c++ - 如何声明一个推导出返回类型的函数？

Question

考虑一下这个 C++1y 代码(LIVE EXAMPLE):

#include <iostream>

auto foo();

int main() {
    std::cout << foo();   // ERROR!
}

auto foo() {
    return 1234;
}

编译器(GCC 4.8.1)慷慨地抛出这个错误:

main.cpp: In function ‘int main()’:
main.cpp:8:18: error: use of ‘auto foo()’ before deduction of ‘auto’
std::cout << foo();
                   ^

我如何在这里转发声明 foo()？ 或者更恰本地说，是否可以转发声明 foo()?

---

我也尝试过编译代码，我试图在 .h 文件中声明 foo()，定义 foo() 只是与 .cpp 文件中的上述类似，在我的 main.cpp 文件中包含 .h，其中包含 int main() 和对 foo() 的调用，并构建它们。

发生了同样的错误。

Answer 1

根据提出的论文，N3638 , 这样做是明确有效的。

相关片段:

auto x = 5;                // OK: x has type int
const auto *v = &x, u = 6; // OK: v has type const int*, u has type const int
static auto y = 0.0;       // OK: y has type double
auto int r;                // error: auto is not a storage-class-specifier
auto f() -> int;           // OK: f returns int
auto g() { return 0.0; }   // OK: g returns double
auto h();                  // OK, h's return type will be deduced when it is defined

但它继续说:

If the type of an entity with an undeduced placeholder type is needed to determine the type of an expression, the program is ill-formed. But once a return statement has been seen in a function, the return type deduced from that statement can be used in the rest of the function, including in other return statements.

auto n = n;            // error, n's type is unknown
auto f();
void g() { &f; }       // error, f's return type is unknown
auto sum(int i) {
  if (i == 1)
    return i;          // sum's return type is int
  else
    return sum(i-1)+i; // OK, sum's return type has been deduced
}

因此，您在定义之前使用它的事实会导致它出错。