如何从QList中删除元素，同时使用foreach迭代它？

问题描述：

我是Qt的新手，试图学习习语。

I'm new to Qt and trying to learn the idioms.

foreach 文档说：

Qt在进入foreach循环时自动获取容器的副本。如果你在迭代时修改容器，这不会影响循环。

Qt automatically takes a copy of the container when it enters a foreach loop. If you modify the container as you are iterating, that won't affect the loop.

但它不说如何在使用 foreach 进行迭代时删除元素。我最好的猜测是：

But it doesn't say how to remove an element while iterating with foreach. My best guess is something like:

int idx = 0;
foreach (const Foo &foo, fooList) {
  if (bad(foo)) {
    fooList.removeAt(idx);
  }
  ++idx;
}

看起来很丑陋，必须有 idx 外部循环（并且必须保持一个单独的循环计数器）

Seems ugly to have to scope idx outside the loop (and to have to maintain a separate loop counter at all).

~~此外，我知道 foreach 生成 QList 的一个副本，它很便宜，但是当我删除一个元素时会发生什么 - 仍然很便宜，昂贵的复制副本正在进行？~~是的，深度复制发生。

~~Also, I know that foreach makes a copy of the QList, which is cheap, but what happens once I remove an element -- is that still cheap or is there an expensive copy-on-modify going on?~~ Yes, deep copy happens.

编辑：这似乎并不像惯用的Qt。

EDIT : This doesn't seem like idiomatic Qt either.

for (int idx = 0; idx < fooList.size(); ) {
  const Foo &foo = fooList[idx];
  if (bad(foo)) {
    fooList.removeAt(idx);
  }
  else ++idx;
}

答

你应该更好地使用迭代器：

// Remove all odd numbers from a QList<int> 
QMutableListIterator<int> i(list);
while (i.hasNext()) {
    if (i.next() % 2 != 0)
        i.remove();
}

如何从QList中删除元素，同时使用foreach迭代它？

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