如何在迭代时从地图中删除？

如何在迭代时从地图中删除？喜欢：

std::map<K, V> map;
for(auto i : map)
    if(needs_removing(i))
        // remove it from the map

如果我使用map.erase它将使迭代器无效

标准的关联容器擦除习惯用法：

for (auto it = m.cbegin(); it != m.cend() /* not hoisted */; /* no increment */)
{
  if (must_delete)
  {
    m.erase(it++);    // or "it = m.erase(it)" since C++11
  }
  else
  {
    ++it;
  }
}

请注意for，由于要修改容器本身，因此我们确实希望在此处进行普通循环。基于范围的循环应严格保留给我们只关心元素的情况。RBFL的语法甚至没有将容器暴露在循环体内，从而使这一点很清楚。

编辑。在C ++ 11之前的版本中，您无法删除构造函数。在那里你不得不说：

for (std::map<K,V>::iterator it = m.begin(); it != m.end(); ) { /* ... */ }

从容器中删除元素与元素的恒定性并不矛盾。通过类比，它一直是完全合法delete p这里p是一个指针到常量。坚固性不会限制寿命。C ++中的const值仍然可以停止存在。

我个人更喜欢这种模式，它更清晰，更简单，但会增加一个变量：

for (auto it = m.cbegin(), next_it = it; it != m.cend(); it = next_it)
{
  ++next_it;
  if (must_delete)
  {
    m.erase(it);
  }
}

这种方法的优点：

• for循环增量器作为增量器是有意义的；
• 擦除操作是简单的擦除，而不是与增量逻辑混合在一起；
• 在循环体的第一行之后，it和的含义next_it在整个迭代过程中保持不变，从而使您可以轻松地添加引用它们的附加语句，而不必担心它们是否会按预期工作（当然，除非it删除后不能使用） 。

简而言之，“如何在迭代时从地图中删除它？”

• 使用旧地图暗示：您不能
• 带有新的地图隐含内容：几乎与@KerrekSB建议的一样。但是他发布的内容中存在一些语法问题。

从GCC映射中显示（请注意GXX_EXPERIMENTAL_CXX0X）：

#ifdef __GXX_EXPERIMENTAL_CXX0X__
      // _GLIBCXX_RESOLVE_LIB_DEFECTS
      // DR 130. Associative erase should return an iterator.
      /**
       *  @brief Erases an element from a %map.
       *  @param  position  An iterator pointing to the element to be erased.
       *  @return An iterator pointing to the element immediately following
       *          @a position prior to the element being erased. If no such 
       *          element exists, end() is returned.
       *
       *  This function erases an element, pointed to by the given
       *  iterator, from a %map.  Note that this function only erases
       *  the element, and that if the element is itself a pointer,
       *  the pointed-to memory is not touched in any way.  Managing
       *  the pointer is the user's responsibility.
       */
      iterator
      erase(iterator __position)
      { return _M_t.erase(__position); }
#else
      /**
       *  @brief Erases an element from a %map.
       *  @param  position  An iterator pointing to the element to be erased.
       *
       *  This function erases an element, pointed to by the given
       *  iterator, from a %map.  Note that this function only erases
       *  the element, and that if the element is itself a pointer,
       *  the pointed-to memory is not touched in any way.  Managing
       *  the pointer is the user's responsibility.
       */
      void
      erase(iterator __position)
      { _M_t.erase(__position); }
#endif

新旧样式示例：

#include <iostream>
#include <map>
#include <vector>
#include <algorithm>

using namespace std;
typedef map<int, int> t_myMap;
typedef vector<t_myMap::key_type>  t_myVec;

int main() {

    cout << "main() ENTRY" << endl;

    t_myMap mi;
    mi.insert(t_myMap::value_type(1,1));
    mi.insert(t_myMap::value_type(2,1));
    mi.insert(t_myMap::value_type(3,1));
    mi.insert(t_myMap::value_type(4,1));
    mi.insert(t_myMap::value_type(5,1));
    mi.insert(t_myMap::value_type(6,1));

    cout << "Init" << endl;
    for(t_myMap::const_iterator i = mi.begin(); i != mi.end(); i++)
        cout << '\t' << i->first << '-' << i->second << endl;

    t_myVec markedForDeath;

    for (t_myMap::const_iterator it = mi.begin(); it != mi.end() ; it++)
        if (it->first > 2 && it->first < 5)
            markedForDeath.push_back(it->first);

    for(size_t i = 0; i < markedForDeath.size(); i++)
        // old erase, returns void...
        mi.erase(markedForDeath[i]);

    cout << "after old style erase of 3 & 4.." << endl;
    for(t_myMap::const_iterator i = mi.begin(); i != mi.end(); i++)
        cout << '\t' << i->first << '-' << i->second << endl;

    for (auto it = mi.begin(); it != mi.end(); ) {
        if (it->first == 5)
            // new erase() that returns iter..
            it = mi.erase(it);
        else
            ++it;
    }

    cout << "after new style erase of 5" << endl;
    // new cend/cbegin and lambda..
    for_each(mi.cbegin(), mi.cend(), [](t_myMap::const_reference it){cout << '\t' << it.first << '-' << it.second << endl;});

    return 0;
}

印刷品：

main() ENTRY
Init
        1-1
        2-1
        3-1
        4-1
        5-1
        6-1
after old style erase of 3 & 4..
        1-1
        2-1
        5-1
        6-1
after new style erase of 5
        1-1
        2-1
        6-1

Process returned 0 (0x0)   execution time : 0.021 s
Press any key to continue.