Reputation: 3
I'm trying to use result_of for the case when Callable returns template type and get the following error (clang++). I also included a simple case where everything works fine.
Error:
main.cpp:22:50: note: candidate template ignored: could not match '<type-parameter-0-1>' against 'std::__1::shared_ptr<int> (*)()'
typename std::result_of<FunctionType<T>()>::type submit(FunctionType<T> f) {
Code:
int f() {
int x = 1;
return x;
}
template<typename T>
std::shared_ptr<T> g() {
std::shared_ptr<T> x;
return x;
}
template <template<typename> class FunctionType, typename T>
typename std::result_of<FunctionType<T>()>::type submit(FunctionType<T> f) {
using result_type = typename std::result_of<FunctionType<T>()>::type;
result_type x;
return x;
}
template<typename FunctionType>
typename std::result_of<FunctionType()>::type submit2(FunctionType f) {
using result_type = typename std::result_of<FunctionType()>::type;
result_type x;
return x;
}
int main()
{
submit(g<int>); // error
submit2(f); // ok
return 0;
}
Upvotes: 0
Views: 316
Reputation: 96810
g<int>
is of type shared_ptr<int>()
which when deduced by the function decays to a pointer to that type (shared_ptr<int>(*)()
). FunctionType
in submit
is therefore not a template and you can't use template arguments on it.
If you could be more clear about what you're trying to do we can figure out a solution to your main issue.
Upvotes: 1