是否可以从尾到头迭代一个向量?
Is it possible to iterate a vector from the end to the beginning?
for (vector<my_class>::iterator i = my_vector.end();
i != my_vector.begin(); /* ?! */ ) {
}
或者只有这样的事情才有可能:
Or is that only possible with something like that:
for (int i = my_vector.size() - 1; i >= 0; --i) {
}
一种方法是:
for (vector<my_class>::reverse_iterator i = my_vector.rbegin();
i != my_vector.rend(); ++i ) {
}
rbegin()
/rend()
专为此目的而设计.(是的,增加 reverse_interator
会将其向后移动.)
rbegin()
/rend()
were especially designed for that purpose. (And yes, incrementing a reverse_interator
moves it backward.)
现在,理论上,您的方法(使用 begin()
/end()
& --i
)会起作用,std::vector
的迭代器是双向的,但请记住,end()
不是最后一个元素 - 它是最后一个元素之后的一个,因此您必须递减首先,当你到达 begin()
时你就完成了——但你仍然需要做你的处理.
Now, in theory, your method (using begin()
/end()
& --i
) would work, std::vector
's iterator being bidirectional, but remember, end()
isn't the last element — it's one beyond the last element, so you'd have to decrement first, and you are done when you reach begin()
— but you still have to do your processing.
vector<my_class>::iterator i = my_vector.end();
while (i != my_vector.begin())
{
--i;
/*do stuff */
}
更新:在将 for()
循环重写为 while()
循环时,我显然过于激进.(重要的部分是 --i
在开头.)
UPDATE: I was apparently too aggressive in re-writing the for()
loop into a while()
loop. (The important part is that the --i
is at the beginning.)
这篇关于从头到尾迭代C++向量的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持跟版网!