C++ 向量迭代器在 for 循环中不可解引用
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Vector iterator not dereferencable in for loop
提问by TPOT94
I'm using a loop to count how many times that a word was entered then print the word and how many times it was entered, which works but it never prints the last word, I have it sorted alphabetically. Before the last word is printed it errors out saying the iterator is not dereferencable. Here is my code for the loop:
我正在使用循环来计算输入单词的次数，然后打印该单词以及输入的次数，这可行，但它从不打印最后一个单词，我按字母顺序对其进行了排序。在打印最后一个单词之前，它会出错说迭代器不可解引用。这是我的循环代码：
for (vector<string>::iterator it = v.begin() ; it != v.end(); ++it)
{
if (*it == *(it+1))
{
count++;
}
else if (*it != *(it+1))
{
count++;
cout << *it << " ---- " << count << endl;
count=0;
}
}
回答by billz
Your code has undefined behavior - imagine itis pointing to the last element of v, then you are trying to dereference v.end()in *(it+1)
你的代码有不确定的行为-想象it指向的最后一个元素v，那么您要提领v.end()的*(it+1)
if (*it != *(it+1)
STL iterator, end doesn't point to last element; end() returns an iterator that represents the end of the elements in the container. The end is the position behindthe last element. Such an iterator is also called a past-the-end iterator.
STL迭代器，end不指向最后一个元素；end() 返回一个迭代器，表示容器中元素的结尾。结尾是最后一个元素后面的位置。这样的迭代器也称为过去的迭代器。
Thus, begin() and end() define a half-openrange that includes the first element but excludesthe last
因此，begin() 和 end() 定义了一个包含第一个元素但不包括最后一个元素的半开范围
--------------------------------
| | | | | | | | |
--------------------------------
/\ /\
begin() end()
For what you are trying to achieve, have a look at std::adjacent_find
对于您想要实现的目标，请查看std::adjacent_find
auto it = std::adjacent_find(v.begin(), v.end());
if (it != v.end())
{
count ++;
}
else
{
cout << *it << " ---- " << count << endl;
}
回答by Jarod42
when it == v.end() - 1, you deference (it+1)so v.end(),
and deference v.end()is undefined behaviour.
when it == v.end() - 1，你尊重(it+1)so v.end()，并且尊重v.end()是未定义的行为。
回答by Gorpik
When you are at the last word and try to execute:
当你说到最后一句话并尝试执行时：
if (*it == *(it+1))
if (*it == *(it+1))
it+1is pointing at v.end(), which is a valid iterator, but is not derefernceable. Hence the error.
it+1指向v.end()，这是一个有效的迭代器，但不可取消引用。因此错误。
回答by Neil Kirk
When it is one before the end iterator, you have a problem here: *(it+1)as this attempts to dereference the end iterator, which is invalid.
当它在结束迭代器之前是 1 时，这里有一个问题：*(it+1)因为这试图取消引用结束迭代器，这是无效的。
I'm not sure what you want your logic to do in this case, but you can check for this with if (it+1 != v.end())before doing your stuff.
我不确定在这种情况下你希望你的逻辑做什么，但你可以if (it+1 != v.end())在做你的事情之前检查这个。
回答by hate-engine
Because when itis near end, it+1is at end, and you trying to dereference it in ifoperator.
因为何时it接近结束，it+1在结束时，您试图在if运算符中取消引用它。
