“a variable declared with an auto specifier cannot appear in its own initializer”

问题

It seems like there's an error when using the trailing return type in the function pointer declaration for Func_ptr. I know I can do it if I put the declaration and initialization in the same statement or simply use the standard declaration by specifying the return type directly, but I want to understand the language's limitations, so can someone please explain what this error means in the code below:

"a variable declared with an auto type specifier cannot appear in its own initializer"

#include <utility>
#include <iostream>

int Func(const std::pair<int, int>& p)
{
    std::cout << p.first << "->" << p.second << std::endl;
    return 1;
}

int main()
{
    auto (*Func_ptr)(const std::pair<int, int>& p) -> int;
    //Error below, Func_ptr underlined, "a variable declared with the auto
    //specifier cannot appear in its own initializer
    Func_ptr = Func;
}

回答1:

Problem is that variable is declared in C++03 style and function format in C++11 way. Make it uniform and it will work.

// the old way
int (*Func_ptr1)(const std::pair<int, int>& p);

// the C++11
auto func_ptr2 = &Func;

Here is example. What is more interesting Clang is able to handle mixture.

来源：https://stackoverflow.com/questions/44857390/a-variable-declared-with-an-auto-specifier-cannot-appear-in-its-own-initializer