错误 C2679：二进制"<<"：未找到采用类型 'mystring' 的右侧操作数的运算符(或者没有可接受的转换)

Question

因此，我尝试使用智能指针重新实现自己的字符串类，以便我可以练习将它们集成到日常使用中。可悲的是，我撞到了一堵墙，我无法弄清楚为什么我无法加入两个" mystring"对象。有什么建议么？另外，如果这种类型的程序不是使用智能指针的适当情况，我也将感谢与此相关的建议。

#include <iostream>
#include <memory>
#include <cstring>
using namespace std;
class mystring {
public:
    mystring() : word(make_unique<char[]>('')), len(0) {}
    ~mystring() { cout << "goodbye objects!";}
    mystring(const char *message) : word(make_unique<char[]>(strlen(message) + 1)), len(strlen(message)) {
        for (int i = 0; i < len; i++) word[i] = message[i];
    }
    mystring(const mystring &rhs) : word(make_unique<char[]>(rhs.len)), len(rhs.len + 1) {
        for (int i = 0; i < len; i++) word[i] = rhs.word[i];
    }
    mystring &operator=(mystring &rhs) {
        if (this != &rhs) {
            char *temp = word.release();
            delete[] temp;
            word = make_unique<char[]>(rhs.len + 1);
            len = rhs.len;
            for (int i = 0; i < len; i++) word[i] = rhs.word[i];
        }
        return *this;
    }
    mystring &operator=(const char *rhs) {
        char *temp = word.release();
        delete[] temp;
        word = make_unique<char[]>(strlen(rhs)+ 1);
        len = strlen(rhs);
        for (int i = 0; i < len; i++) word[i] = rhs[i];
        return *this;
    }
    friend mystring operator+(const mystring& lhs, const mystring& rhs) {
        mystring Result;
        int lhsLength = lhs.len, rhsLength = rhs.len;
        Result.word = make_unique<char[]>(lhsLength + rhsLength + 1);
        Result.len = lhsLength + rhsLength;
        for (int i = 0; i < lhsLength; i++) Result.word[i] = lhs.word[i];
        for (int i = lhsLength; i < lhsLength + rhsLength; i++) Result.word[i] = rhs.word[i];
        return Result;
    }
    friend ostream &operator<<(ostream &os, mystring &message) {
        for (int i = 0; i < message.len; i++) os << message.word[i];
        return os;
    }
private:
    int len;
    unique_ptr<char[]> word;
};
int main()
{
    mystring word1 = "Darien", word2 = "Miller", word3;
    cout << word1 + word2;//error message: no binary '<' found
    word3 = word1 + word2;//error message: no binary '=' found
    cout << word3;
    return 0;
}

Answer 1

将 message的 operator<<类型从 mystring &(引用非const(到 const mystring &(引用const(：

friend ostream &operator<<(ostream &os, const mystring &message) {
    for (int i = 0; i < message.len; i++) os << message.word[i];
    return os;
}

operator+按值返回，因此它返回的是暂时的，不能绑定到非const。

请注意，您不仅应该解决此问题；message不应参考非const传递，因为该参数不应在operator<<中进行修改。