清除已经空的向量会导致未定义的行为吗？

Question

如果我发生什么clear一个vector这是已经清除之前？我在Visual Studio中尝试过,它没有导致任何运行时错误.但我正在寻找一个可能的例外原因(下图)并想知道这是否是原因？

*** glibc detected *** process.out: double free or corruption (out): 0x01a0b588 *** 

更新代码:

bool myclass::Sort(SortType enSortOption)
{

  bool bRet = TRUE;


  m_oVecOfIntsOrig.clear();
  m_oVecOfIntsSorted.clear();

  for(int i = 0;i<m_oList.listsize();i++)
  {
    m_oVecOfIntsOrig.push_back(i);
  }
  m_oVecOfIntsSorted = m_oVecOfIntsOrig; 
  //Just sort the indices     
  switch(enSortOption)
  {
  case Alpha:
    { 
      t_SortStructAlpha sSortAlpha(this);
      sort( m_oVecOfIntsSorted.begin(), m_oVecOfIntsSorted.end(), sSortAlpha );        
    }
    break;
  case Dist:
    {        
      t_SortStructDist sSortDist(this);
      sort( m_oVecOfIntsSorted.begin(), m_oVecOfIntsSorted.end(), sSortDist );
    }
    break;
  case none:  
    {

      //Nothing to do
      return TRUE;
    }
    break;  
  default:
    {
      bRet = FALSE;

    }
    break;
  }

  //Update the list based on sorted index vector
  ListElem oSortedListElements;
  //this clear function simply clears all the 7 constituent vectors. 
  oSortedListElements.vClear();
  if(TRUE == bGetSortedList(oSortedListElements))
  {
    //If successfully copied to temp object, then update member variable
    //m_oList is what I'm intending to display
    m_oList.vClear();
    m_oList = oSortedListElements;
  }
  else
  {
    //copy failed
    bRet = FALSE;

  }
  return bRet;
}


bool myclass::bGetSortedList(ListElem& oSortedListElements)
{

  bool bRet = TRUE;

  //Creating object from beginning, so clear old contents if any
  oSortedListElements.vClear();
  /* Sort done. Now update the list based on new indices*/
  for(int u16Index = 0; u16Index<m_oVecOfIntsSorted.size(); u16Index++)
  {
    if(
      (m_oVecOfIntsSorted.at(u16Index) >= m_oListConstituents.oVec1.size())
      ||
      (m_oVecOfIntsSorted.at(u16Index) >= m_oListConstituents.oVec2.size())
      ||
      (m_oVecOfIntsSorted.at(u16Index) >= m_oListConstituents.oVec3.size())
      ||
      (m_oVecOfIntsSorted.at(u16Index) >= m_oListConstituents.oVec4.size())
      ||
      (m_oVecOfIntsSorted.at(u16Index) >= m_oListConstituents.oVec5.size())
      ||
      (m_oVecOfIntsSorted.at(u16Index) >= m_oListConstituents.oVec6.size())
      ||
      (m_oVecOfIntsSorted.at(u16Index) >= m_oListConstituents.oVec7.size())
      )
    {
      //out of bounds check
      return FALSE;
    }
    //m_oListConstituents contains overall list
    oSortedListElements.oVec1.push_back(m_oListConstituents.oVec1.at(m_oVecOfIntsSorted.at(u16Index)));
    oSortedListElements.oVec2.push_back(m_oListConstituents.oVec2.at(m_oVecOfIntsSorted.at(u16Index)));
    oSortedListElements.oVec3.push_back(m_oListConstituents.oVec3.at(m_oVecOfIntsSorted.at(u16Index)));
    oSortedListElements.oVec4.push_back(m_oListConstituents.oVec4.at(m_oVecOfIntsSorted.at(u16Index)));
    oSortedListElements.oVec5.push_back(m_oListConstituents.oVec5.at(m_oVecOfIntsSorted.at(u16Index)));
    oSortedListElements.oVec6.push_back(m_oListConstituents.oVec6.at(m_oVecOfIntsSorted.at(u16Index)));
    oSortedListElements.oVec7.push_back(m_oListConstituents.oVec7.at(m_oVecOfIntsSorted.at(u16Index)));     

  }

  return bRet;

}

Answer 1

如果你做了什么相当于

v.clear();
v.clear();

然后不,没有UB:其中一个后置条件v.clear()是v.size() == 0,所以第二次调用clear应该是无操作.