问题描述
我正在尝试使用 erase
和列表迭代器从 C++ 链接列表中删除项目:
I'm trying to remove items from a C++ linked list using erase
and a list iterator:
#include <iostream>
#include <string>
#include <list>
class Item
{
public:
Item() {}
~Item() {}
};
typedef std::list<Item> list_item_t;
int main(int argc, const char *argv[])
{
// create a list and add items
list_item_t newlist;
for ( int i = 0 ; i < 10 ; ++i )
{
Item temp;
newlist.push_back(temp);
std::cout << "added item #" << i << std::endl;
}
// delete some items
int count = 0;
list_item_t::iterator it;
for ( it = newlist.begin(); count < 5 ; ++it )
{
std::cout << "round #" << count << std::endl;
newlist.erase( it );
++count;
}
return 0;
}
我得到了这个输出,但似乎无法追踪原因:
I get this output and can't seem to trace the reason:
added item #0
added item #1
added item #2
added item #3
added item #4
added item #5
added item #6
added item #7
added item #8
added item #9
round #0
round #1
Segmentation fault
我可能做错了,但无论如何都希望得到帮助.谢谢.
I'm probably doing it wrong, but would appreciate help anyway. thanks.
推荐答案
这里的核心问题是您在调用 erase
之后使用迭代器值 it
代码>就可以了.erase
方法使迭代器无效,因此继续使用它会导致不良行为.相反,您希望使用 erase
的返回值来获取擦除值之后的下一个有效迭代器.
The core problem here is you're using at iterator value, it
, after you've called erase
on it. The erase
method invalidates the iterator and hence continuing to use it results in bad behavior. Instead you want to use the return of erase
to get the next valid iterator after the erased value.
it = newList.begin();
for (int i = 0; i < 5; i++) {
it = newList.erase(it);
}
检查 newList.end()
以解决 list
中至少没有 5 个元素的情况也没有坏处.
It also doesn't hurt to include a check for newList.end()
to account for the case where there aren't at least 5 elements in the list
.
it = newList.begin();
for (int i = 0; i < 5 && it != newList.end(); i++) {
it = newList.erase(it);
}
正如 Tim 所指出的,这里是 erase
As Tim pointed out, here's a great reference for erase
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