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

Question

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;
}

Are you referring to a compiler error or just an intellisense error? — Retired Ninja, Jul 01 '17 at 05:10
Name your compiler. GCC and Clang both just accepted it for me. — StoryTeller - Unslander Monica, Jul 01 '17 at 05:16
I also fixed up your code sample for you this time. But you should make your code a [mcve] on your own next time. — StoryTeller - Unslander Monica, Jul 01 '17 at 05:17
The compiler is visual c++ 14.0! Thank you for the feedback everyone. I'm pretty sure it's just an intelliSense error, I just learned something new (didn't know what intelliSence was). @StoryTeller What should I have changed/added to the code sample? — Quoc-Minh, Jul 01 '17 at 18:28
I've got the same issue while *compiling* open-source(tensorflow). But seems like no one knows how to solve it. — 김선달, Mar 30 '20 at 06:11
@김선달 i think this happens because to determine which return type has the function the compiler must look inside it, and if inside the function there is call to itself everything breaks — Giuppox, Feb 06 '21 at 08:41

score 1 · Answer 1 · answered Jul 01 '17 at 06:58

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.

answered Jul 01 '17 at 06:58

Marek R

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All compilers I know handle this "mixture" without any problems. There's nothing unusual in it. Why do you see it as "more interesting"? Also, your initial statement is backwards. – AnT stands with Russia Jul 01 '17 at 07:03
Yes, I know your syntax works, I was just trying out different syntax possibilities to better understand C++ boundaries. Thank you for taking the time to help! – Quoc-Minh Jul 01 '17 at 18:38

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

1 Answers1

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