本文介绍了CAtlList::RemoveAt 是否会使现有位置无效?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我在看这个,其中 m_Rows 是一个 CAtlList:
I'm looking at this, where m_Rows is a CAtlList:
void CData::RemoveAll()
{
size_t cItems = m_Rows.GetCount();
POSITION Pos = m_Rows.GetHeadPosition();
while(Pos != 0)
{
CItem* pItem = m_Rows.GetAt(Pos);
if (pItem != 0)
delete pItem;
POSITION RemoveablePos = Pos;
pItem = m_Rows.GetNext(Pos);
m_Rows.RemoveAt(RemoveablePos);
}
}
我想知道 RemoveAt 调用是否有可能使 Pos 失效?
and am wondering if there's potential that the RemoveAt call may invalidate Pos?
推荐答案
根据 文档,CAtlList 的行为类似于双链表,因此删除一个列表项应该不会使指向其他项的指针无效.POSITION
类型直接引用列表项的内存位置:
According to the documentation, CAtlList behaves like a double linked list, so removing one list item should not invalidate the pointers to other items. The POSITION
type references the memory location of a list item directly:
大多数 CAtlList 方法使用位置值.方法使用此值来引用存储元素的实际内存位置,不应直接计算或预测.
在 atlcoll.h 中似乎不是这种情况:
It seems this is not the case in atlcoll.h:
template< typename E, class ETraits >
void CAtlList< E, ETraits >::RemoveAt( POSITION pos )
{
ATLASSERT_VALID(this);
ATLENSURE( pos != NULL );
CNode* pOldNode = (CNode*)pos;
// remove pOldNode from list
if( pOldNode == m_pHead )
{
m_pHead = pOldNode->m_pNext;
}
else
{
ATLASSERT( AtlIsValidAddress( pOldNode->m_pPrev, sizeof(CNode) ));
pOldNode->m_pPrev->m_pNext = pOldNode->m_pNext;
}
if( pOldNode == m_pTail )
{
m_pTail = pOldNode->m_pPrev;
}
else
{
ATLASSERT( AtlIsValidAddress( pOldNode->m_pNext, sizeof(CNode) ));
pOldNode->m_pNext->m_pPrev = pOldNode->m_pPrev;
}
FreeNode( pOldNode );
}
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