. If a template parameter is used only in non-deduced contexts and is not explicitly specified, template argument deduction fails.
In your question, P is std::enable_if<is_allowed<T>::value, T>:type*.
In some cases, :: may be used to access the internal name using :: such as ::type with In the following cases.
In other words, because you used ::type in P, it is included in the case of non-deduced contexts, which is the error you mentioned because of the rule that you do not make type inference.
std::enable_if<is_allowed<T>::value, T>:type* is not an inference, so in order to replace it with A* you need to modify it so that T can be inferred.
Therefore, T can be changed as below to make it simple to infer.
template< typename T >
void f( T* Parent, std::enable_if_t< is_allowed<T>::value , T >* = 0 )
{
}
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