抛出std :: bad_alloc的实例后调用Terminate。使用两个类，其中一个指向另一个指针

Question

当我使用包含成员函数的普通函数时，我的程序不断收到错误的alloc错误。

该程序是关于从命令行获取一些特定的输入并打印指针数组的元素。这必须通过指针数组完成。

首先，我创建了一个需要2个字符串的类。一个用于名称，另一个用于房间。然后，我创建了另一个具有大小和指向我的第一个类的指针，以创建一个数组。

我的主要是在最后，以上主要是2正常功能。这段代码有什么问题？当我第一次输入命令时，它会运行，直到我输入一个连接到普通函数的命令。那里可能有些问题，但我似乎无法找到它。

#include <iostream> 
#include <string> 
using namespace std; 

class Address 
{ 
    private: 
     string name; 
     string room; 
    public: 
     Address(){}; 
     Address(string, string); 
     string get_name(); 
     string get_room(); 
     void change_room(string); 
}; 

Address::Address (string n, string r) 
{ 
    name = n; 
    room = r; 
} 


string Address::get_name() 
{ 
    return name; 
} 

string Address::get_room() 
{ 
    return room; 
} 

void Address::change_room(string change) 
{ 
    room = change; 
} 


//end of Address class 


class Address_Book 
{ 
    private: 
     int size; 
     Address* addresses; 
    public: 
     Address_Book(); 
     ~Address_Book(){ delete[] addresses;} 
     void add(Address); 
     void move(string, string); 
     int get_size(); 
     Address location(int); 
     int find(string); 
     void clear(); 
     void remove_address(string); 
     int exists(string); 
     void sort(); 
}; 


Address_Book::Address_Book() 
{ 
    int s = 0; 
    size = s; 
    addresses = new Address[s]; 
} 

void Address_Book::add(Address add) 
{ 
    Address* temp = new Address [size + 1]; 
    for (int i = 0; i < size; i++) 
    { 
     temp[i] = addresses[i]; 
    } 
    temp[size] = add; 
    delete[] addresses; 
    addresses = temp; 
    size ++; 

} 

void Address_Book::move(string name, string newroom) 
{ 
    for (int i = 0; i < size ; i++) 
    { 
     if (addresses[i].get_name() == name) 
     { 
      addresses[i].change_room(newroom); 
     } 
    } 
} 

void Address_Book::remove_address(string name) 
{ 
    Address* temp = new Address [size - 1]; 

    for (int i = 0; i < size; i++) 
    { 
     if (addresses[i].get_name() != name) 
     { 
      temp[i] = addresses[i]; 
     } 
     else if (addresses[i].get_name() == name) 
     { 
      for (int j = i + 1; j < size; j++) 
      { 
       temp[i] = addresses[j]; 
       i++; 
      } 
      break; 
     } 
    } 
    delete[] addresses; 
    addresses = temp; 
    size--; 
} 


int Address_Book::get_size() 
{ 
    return size; 
} 


Address Address_Book::location(int index) 
{ 
    return addresses[index]; 
} 


void Address_Book::sort() 
{ 
    Address temp; 
    for (int i = 0; i < size; i++) 
    { 
     for(int j = 0; j < size - 1; j++) 
     { 
      if (addresses[j].get_room() > addresses[j + 1].get_room()) 
      { 
       temp = addresses[j]; 
       addresses[j] = addresses[j + 1]; 
       addresses[j + 1] = temp; 
      } 
     } 
    } 
    for (int i = 0; i < size; i++) 
    { 
     if (addresses[i].get_room() == addresses[i + 1].get_room()) 
     { 
      if (addresses[i].get_name() > addresses[i + 1].get_name()) 
      { 
       temp = addresses[i]; 
       addresses[i] = addresses[i + 1]; 
       addresses[i + 1] = temp; 
      } 
     } 
    } 
} 

void Address_Book::clear() 
{ 
    Address * temp = new Address[0]; 
    delete[] addresses; 
    addresses = temp; 
    size = 0; 
} 


int Address_Book::find(string name) 
{ 
    for (int i = 0; i < size; i++) 
    { 
     if (addresses[i].get_name() == name) 
     { 
      return i; 
     } 
    } 
    return -1; 
} 



