Question

我一直在研究这段代码，试图用test bed main测试我的公司类。我得到了我想要的所有结果，但是在完成主要功能后，程序崩溃了 this message: （我道歉但我是新来的，我还不能直接发布图片。）我一直在处理这个错误大约需要4-5个小时，而且我发现的一切似乎都不起作用。从我环顾四周的情况来看，我觉得更好地理解指针，但我似乎无法找到问题所在。

//COMPANY.H

#ifndef _COMPANY_H
#define _COMPANY_H
#include <stddef.h>
#include <fstream>
#include <iomanip>
enum { PHONE_LEN = 10 };
enum { MAX_NAME_LEN = 30 };

#define _TESTING_COMP
#ifdef _TESTING_COMP
#include <iostream>
using namespace std;
#endif


class Company
{
public:
   Company();
   ~Company();
   Company &operator =(Company const & c);
   bool operator==(const Company & c)const {return (*name == *c.name);}
   bool operator!=(const Company & c)const {return (*name != *c.name);}
   bool operator<(const Company & c)const {return (*name < *c.name);}
   friend istream & operator>>(istream & in, Company & c);
   friend ostream & operator<<(ostream & out, const Company & c);
private:
   char *name; // allocate memory (use new) - zero-terminated
   char phone[PHONE_LEN];     // NOT zero-terminated (be careful!)
};
#endif

这是我的头文件，包含函数和数据声明。

//COMPANY.CPP

#include "Company.h"

Company::Company() 
{
   name = new char [MAX_NAME_LEN];
}

Company::~Company()
{
   delete [] name;
}

Company &Company::operator =(Company const & c)
{
   for (int i = 0; i < MAX_NAME_LEN; i++)
      *(name + i) = *(c.name + i);
   for (int i = 0; i < PHONE_LEN; i++)
      phone[i] = c.phone[i];
   return *this;
}

istream & operator>>(istream & in, Company & c)
{
   in >> c.name;
   for (int i = 0; i < PHONE_LEN; i++)
      in >> c.phone[i];
    return in;
}

ostream & operator<<(ostream & out, const Company & c)
{
    //because value is a pointer the * is needed to input a value
    out << c.name << setiosflags(ios::left) << setw(MAX_NAME_LEN) << "";
   for (int i = 0; i < PHONE_LEN; i++)
      out << c.phone[i];
   out << endl;
   return out;
}



#ifdef _TESTING_COMP
void main ()
{
   char end;
   Company test1, test2;
   cin >> test1;
   cout << test1;
   cin >> test2;
   cout << test2;
   cout << "testing comp1 < comp2: expecting (0)" << (test1 < test2) << endl;
   cout << "testing comp2 < comp1: expecting (1)" << (test2 < test1) << endl;
   cout << "testing comp2 == comp1: expecting (0)" << (test2 == test1) << endl;
   cout << "testing comp2 != comp1: expecting (1)" << (test2 != test1) << endl;
   cout << "set test1 to = test2" << endl;
   test1 = test2;
   cout << test1;
   cout << test2;
   cout << "testing comp1 < comp2: expecting (0)" << (test1 < test2) << endl;
   cout << "testing comp2 < comp1: expecting (0)" << (test2 < test1) << endl;
   cout << "testing comp2 == comp1: expecting (1)" << (test2 == test1) << endl;
   cout << "testing comp2 != comp1: expecting (0)" << (test2 != test1) << endl;
   test1.~Company();
   test2.~Company();
   cin >> end;
}
#endif

这是保存额外代码和testbed main的cpp文件。

如果你们可以帮助我解决这个用户错误，那将是非常有必要的。即使你不能感谢时间。我不完全确定我应该寻找什么 these are the data values at the end of the program

如果你们需要，我可以发帖更多，谢谢你。

Answer 1

这么多的样板是不必要的。通过动态分配name，你可以毫无理由地做一大堆家务管理（最终导致你的崩溃）。此外，您的所有比较功能都会给您带来惊人的效果。

这是你的类重构为cannonical C ++：

#include <string>

class Company
{
public:
   const std::string& get_name() const { return name; }
   const std::string& get_phone() const { return phone; }

private:
   std::string name;
   std::string phone;

   friend istream & operator>>(istream & in, Company & c);
   friend ostream & operator<<(ostream & out, const Company & c);
};

bool operator==(const Company& lhs, const Company& rhs) { return lhs.name == rhs.name; }
bool operator!=(const Company& lhs, const Company& rhs) { return !(lhs == rhs); }
bool operator<(const Company& lhs, const Company& rhs) { return lhs.name < rhs.name; }

表达式：_BLOCK_TYPE_IS_VALID删除指针时出现问题

1 个答案: