c++ - 为什么在隐式共享容器上使用迭代器时必须先复制？

Question

Qt 的文档说明如下:

Thanks to implicit sharing, it is very inexpensive for a function to return a container per value. The Qt API contains dozens of functions that return a QList or QStringList per value (e.g., QSplitter::sizes()). If you want to iterate over these using an STL iterator, you should always take a copy of the container and iterate over the copy. For example:

// RIGHT
const QList<int> sizes = splitter->sizes();
QList<int>::const_iterator i;
for (i = sizes.begin(); i != sizes.end(); ++i)
    ...

// WRONG
QList<int>::const_iterator i;
for (i = splitter->sizes().begin();
        i != splitter->sizes().end(); ++i)
    ...

如果应用“错误”方法会发生什么？

Answer 1

对 splitter->sizes() 的两次调用产生了容器的两个不同的拷贝。由于 begin() 来自一个，而 end() 来自另一个，因此它们不构成有效范围。然后循环会离开第一个容器的末尾，进入未定义行为的领域。

虽然基于范围的循环会工作得很好:for (int size: splitter->sizes()) { ... }