对于我使用的任何 STL 容器，如果我使用迭代器的默认构造函数声明一个迭代器(该特定容器类型的)，迭代器将被初始化为什么?
For any STL container that I'm using, if I declare an iterator (of this particular container type) using the iterator's default constructor, what will the iterator be initialised to?
例如，我有:
std::list<void*> address_list;
std::list<void*>::iterator iter;
iter 将被初始化为什么?
What will iter be initialised to?
约定为容器的NULL迭代器"，用于表示无结果，比较等于container.end().
By convention a "NULL iterator" for containers, which is used to indicate no result, compares equal to the result of container.end().
std::vector<X>::iterator iter = std::find(my_vec.begin(), my_vec.end(), x);
if (iter == my_vec.end()) {
//no result found; iter points to "nothing"
}
然而，由于默认构造的容器迭代器不与任何特定容器相关联，因此它没有任何价值.因此，它只是一个未初始化的变量，唯一合法的操作是为其分配一个有效的迭代器.
However, since a default-constructed container iterator is not associated with any particular container, there is no good value it could take. Therefore it is just an uninitialized variable and the only legal operation to do with it is to assign a valid iterator to it.
std::vector<X>::iterator iter; //no particular value
iter = some_vector.begin(); //iter is now usable
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对于其他类型的迭代器，这可能不是真的.例如，在 istream_iterator 的情况下，默认构造的迭代器表示(比较等于)一个 istream_iterator 已经达到输入流的 EOF.
For other kinds of iterators this might not be true. E.g in case of istream_iterator, a default-constructed iterator represents (compares equal to) an istream_iterator which has reached the EOF of an input stream.
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