SFINAE works differently in cases of type and non-type template parameters
Why does this code work:
template<
typename T,
std::enable_if_t::value, T>* = nullptr>
void Add(T) {}
templa
-
SFINAE is about substitution. So let us substitute!
template< typename T, std::enable_if_tBecomes:
template< class T=int, int* = nullptr> void Add(int) {} template< class T=int, Substitution failure* = nullptr> void Add(int) { template< class T=double, Substitution failure* = nullptr> void Add(double) {} template< class T=double double* = nullptr> void Add(double) {}Remove failures we get:
template< class T=int, int* = nullptr> void Add(int) {} template< class T=double double* = nullptr> void Add(double) {}Now remove template parameter values:
template< class T, int*> void Add(T) {} template< class T double*> void Add(T) {}These are different templates.
Now the one that messes up:
template< typename T, typename = typename std::enable_ifBecomes:
template< typename T=int, typename =int> void Add(int) {} template< typename int, typename = Substitution failure > void Add(int) {} template< typename T=double, typename = Substitution failure > void Add(double) {} template< typename T=double, typename = double> void Add(double) {}Remove failures:
template< typename T=int, typename =int> void Add(int) {} template< typename T=double, typename = double> void Add(double) {}And now template parameter values:
template< typename T, typename> void Add(T) {} template< typename T, typename> void Add(T) {}These are the same template signature. And that is not allowed, error generated.
Why is there such a rule? Beyond the scope of this answer. I'm simply demonstrating how the two cases are different, and asserting that the standard treats them differently.
When you use a non-type template parameter like the above, you change the template signature not just the template parameter values. When you use a type template parameter like the above, you only change the template parameter values.
加载中...
- 热议问题