[英]multimap upper_bound return confusion
I have a multimap filled with pairings. 我有一个填充配对的多图。 I want to iterate through a range.
我想迭代一个范围。 upper_bound does not return an iterator pointing to an element unless it find the first element whose key is greater than the value passed to upper_bound.
upper_bound不返回指向元素的迭代器,除非它找到第一个元素,其键大于传递给upper_bound的值。
How can I tell if there was no value returned from upper_bound because there's nothing greater than the passed value? 如何判断upper_bound是否没有返回值,因为没有任何值超过传递的值?
Thanks! 谢谢!
See reference and msdn 参见参考和msdn
upper_bound returns the first element that is greater than your key value, if there is no element then it will return the same as end()
upper_bound返回大于键值的第一个元素,如果没有元素,则返回与
end()
相同的元素
So you can just compare this against the end of your multimap 因此,您可以将其与多图的结尾进行比较
auto it = my_multi_map.upper_bound(some_val);
if (it == my_multi_map.end())
{
// iterator is pointing past end so no value found
}
At least if memory serves, if upper_bound doesn't find an element with a key larger than what you passed it, it'll return container.end()
. 至少如果内存服务,如果upper_bound没有找到一个键大于你传递的键的元素,它将返回
container.end()
。 This is fine for iterating with, as the norm for a pair of iterators is the beginning and one past the end of the range, so you can use container.end()
as the end of a range without any problems. 这对于迭代是很好的,因为一对迭代器的范数是范围的开始和一个结束,所以你可以使用
container.end()
作为范围的结尾而没有任何问题。
upper_bound
is actually perfect for iterating, because it points to the element just after the end of the range - just like end
does. upper_bound
实际上非常适合迭代,因为它指向范围结束后的元素 - 就像end
一样。 So you form your loop similarly. 所以你形成你的循环类似。
for (auto it = mymap.lower_bound(start_key), end = mymap.upper_bound(end_key); it != end; ++it)
If you only want a single key, pass the same value to lower_bound
and upper_bound
, or use equal_range
to get both at once. 如果您只需要一个键,则将相同的值传递给
lower_bound
和upper_bound
,或使用equal_range
同时获取两个键。 If the value doesn't exist in the map at all, lower_bound
and upper_bound
will be equal and the loop won't execute. 如果该值根本不存在于map中,则
lower_bound
和upper_bound
将相等,并且循环将不会执行。
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