如果增加一个等于STL容器的结束迭代器的迭代器，会发生什么 [英] What happens if you increment an iterator that is equal to the end iterator of an STL container
问题描述
当迭代器指向向量的最后一个元素时，如果将迭代器增加2，该怎么办？在这个问题中询问如何通过以下方式调整迭代器到STL容器2个元素提供了两种不同的方法：
What if I increment an iterator by 2 when it points onto the last element of a vector? In this question asking how to adjust the iterator to an STL container by 2 elements two different approaches are offered:
- 使用算术运算符的形式 - + = 2或++两次
- 或者使用std :: advance（）
我用VC ++ 7测试了它们的两个边缘迭代器指向STL容器的最后一个元素或更高的情况：
I've tested both of them with VC++ 7 for the edge case when the iterator points onto the last element of the STL container or beyond:
vector<int> vec;
vec.push_back( 1 );
vec.push_back( 2 );
vector<int>::iterator it = vec.begin();
advance( it, 2 );
bool isAtEnd = it == vec.end(); // true
it++; // or advance( it, 1 ); - doesn't matter
isAtEnd = it == vec.end(); //false
it = vec.begin();
advance( it, 3 );
isAtEnd = it == vec.end(); // false
我见过可能会建议比较vector :: end（）时遍历向量和其他容器：
I've seen may times an advise to compare against vector::end() when traversing the vector and other containers:
for( vector<int>::iterator it = vec.begin(); it != vec.end(); it++ ) {
//manipulate the element through the iterator here
}
显然，如果迭代器超出循环内的最后一个元素，则for循环语句中的比较将计算为false，并且循环将很快继续进入未定义的行为。
Obviously if the iterator is advanced past the last element inside the loop the comparison in the for-loop statement will evaluate to false and the loop will happily continue into undefined behaviour.
我是否认为如果我在迭代器上使用advance（）或任何类型的递增操作并使其指向容器的末尾，我将无法检测到这种情况？如果是这样，最佳做法是什么 - 不使用这些进步？
Do I get it right that if I ever use advance() or any kind of increment operation on an iterator and make it point past the container's end I will be unable to detect this situation? If so, what is the best practice - not to use such advancements?
推荐答案
以下是Nicolai Josuttis的书：
Following is the quote from Nicolai Josuttis book:
请注意，advance（）不会检查
它是否越过
序列的end（）（它可以检查是因为
迭代器通常不知道它们运行的
容器。
因此，调用此函数可能会导致
导致未定义的行为，因为
调用operator ++作为
序列的末尾未定义
Note that advance() does not check whether it crosses the end() of a sequence (it can't check because iterators in general do not know the containers on which they operate). Thus, calling this function might result in undefined behavior because calling operator ++ for the end of a sequence is not defined
换句话说，在范围内维护迭代器的责任完全在于调用者。
In other words, the responsibility of maintaining the iterator within the range lies totally with the caller.
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