[英]allocated memory gets reclaimed
嗨,我想编码自己的列表类,实际上它的工作原理类似于std :: vector类。 问题是当我使用new
为下一个列表分配内存时,它工作正常。 但是当我尝试为目标(数据)分配内存时,当程序到达push_back()范围的末尾时,它会被回收,我不明白为什么这两种方式不一样,以及如何使用分配的内存我的数据没有被破坏?
代码在这里
#include <iostream>
#include <cstdlib>
using namespace std;
struct pos{
int x;
int y;
pos()
{
x = y = 0;
}
pos(int x, int y)
{
this->x = x;
this->y = y;
}
pos& operator=(pos rhs)
{
x = rhs.x;
y = rhs.y;
return *this;
}
bool operator==(const pos& rhs)
{
if(x == rhs.x && y == rhs.y)
return true;
else
return false;
}
~pos()
{
cout << "x =" << x << ", y =" << y << "got distorted!" << endl;
}
};
class list {
private:
pos *target;
list* next;
int index;
public :
list();
list(pos target);
void push_back (int first , int second);
void push_back (const pos target);
pos pop_back();
pos* search(int first , int second);
pos* search(pos target);
int erase(int index);
pos get(int index);
void change(const pos target,int index);
void change(int first,int second,int index);
~list();
};
void print(list lst);
// function declarations
list::~list()
{
cout << "list is destroyed!" << endl;
if(target != NULL)
delete target;
if(next != NULL)
delete next;
}
list::list()
{
target = NULL;
next = NULL;
index = 0;
}
list::list(pos target)
{
this->target = new pos(target);
index = 0;
next = NULL;
}
void list::push_back(const pos target)
{
cout << "push_back() begin" << endl;
list* it = this;
while(it->next != NULL)
{
it = it->next;
}
if(it->target == NULL)
{
it->target = new pos(target);
}
else
{
it->next = new list;
it->next->index = it->index+1;
//option one
it->next->target = new pos(target);
//option two
it->next->target = (pos*)malloc(sizeof(pos));
(*it->next->target) = target;
//it->next->next is already NULL
}
cout << "push_back() end" << endl;
}
void list::push_back(int first , int second)
{
push_back(pos(first,second));
}
pos list::pop_back()
{
print(*this);
list* it = this;
cout << "address of x is" << this << endl;
cout << "this->target is" << this->target << endl;
cout << (*target).x << endl;
if(it->target == NULL)
return *(new pos); // an error is occurred there is not any data to return! must find another solution maybe throw an exception
if(it->next == NULL)
{
pos return_data = *(it->target);
delete it->target;
it->target = NULL;
return return_data;
}
while(it->next->next != NULL)
{
cout << "it->target is" << it->target << endl;
it = it->next;
}
pos return_data = *(it->next->target);
delete it->next;
it->next = NULL;
return return_data;
}
pos* list::search(pos target)
{
list* it = this;
do
{
if(target == *(it->target))
return it->target;
if(it->next != NULL)
it = it->next;
else
return NULL;
}while(1);
}
pos* list::search(int first , int second){
return search(pos(first,second));
}
int list::erase(int index){
if(index < 0)
return 0;
list *it = this , *it_next = this->next;
if(index == 0)
{
if(it->next == NULL)
{
delete it->target;
return 1;
}
while(it_next->next != NULL)
{
it->target = it_next->target;
it = it_next;
it_next = it_next->next;
}//needs to be completed
}
do
{
if(it_next->index == index)
{
it->next = it_next->next;
delete it_next;
return 1;
}
if(it_next->next != NULL)
{
it = it_next;
it_next = it_next->next;
}
else
return 0;
}while(1);
return 1;
}
pos list::get(int index)
{
if(index < 0)
return *(new pos);//error
list* it = this;
do
{
if(it->index == index)
{
return *(it->target);
}
if(it->next != NULL)
it = it->next;
else
return *(new pos);//error , index is bigger than [list size] - 1
}while(1);
}
void list::change(const pos target,int index)
{
if(index < 0)
return ;//error
list* it = this;
do
{
if(it->index == index)
{
*(it->target) = target;
}
if(it->next != NULL)
it = it->next;
else
return;//error , index is bigger than [list size] - 1
}while(1);
}
void list::change(const int first,const int second,int index)
{
change(pos(first,second),index);
}
void print(list lst)
{
int idx = 0;
while(!(lst.get(idx)==pos(0,0)))
{
cout << "index " << idx << " : x = " << lst.get(idx).x << ", y = " << lst.get(idx).y << endl;
idx++;
}
}
int main(int argc, char const *argv[])
{
list x;
cout << "address of x is" << &x << endl;
x.push_back(1,1);
x.push_back(2,2);
x.push_back(3,3);
x.push_back(4,4);
x.push_back(5,5);
print(x);
cout << "--------------------------" << endl;
x.pop_back();
print(x);
cout << "--------------------------" << endl;
cout << x.get(2).x << endl;
x.erase(2);
print(x);
cout << "--------------------------" << endl;
return 0;
}
换句话说,为什么push_back返回时它-> next-> target和/或it-> target被破坏?
正在销毁(或调用析构函数)的是在下一行作为输入参数创建的临时对象:
push_back(pos(first,second));
在
void list::push_back(const pos target)
target
是按值传递的,因此push_back
的target
是一个临时副本。 函数结束时,副本将超出范围并被销毁。 这就是您所看到的。 烦人,但不是您真正的问题。
list
违反三规则 。 这意味着在复制list
时,不会正确复制它。 指针将被复制,而不是指向的项目。 每次复制list
,原件和复制点都位于同一位置。 当副本超出范围并被销毁时,它将随其携带原始数据。
碰巧的是,您只按值传递,因此有很多复制和销毁操作。 例如, print
功能将错误地复制并在返回时清除提供的list
。
解决方案:在list
中添加一个复制构造函数和一个赋值运算符,以遍历列表的方式进行工作,并复制所有链接,并通过引用进行读取。
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