Template function overload resolution

Question

Consider this code:

#include <iostream> void func(int&) { std::cout << "mutable" << std::endl; } void func(const int&) { std::cout << "const" << std::endl; } template<typename T> void tpl_func(T&) { std::cout << "mutable_tpl" << std::endl; } template<typename T> void tpl_func(const T&) { std::cout << "const_tpl" << std::endl; } class number { public: operator int&() { return nb_; } operator const int&() { return nb_; } private: int nb_ = 42; }; int main() { number n; func(n); // This produces: error: call to 'func' is ambiguous tpl_func(n); // This compiles fine }

Tested with clang3.5

Questions:

Why is the overload resolution for the template functions not ambiguous ?
What rules determine which overload is chosen ?

well yes as mentioned below with two answers, it is actually not ambiguous. It would most likely become ambiguous if you called tpl_func<int>(n); — Creris
– Creris, Commented Oct 22, 2014 at 16:49

BeyelerStudios · Accepted Answer · 2014-10-22 16:40:19Z

Because in func(n) there's an implicit function call (the int-conversion operator) which is ambiguous (you could choose any of the two) and in tpl_func(n) you're NOT int-converting, i.e. the template deduces to tpl_func<number>(number &) since n is lvalue.

Jarod42 · Accepted Answer · 2014-10-22 16:41:08Z

func(n); requires a conversion, both func are viable and overload is ambiguous.

tpl_func(n); has an exact match (template<typename T> void tpl_func(T&)).

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