如何声明一个返回类型被推导的函数？ [英] How do I declare a function whose return type is deduced?

问题描述

请考虑此 C ++ 1y 代码（ LIVE示例） ）：

Consider this C++1y code (LIVE EXAMPLE):

#include <iostream>

auto foo();

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

auto foo() {
    return 1234;
}

编译器（GCC 4.8.1） p>

The compiler (GCC 4.8.1) generously shoots out this error:

main.cpp：在函数'int main（）'中：

main.cpp：8：18：在扣除'auto'之前使用'auto foo（）'

std :: cout< foo（）;

                 

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

如何转发声明 foo（） 或者更合适的是，可以转发声明 foo（）？

How do I forward-declare foo() here? Or maybe more appropriately, is it possible to forward-declare foo()?

我也试过编译代码，我试图在<$ c $中声明 foo（） c> .h 文件，定义 foo（），就像上面的 .cpp 文件，包含 .h 在我的 main.cpp 文件包含 int main ）和对 foo（）的调用，并构建它们。

I've also tried compiling code where I tried to declare foo() in the .h file, defined foo() just like the one above in a .cpp file, included the .h in my main.cpp file containing int main() and the call to foo(), and built them.

错误。

推荐答案

根据提出的文件， N3638 ，它显然有效。

According to the paper it was proposed in, N3638, it is explicitly valid to do so.

相关片段：

[ Example:

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
— end example ]

但是它继续说：

如果需要具有未减少的占位符类型的实体类型来确定表达式的类型，程序是不成形的。但是一旦在函数中看到返回语句，从该语句推导出的返回类型可以在函数的其余部分中使用，包括在其他返回语句中。

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.

[ Example:

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
}
—end example]

因此，你在定义之前使用它的事实导致它的错误。

So the fact that you used it before it was defined causes it to error.

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