Question

所以我尝试使用智能指针重新实现我自己的字符串类，这样我就可以练习将它们集成到我的日常使用中。可悲的是，我已经撞墙了，我无法弄清楚为什么我无法连接两个“mystring”对象。有什么建议？此外，如果这种类型的程序不适合使用智能指针，我也会很感激与此相关的建议。

#include <iostream>
#include <memory>
#include <cstring>

using namespace std;

class mystring {
public:
    mystring() : word(make_unique<char[]>('\0')), 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

将参数message的{{1}}类型从operator<<（对非const引用）更改为mystring &（对const的引用）：

const mystring &

friend ostream &operator<<(ostream &os, const mystring &message) { for (int i = 0; i < message.len; i++) os << message.word[i]; return os; }按值返回，因此返回的是临时的，不能绑定引用非const。

请注意，您应该这样做不仅是为了解决这个问题; operator+不应该通过引用非const来传递，因为参数不应该在message内修改。

错误C2679：二进制'＆lt;＆lt;'：找不到运算符，它接受'mystring'类型的右手操作数（或者没有可接受的转换）

1 个答案: