-1
當我使用包含成員函數的普通函數時，我的程序不斷收到錯誤的alloc錯誤。拋出std :: bad_alloc的實例後調用Terminate。使用兩個類，其中一個指向另一個指針
該程序是關於從命令行獲取一些特定的輸入並打印指針數組的元素。這必須通過指針數組完成。
首先，我創建了一個需要2個字符串的類。一個用於名稱，另一個用於房間。然後，我創建了另一個具有大小和指向我的第一個類的指針，以創建一個數組。
我的主要是在最後，以上主要是2正常功能。這段代碼有什麼問題？當我第一次輸入命令時，它會運行，直到我輸入一個連接到普通函數的命令。那裏可能有些問題，但我似乎無法找到它。
#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;
}
將有助於瞭解你使用的是什麼操作系統，編譯器等。 導致問題的輸入示例會很有幫助。 此外，請讓我們知道什麼是在屏幕上打印。 – CplusPuzzle
我必須重新創建這個向量。它是必須使用的。我需要了解記憶是如何工作的。 –
操作系統是mac。當我輸入'add'後跟着名字和一個房間時，他們直接進入我創建的地址數組。當我輸入'list'時，它會打印整個數組。但是，當我想刪除或找到它只有第一次。那麼它會給我一個糟糕的分配（）。這是怎麼回事？ –