重载运算符<<模板类[重复]

Question

我正在使用 Microsoft Visual Studio 作为源代码编辑器，当我尝试编译以下代码时出现链接器错误：

LNK2019 未解析的外部符号 "class std::basic_ostream > & __cdecl operator > &,类 X &)" (??6@YAAAV?$basic_ostream@DU?$char_traits@D@std@@@std@@AAV01@AAV?$X@H@@@Z) 在函数 _main Zadatak7 中引用.19.04.2021 C:\Users\AT95\Desktop\Zadatak7.19.04.2021\Zadatak7.19.04.2021\main.obj 1

我必须怎样做才能使这个程序正常工作？

我想让你看一下重载operator<<的定义，因为 该方法导致错误。

#include <iostream>

template <class T>
class X
{

private:

    int index, capacity;

    T* collection{};
        
public:

    X(const int nmb, const T* array)
    {
        index = 0;

        capacity = nmb;

        collection = new(T[nmb]);

        for (int i = 0; i < nmb; i++)
            collection[index++] = array[i];
    }

    X(const X& other)
    {
        index = other.index;
        capacity = other.capacity;

        collection = new(T[other.capacity]);

        if(other.index < other.capacity)
            for (int i = 0; i < other.index; i++)
                collection[i] = other.collection[i];
    }

    ~X() { delete[] collection; }

    friend std::ostream& operator<<(std::ostream&, X<T>&);

};
        
template <class T>
std::ostream& operator<<(std::ostream& out, X<T>& other)
{
    for (int i = 0; i < other.index; i++)
        out << i + 1 << ". " << other.collection[i];

    return out;
}

int main()
{
    int array[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }, n = sizeof(array)/sizeof(int);

    X<int> x(n, array), b(x);

    std::cout << x;

    return 0;
}

Answer 1

你的朋友声明是错误的。

    friend std::ostream& operator<<(std::ostream&, X<T>&);

这是一个非模板operator<< 的朋友声明。

应该是这样的：

    template<typename Z>
    friend std::ostream& operator<<(std::ostream&, X<Z>&);

但是有一个更好的方法来声明它。将函数体移动到类中，就不需要声明模板部分了。

template <class T>
class X
{

    // STUFF AS BEFORE


    // Friend class definition.
    friend std::ostream& operator<<(std::ostream& out, X& other)
    {
        for (int i = 0; i < other.index; i++)
            out << i + 1 << ". " << other.collection[i];

        return out;
    }
}

次要说明。在输出运算符中，被输出的对象通常被声明为 const（因为输出它不应该修改它）。

friend std::ostream& operator<<(std::ostream& out, X const& other)
                                                        ^^^^^

Answer 2

你需要在friend函数的开头写上template关键字，像这样：

#include <iostream>

template <class T>
class X
{

private:

    int index, capacity;

    T* collection{};

public:

    X(const int nmb, const T* array)
    {
        index = 0;

        capacity = nmb;

        collection = new(T[nmb]);

        for (int i = 0; i < nmb; i++)
            collection[index++] = array[i];
    }

    X(const X& other)
    {
        index = other.index;
        capacity = other.capacity;

        collection = new(T[other.capacity]);

        if(other.index < other.capacity)
            for (int i = 0; i < other.index; i++)
                collection[i] = other.collection[i];
    }

    ~X() { delete[] collection; }

    template<class U> friend std::ostream& operator<<(std::ostream&, X<U>&);

};

template <class T>
std::ostream& operator<<(std::ostream& out, X<T>& other)
{
    for (int i = 0; i < other.index; i++)
        out << i + 1 << ". " << other.collection[i];

    return out;
}

int main()
{
    int array[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }, n = sizeof(array)/sizeof(int);

    X<int> x(n, array), b(x);

    std::cout << x;

    return 0;
}