7 ms·

In the example the return type of add does not depend on the value of the arguments. So it is not dependent types.

• Davidbrcz 7y ago
  In this piece of code, the return type is explicitly dependent of the type of the argument (it's path related template #include <iostream> #include <string> struct A {using ret = std::string;}; struct B {using ret = int;}; template <class T> auto f(T t) -> typename T::ret { return {}; } int main(int argc, char *argv[]) { auto s = f(A{}); std::cout << s.size() << "\n"; auto i = f(B{}); std::cout << i+1 << "\n"; return 0; } There is also stuff like this where the return type is defined as the return type of builder.makeObject(), builder which is the argument: template <typename Builder> auto makeAndProcessObject (const Builder& builder) -> decltype(builder.makeObject() ) { auto val = builder.makeObject(); // do stuff with val return val; }
  • Retra 7y ago
    The return type has to be dependent on the value of the input, not its type. Like it'll return an int[3] if you provide 3 as in input and int[4] if you provide 4. Note that both 3 and 4 should have the same type, but int[3] and int[4] are different types.