4 ms·
Hm, wouldn't any use of if constexpr break that definition? e.g. template<typename T> void f() { if constexpr (is_int<T>) { return 0; }
by jcelerier 2y ago
Hm, wouldn't any use of if constexpr break that definition?
e.g.
template<typename T>
void f() {
if constexpr (is_int<T>) { return 0; }
else ...