Function Templates vs. Auto Keyword

问题:

Can the auto keyword in C++11 replace function templates and specializations? If yes, What are the advantages of using template functions and specializations over simply typing a function parameter as auto. If not, I fully expect to be downvoted into hell, but I'm new to C++ but so please go easy on me.

template <typename T>
void myFunction(T &arg)
{
    // ~
}

vs.

void myFunction(auto &arg)
{
    // ~
}

回答1:

In a nutshell, auto cannot be used in an effort to omit the actual types of function arguments, so stick with function templates and/or overloads. auto is legally used to automatically deduce the types of variables:

auto i=5;

Be very careful to understand the difference between the following, however:

auto x=...
auto &x=...
const auto &x=...
auto *px=...; // vs auto px=... (They are equivalent assuming what is being 
              //                 assigned can be deduced to an actual pointer.)
// etc...

It is also used for suffix return types:

template <typename T, typename U>
auto sum(const T &t, const U &u) -> decltype(t+u)
{
  return t+u;
}

回答2:

Can the auto keyword in C++11 replace function templates and specializations?

No. There are proposals to use the keyword for this purpose, but it's not in C++11, and I think C++14 will only allow it for polymorphic lambdas, not function templates.

If yes, What are the advantages of using template functions and specializations over simply typing a function parameter as auto.

You might still want a named template parameter if you want to refer to the type; that would be more convenient than std::remove_reference<decltype(arg)>::type or whatever.