Trailing Return Types¶

Motivation¶

When the return type depends on parameter types, place it after parameters.

template<typename T, typename U>
auto multiply(const T& a, const U& b) -> decltype(a * b) {
  return a * b;
}

For simple cases, let auto deduce the return:

template<typename T, typename U>
auto add(T a, U b) { return a + b; }

Note: plain auto drops top-level const and references. Use decltype(auto) or an explicit trailing return to preserve qualifiers.