C++11: How to alias a function? [duplicate]

Solution 1:

You can define a function alias (with some work) using perfect forwarding:

template <typename... Args>
auto g(Args&&... args) -> decltype(f(std::forward<Args>(args)...)) {
  return f(std::forward<Args>(args)...);
}

This solution does apply even if f is overloaded and/or a function template.

Solution 2:

The constexpr function pointer can be used as a function alias.

namespace bar
{
    int f();
}

constexpr auto g = bar::f;

It is highly likely (but not guaranteed by the language) that using g uses bar::f directly. Specifically, this depends on compiler version and optimization level.

In particular, this is the case for:

• GCC 4.7.1+, without optimization,
• Clang 3.1+, without optimization,
• MSVC 19.14+, with optimization.

See assembly generated by these compilers.

Solution 3:

Classes are types, so they can be aliased with typedef and using (in C++11).

Functions are much more like objects, so there's no mechanism to alias them. At best you could use function pointers or function references:

void (*g)() = &bar::f;
void (&h)() = bar::f;

g();
h();

In the same vein, there's no mechanism for aliasing variables (short of through pointers or references).