如何调整vector迭代器以返回struct？ [英] How to adjust vector iterator to return struct?

问题描述

作为我将C ++项目从Unix转换为Linux的持续任务的一部分，我现在遇到以下错误：

As a part of my ongoing task of converting a C++ project from Unix to Linux, I now have the following error:

jh205 .C：在成员函数'FVSearchLogical_t *
FVLogical :: getFirst（long int）'：jh205.C：9615：错误：来自
的无效强制类型'__gnu_cxx :: __ normal_iterator >>'
键入'FVSearchLogical_t *'jh205.C：在成员函数
'FVSearchLogical_t * FVLogical :: getNext（long int）'：jh205.C：9630：
错误：无法转换
' __gnu_cxx :: __ normal_iterator >>'
到'FVSearchLogical_t *'作为回报​​jh205.C：在成员函数'void
FVLogical :: updateTable（）'：jh205.C：9656：错误：无效的强制转换来自键入
'__ gn_cxx :: __ normal_iterator >>'
键入'void *'

jh205.C: In member function ‘FVSearchLogical_t* FVLogical::getFirst(long int)’: jh205.C:9615: error: invalid cast from type ‘__gnu_cxx::__normal_iterator > >’ to type ‘FVSearchLogical_t*’ jh205.C: In member function ‘FVSearchLogical_t* FVLogical::getNext(long int)’: jh205.C:9630: error: cannot convert ‘__gnu_cxx::__normal_iterator > >’ to ‘FVSearchLogical_t*’ in return jh205.C: In member function ‘void FVLogical::updateTable()’: jh205.C:9656: error: invalid cast from type ‘__gnu_cxx::__normal_iterator > >’ to type ‘void*’

它来自于此代码：

FVSearchLogical_t * FVLogical::getFirst(long sensorId) {

  // loop through the Search vector to find the first one for the sensor
  m_searchLogical_it =  m_FVSearchVector.begin();
  for(int i=0; i < m_FVSearchVector.size(); i++){

    // as soon as we find the first one return it
    if(m_searchLogical_it->ml_sensorId == sensorId) {
      return m_searchLogical_it;
    }
    m_searchLogical_it++;
  }

  return NULL;
}

它的结构：

typedef struct {
    long ml_sensorId;
    char mc_startDate[10];
    char mc_startTime[10];
    char mc_endDate[10];
    char mc_endTime[10];
    long ml_startBlk;
    long ml_endBlk;
    long ml_sendUnit;
} FVSearchLogical_t;

有关如何使用项目中最少的代码更改执行此操作的任何建议？

Any suggestions on how to do this with the least amount of code changes in the project?

添加信息：

FVLogical::~FVLogical(){
  m_FVSearchVector.clear();
  m_FVInsertVector.clear();
  m_rptDataVector.clear();
  m_rptHeaderVector.clear();
  m_rptFooterVector.clear();
}

推荐答案

你是尝试返回一个迭代器，就像警告说你在行中一样：

You are trying to return an iterator exactly as the warning says you are in the line:

return m_searchLogical_it;

获取指向元素的原始指针，返回类型为 getFirst ，你需要获得指向对象 m_searchLogical_it 指向的对象。要做到这一点，你需要取消引用迭代器来获取对象，然后获取对象的地址：

To get a raw pointer to the element, which is the return type of getFirst, you'll need to get a pointer to the object m_searchLogical_it points to. To do that you'll need to dereference the iterator to get the object, then take the address of the object:

return &*m_serchLogical_it;

---

如果我可以另外建议;你正在使用迭代器（ m_searchLogical_it ）和一个循环计数器（ i ），当你需要使用的只是迭代器：

---

If I may additionally suggest; you're using an iterator (m_searchLogical_it) and a loop counter (i), when all you need to use is the iterator:

for(m_searchLogical_it = begin(m_FVSearchVector); m_searchLogical_it != end(m_FVSearchVector); ++m_searchLogical_it) {
    if(m_searchLogical_it->ml_sensorId == sensorId) {
        return &*m_searchLogical_it;
    }
}

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