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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 S, typename T> void foo(typename T::type s) { printf("a\n"); } template<typename S, typename T> void foo(int s) { printf("b\n"); } Compiler error (Clang): test.cpp:26:2: error: call to 'foo' is ambiguous foo<std::enable_if<true, int>>(3); ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ test.cpp:16:6: note: candidate function [with T = std::__1::enable_if<true, int>] void foo(typename T::type s) { ^ test.cpp:21:6: note: candidate function [with T = std::__1::enable_if<true, int>] void foo(int s) { ^ 1 error generated. 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 m

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