Template function overloading with identical signatures, why does this work?
- by user1843978
Minimal program:
#include <stdio.h>
#include <type_traits>
template<typename S, typename T>
int foo(typename T::type s) {
return 1;
}
template<typename S, typename T>
int foo(S s) {
return 2;
}
int main(int argc, char* argv[]) {
int x = 3;
printf("%d\n", foo<int, std::enable_if<true, int>>(x));
return 0;
}
output:
1
Why doesn't this give a compile error? When the template code is generated, wouldn't the functions int foo(typename T::type search) and int foo(S& search) have the same signature?
If you change the template function signatures a little bit, it still works (as I would expect given the example above):
template<typename S, typename T>
void foo(typename T::type s) {
printf("a\n");
}
template<typename S, typename T>
void foo(S s) {
printf("b\n");
}
Yet this doesn't and yet the only difference is that one has an int signature and the other is defined by the first template parameter.
template<typename T>
void foo(typename T::type s) {
printf("a\n");
}
template<typename T>
void foo(int s) {
printf("b\n");
}
I'm using code similar to this for a project I'm working on and I'm afraid that there's a subtly to the language that I'm not understanding that will cause some undefined behavior in certain cases. I should also mention that it does compile on both Clang and in VS11 so I don't think it's just a compiler bug.
