在头文件中有全局函数定义,避免重复的符号链接错误

Question

我在标题文件中有以下代码.

#pragma once

class error_code {
public:
    unsigned __int64 hi;
    unsigned __int64 lo;    
};

std::ostream& operator<< (std::ostream& o, const error_code& e) {
    return o << "[" << e.hi << "," << e.lo << "]";
}

我得到链接错误,当项目中有2个cpp包含此头文件时.

错误LNK2005:"类error_code __cdecl运算符|(类error_code const&,类ViTrox :: error_code const&)"(?? U @@ YA？AVerror_code @ 0 @ ABV10 @ 0 @ Z)已在xxx.obj中定义

我知道如果我将定义移动operator<<到cpp文件或DLL文件,我可以解决这个问题.

但是,我只想将它们放在SINGLE头文件中.有没有什么技术可以做到这一点？或者我必须将定义分离到另一个文件？

Answer 1

使用inline关键字.

inline std::ostream& operator<< (std::ostream& o, const error_code& e) {
    return o << "[" << e.hi << "," << e.lo << "]";
}

Answer 2

要么使功能inline:

inline std::ostream& operator<< (std::ostream& o, const error_code& e) {
    return o << "[" << e.hi << "," << e.lo << "]";
}

或使其成为模板功能:

template<class Ch, class Tr>
std::basic_ostream<Ch,Tr>& operator<< (std::basic_ostream<Ch,Tr>& o,
                                       const error_code& e) {
    return o << "[" << e.hi << "," << e.lo << "]";
}