是否可以在带有初始化的函数指针声明中使用 `auto` 关键字作为 return 类型？

Question

下面的代码compiles successfully both with clang++ 3.8.0 and g++ 7.2.0（编译标志是-std=c++14 -O0 -Wall -Wextra -Werror -pedantic-errors）：

#include <iostream>


int foo_int(int)
{
    std::cout << "int foo(int)" << std::endl;

    return 0;
}

void foo_void(int)
{
    std::cout << "void foo(int)" << std::endl;
}

auto foo_auto_int(int)
{
    std::cout << "auto foo(int), auto == int" << std::endl;

    return 0;
}

auto foo_auto_void(int)
{
    std::cout << "auto foo(int), auto == void" << std::endl;

    return void();
}


int main()
{
    auto (*fi)(int) = foo_int;
    auto (*fv)(int) = foo_void;
    auto (*fai)(int) = foo_auto_int;
    auto (*fav)(int) = foo_auto_void;

    (void)fi(0);
    fv(0);
    (void)fai(0);
    fav(0);
}

它是有效的 C++ 代码吗？

请注意 decltype(auto) 是 rejected both by clang++ and g++ in the same situation。

Answer 1

编译器运行正常。

来自 [dcl.spec.auto]

The auto and decltype(auto) type-specifiers are used to designate a placeholder type that will be replaced later by deduction from an initializer.

[...]

auto or decltype(auto) shall appear as one of the decl-specifiers in the decl-specifier-seq and the decl-specifier-seq shall be followed by one or more declarators, each of which shall be followed by a non-empty initializer.

也就是说 auto 和 decltype(auto) 只能用你在声明前面写的说明符来写（static、virtual 等），用它们的类型是从紧随其后的 声明符 和 初始化程序 .

推导出来的

案例auto

声明符在auto (*fi)(int) = foo_int;的情况下是(*fi)(int)，其形式为

( ptr-operator declarator-id ) ( parameter-declaration-clause )

因此 auto (*fi)(int) = foo_int; 是有效的，前提是 deduction succeeds 确实如此。其他几个也一样。

案例decltype(auto)

来自 [dcl.spec.auto.deduct]，给定类型 T 包含占位符类型

If the placeholder is the decltype(auto) type-specifier, T shall be the placeholder alone.

这意味着添加任何其他内容都是非法的

int i;
decltype(auto)* p = &i;  // error, declared type is not plain decltype(auto)

因此decltype(auto) (*fi)(int) = foo_int;是非法的。