//end of Address_Book class 



void find(string name, Address_Book addbook) 
{ 
    int index = addbook.find(name); 
    cout << index << endl; 

    if (index > -1) 
    { 
     cout << addbook.location(index).get_name() << " is in room " << 
     addbook.location(index).get_room() << endl; 
    } 
    else 
    { 
     throw runtime_error("entry does not exist."); 
    } 
} 


void remove_add(string name, Address_Book book) 
{ 
    int exist = book.find(name); 

    if (exist > -1) 
    { 
     book.remove_address(name); 
    } 
    else 
    { 
     throw runtime_error("entry does not existt."); 
    } 

} 






int main() 
{ 
    Address_Book addbook; 
    string action, in_name, in_room; 
    do 
    { 
     try 
     { 
      cout << "> "; 
      cin >> action; 

      if (action == "add") 
      { 
       cin >> in_name >> in_room; 
       Address newadd(in_name, in_room); 
       addbook.add(newadd); 
      } 
      else if (action == "move") 
      { 
       cin >> in_name >> in_room; 
       addbook.move(in_name, in_room); 
      } 
      else if (action == "remove") 
      { 
       cin >> in_name; 
       remove_add(in_name, addbook); 
      } 
      else if (action == "find") 
      { 
       cin >> in_name; 
       find(in_name, addbook); 
      } 
      else if (action == "list") 
      { 
       addbook.sort(); 

       for (int i = 0; i < addbook.get_size(); i++) 
       { 
       cout << addbook.location(i).get_name() << " is in room 
       " << addbook.location(i).get_room() << endl; 
       } 
      } 
      else if (action == "clear") 
      { 
       addbook.clear(); 
      } 
      else 
      { 
       throw runtime_error("input mismatch."); 
      } 
     } 
     catch (runtime_error& e) 
     { 
      cerr << "error: " << e.what() << endl; 
     } 
    }while (action != "exit"); 

    return 0; 
} 

Answer 1

函数remove_add需要通过引用或指针获取地址簿对象。 它现在的样子，从地址簿的副本中删除。

它应该是这样的：

void remove_add(string name, Address_Book& book) 
{ 
    int exist = book.find(name); 

    if (exist > -1) 
    { 
     book.remove_address(name); 
    } 
    else 
    { 
     throw runtime_error("entry does not existt."); 
    } 

} 

此外，你应该做的事情，在下面的功能的情况下size == 1不同。例如如果您的编译器支持它，请将addresses设置为NULL,0或nullptr。

void Address_Book::remove_address(string name) 
{ 
    Address* temp = new Address[size - 1]; 

    for (int i = 0; i < size; i++) 
    { 
     if (addresses[i].get_name() != name) 
     { 
      temp[i] = addresses[i]; 
     } 
     else if (addresses[i].get_name() == name) 
     { 
      for (int j = i + 1; j < size; j++) 
      { 
       temp[i] = addresses[j]; 
       i++; 
      } 
      break; 
     } 
    } 
    delete[] addresses; 
    addresses = temp; 
    size--; 
} 

玩得开心学习语言和好运:)

Answer 2

导致你的问题的确切命令你的问题没有规定，所以我周围戳一点点，直到代码段故障而坠毁。

Valgrind和Dr. Memory是找到这些问题的根本原因很棒的工具。你的情况：

$ g++ -g 46865300.cpp 
$ valgrind ./a.out 
> add foo bar 
> list 
==102== Invalid read of size 8 
==102== at 0x4EF4EF8: std::basic_string<char, std::char_traits<char>, std::allocator<char> >::basic_string(std::string const&) (in /usr/lib64/libstdc++.so.6.0.19) 
==102== by 0x401354: Address::get_room() (46865300.cpp:33) 
==102== by 0x401C05: Address_Book::sort() (46865300.cpp:152) 
==102== by 0x4026A3: main (46865300.cpp:262) 
==102== Address 0x5a17410 is 8 bytes after a block of size 24 alloc'd 
==102== at 0x4C2A8A8: operator new[](unsigned long) (vg_replace_malloc.c:423) 
==102== by 0x4014BF: Address_Book::add(Address) (46865300.cpp:74) 
==102== by 0x40245C: main (46865300.cpp:243) 

它说，下面的代码进行了界外接入：

150  for (int i = 0; i < size; i++) 
151  { 
152   if (addresses[i].get_room() == addresses[i + 1].get_room()) 
153   { 
154    if (addresses[i].get_name() > addresses[i + 1].get_name()) 

我猜循环条件应该用“大小 - 1”而不是“大小”。