C++: Pass string literal or variable to function

Question

I have a function f that takes a string as input. I usually want to provide a string literal, eg, f("hello") . However, I want to implement another function g that builds upon f :

std::string f(const std::string&& x) {
  return x + " world";
}

std::string g(const std::string&& x) {
  std::string res = f(x);  // problem: rvalue reference to std::string cannot bind to lvalue of type std::string
  res += "!";
  return res;
}

int main() {
  std::string res_a = f("hello");
  std::string res_b = g("world");
  return 0;
}

How can I achieve this in C++11/14 in a way that I can use f with string literals as well as variables?

Answer 1

A generic way of solving the problem of a function taking both l-value and r-value references is to use templated functions like so-

template <typename T>
T f(T&& val) {
}

template <typename T>
T g(T&& val) {
  T some_val = f(std::forward<T>(val));
}

std::foward<T>(val) forwards an l-value as an l-value and an r-value as an r-value, just as its name implies.

By templating the function, you ensure that this logic works for any type and not just strings.

Answer 2

The traditional way to take a read-only parameter is by const lvalue reference.

std::string f(const std::string& x)

This rule of thumb applies to many types, not just std::string . The primary exceptions are types that are not bigger than a pointer (eg a char ).

It's rather unusual for a function to have a const rvalue reference. As you discovered, that adds difficulty when trying to pass a variable as the argument. A non- const rvalue reference has value, but a const rvalue reference is inferior to a const lvaue reference in most cases. See also Do rvalue references to const have any use?