如何聲明返回類型被推導出來的函數？

Question

考慮這個C++ 1Y代碼（LIVE EXAMPLE）：

#include <iostream> 

auto foo(); 

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

auto foo() { 
    return 1234; 
} 

編譯器（GCC 4.8.1）慷慨地拍攝出這樣的錯誤：

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

我如何前瞻性聲明foo()在這裏？或者更合適，是否可以提前申報foo()？

---

我也試着編譯代碼，我試圖在.h文件中聲明foo()定義foo()就像一個.cpp文件上面的一個，包括在我main.cpp文件中包含.h和int main()通話到foo()，並構建它們。

發生同樣的錯誤。

Answer 1

根據它在，N3638提出的紙，它是明確有效的這樣做。

相關片段：

[ 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 ] 

但是它接着說：

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] 

所以，你使用它，它被定義之前的事實導致其錯誤。