Question

我可能在这里遗漏了一些明显的东西。这是我的代码（我只是学习真正的C ++，我希望得到一些练习）：

#include <iostream>
#include <cstring>

using namespace std;

class String {
private:
    char * value;
    int len;
    friend ostream & operator<<(ostream & os, String s);
public:
    String();
    String(const char * base);
    ~String();
    String operator+(String s);
    String operator*(int n);
    int length();
};

String::String() {
    this->value = new char[0];
    this->len = 0;
}

String::String(const char * base) {
    this->value = new char[this->len = strlen(base)];
    strcpy(this->value, base);
}

String::~String() {
    delete [] this->value;
}

int String::length() {
    return this->len;
}

String String::operator+(String s) {
    String n;
    delete [] n.value;

    cout << "Upon entering, I am: \"" << *this << "\"\n";

    n.value = new char[this->len + s.len];
    for(int i = 0; i < this->len; i++) {
        n.value[i] = this->value[i];
    }
    for(int i = 0; i < s.len; i++) {
        n.value[i + this->len] = s.value[i];
    }

    n.len = this->len + s.len;

    cout << "String::operator+(" << *this << ", " << s << ") succeeded with new value = \"" << n << "\"\n";

    return n;
}

String String::operator*(int n) {
    String s;
    delete [] s.value;

    s.value = new char[this->len * n];

    for(int i = 0; i < this->len * n; i++) {
        s.value[i] = this->value[i % this->len];
    }

    cout << "String::operator* succeeded with new value = \"" << s << "\"\n";

    return s;
}

ostream & operator<<(ostream & os, String s) {
    return os << s.value;
}

int main() {
    String s("Hello, world!");
    cout << s << "\nLength = " << s.length() << "\n\n";
    cout << (s + String("\n")) * 5;
    return 0;
}

字符串初始化并正确显示，但我的输出真的很奇怪;似乎进入运营商+，“你好，世界！”突然变成“x％r”？

C:\Users\Ryan\Documents\My Dropbox\C++ Projects>strings
Hello, world!
Length = 13

Upon entering, I am: "x%r"
String::operator+(x%r,
) succeeded with new value = "x%r"
String::operator* succeeded with new value = "╚%r"
─

Answer 1

试试这个：

ostream & operator<<(ostream & os, const String& s) {
    return os << s.value;
}

否则你应该为你的String类定义复制构造函数。

Answer 2

您需要提供复制构造函数和赋值运算符。

Answer 3

您的代码存在很多问题。

你在管理自己的记忆。如果可能的话，你应该避免这样做。
你总是忘记字符串有一个空终止符。为了容纳strin Hello, world!，你需要一个14字节而不是13字节的字符缓冲区。
你有一个len成员变量与strlen函数有效地完成相同的事情，除了对上面＃1的不一致考虑。
您的字符串类没有复制构造函数，这会导致狂野指针并最终崩溃。

以下是您的代码的重构，非常有效。

#define _CRT_SECURE_NO_WARNINGS

#include <iostream>
#include <cstring>

using namespace std;

class String {
private:
    char * value;
//    size_t len;
    friend ostream & operator<<(ostream & os, String s);
public:
    String();
    String(const char * base);
    String(const String& rhs) 
    {
        value = new char[strlen(rhs.value)+1];
        strcpy(value,rhs.value);
    }
    ~String();
    String operator+(String s);
    String operator*(int n);
    size_t length();
};

String::String() {
    this->value = new char[0];
}

String::String(const char * base) {
    this->value = new char[strlen(base)+1];
    strcpy(this->value, base);
}

String::~String() {
    delete [] this->value;
}

size_t String::length() {
    return strlen(value);
}

String String::operator+(String s) {
    String n;
    delete [] n.value;

    cout << "Upon entering, I am: \"" << *this << "\"\n";

    n.value = new char[strlen(value)+strlen(s.value)+1];
    for(int i = 0; i < strlen(value); i++) {
        n.value[i] = this->value[i];
    }
    for(int i = 0; i < strlen(s.value); i++) {
        n.value[i + strlen(value)] = s.value[i];
    }
    n.value[strlen(value)+strlen(s.value)] = '\0';

    cout << "String::operator+(" << *this << ", " << s << ") succeeded with new value = \"" << n << "\"\n";

    return n;
}

String String::operator*(int n) {
    String s;
    delete [] s.value;

    s.value = new char[(strlen(value)*n)+1];

    for(int i = 0; i < strlen(value) * n; i++) {
        s.value[i] = this->value[i % strlen(value)];
    }
    s.value[strlen(value)*n] = '\0';

    cout << "String::operator* succeeded with new value = \"" << s << "\"\n";

    return s;
}

ostream & operator<<(ostream & os, String s) {
    return os << s.value;
}

int main() {
    String s("Hello, world!");
    cout << s << "\nLength = " << s.length() << "\n\n";
    cout << (s + String("\n")) * 5;
    return 0;
}

进入运营商时程序的奇数输出？

3 个答案: