I wonder what is the difference between this code that works :
#include <type_traits>
#include <iostream>
template<typename T> using is_ref = std::enable_if_t<std::is_reference_v<T>, bool>;
template<typename T> using is_not_ref = std::enable_if_t<!std::is_reference_v<T>, bool>;
template<typename T, is_ref<T> = true>
void foo(T&&) {
std::cout << "ref" << std::endl;
}
template<typename T, is_not_ref<T> = true>
void foo(T&&) {
std::cout << "not ref" << std::endl;
}
int main() {
int a = 0;
foo(a);
foo(5);
}
and this one that does not work :
#include <type_traits>
#include <iostream>
template<typename T> using is_ref = std::enable_if_t<std::is_reference_v<T>, bool>;
template<typename T> using is_not_ref = std::enable_if_t<!std::is_reference_v<T>, bool>;
template<typename T, typename = is_ref<T>>
void foo(T&&) {
std::cout << "ref" << std::endl;
}
template<typename T, typename = is_not_ref<T>>
void foo(T&&) {
std::cout << "not ref" << std::endl;
}
int main() {
int a = 0;
foo(a);
foo(5);
}
What is the true difference between typename = is_ref<T>
and is_ref<T> = true
?
If you remove the default template arguments, it'll become clear what the difference is. Function declarations cannot differ in just their defaults. This is ill-formed:
void foo(int i = 4);
void foo(int i = 5);
And likewise this is ill-formed:
template <typename T=int> void foo();
template <typename T=double> void foo();
With that in mind, your first case:
template<typename T, is_ref<T>>
void foo(T&&);
template<typename T, is_not_ref<T>>
void foo(T&&);
The two declarations here are unique because the second template parameter differs between the two declarations. The first has a non-type template parameter whose type is std::enable_if_t<std::is_reference_v<T>, bool>
and the second has a non-type template parameter whose type is std::enable_if_t<!std::is_reference_v<T>, bool>
. Those are different types.
Whereas, your second case:
template<typename T, typename>
void foo(T&&);
template<typename T, typename>
void foo(T&&)
This is clearly the same signature - but we've just duplicated it. This is ill-formed.
See also cppreference's explanation.
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