c ++模板特化 - 链接器错误多个定义

Question

我今天在这里提出第三个问题;-),但我对c ++模板编程和运算符重载都很陌生.

我正在尝试以下方法:

terminallog.hh

//snipped code

class Terminallog {
public:

    Terminallog();
    Terminallog(int);
    virtual ~Terminallog();

    template <class T>
    Terminallog & operator<<(const T &v);
    template <class T>
    Terminallog & operator<<(const std::vector<T> &v);
    template <class T>
    Terminallog & operator<<(const T v[]);
    Terminallog & operator<<(const char v[]);

//snipped code
};

//snipped code
template <class T>
Terminallog &Terminallog::operator<<(const T &v) {
    std::cout << std::endl;
    this->indent();
    std::cout << v;
    return *this;
}

template <class T>
Terminallog &Terminallog::operator<<(const std::vector<T> &v) {
    for (unsigned int i = 0; i < v.size(); i++) {
        std::cout << std::endl;
        this->indent();
        std::cout << "Element " << i << ": " << v.at(i);
    }
    return *this;
}

template <class T>
Terminallog &Terminallog::operator<<(const T v[]) {
    unsigned int elements = sizeof (v) / sizeof (v[0]);
    for (unsigned int i = 0; i < elements; i++) {
        std::cout << std::endl;
        this->indent();
        std::cout << "Element " << i << ": " << v[i];
    }
    return *this;
}

Terminallog &Terminallog::operator<<(const char v[]) {
    std::cout << std::endl;
    this->indent();
    std::cout << v;
    return *this;
}
//snipped code

尝试编译此代码,它给我一个链接器错误:

g++ src/terminallog.cc inc/terminallog.hh testmain.cpp -o test -Wall -Werror 
/tmp/ccifyOpr.o: In function `Terminallog::operator<<(char const*)':
testmain.cpp:(.text+0x0): multiple definition of `Terminallog::operator<<(char const*)'
/tmp/cccnEZlA.o:terminallog.cc:(.text+0x0): first defined here
collect2: ld returned 1 exit status

因此,我目前正在撞墙,寻找解决方案.但我在墙上找不到一个......

如果有人能告诉我我的错误,那就太好了

非常感谢

ftiaronsem

Answer 1

这个功能

Terminallog &Terminallog::operator<<(const char v[]) {
    std::cout << std::endl;
    this->indent();
    std::cout << v;
    return *this;
}

不是模板,而是常规成员函数.如果这个.h文件包含在几个.cpp过滤器中,它将导致您看到的多重定义错误.如果您声明它inline,编译器将允许多个定义,并且您的错误应该被修复.

inline
Terminallog &Terminallog::operator<<(const char v[]) {