如何返回严格小于 std::multimap 中给定键的最大键？
multimap 提供了lower_bound 和upper_bound 方法。两者都可能返回一个迭代器，该迭代器的键值大于所需值，而 lower_bound 可能恰好产生所需值。
现在我想要一个值的迭代器，其中键严格小于所请求的值。如果它是 map 而不是 multimap，则实现起来相对简单，如下所述： 返回严格小于的最大密钥比 C++ 映射中的给定键。 但在 multimap 中，递减迭代器并不能保证使其指向严格较小的键。所以我需要反复递减，直到找到更小的密钥。不是特别好。
有没有更优雅的方法来做到这一点？
键通常是浮点型的。
My apologies, it turns out that you can actually do it with a single decrement. I just placed it wrong in my program, that was the real error.
multimap offers the methods lower_bound and upper_bound. Both may return an iterator to a value with key greater than the desired, with lower_bound possibly yielding exactly the desired.
Now I want an iterator to a value where the key is strictly less the requested. If it were a map rather than multimap, this would be relatively simple to achieve as described here:
Returning the greatest key strictly less than the given key in a C++ Map.
But in a multimap, decrementing an iterator is not guaranteed to make it point to a strictly smaller key. So I would need to decrement repeatedly, until a smaller key is found. Not particularly nice.
Is there a more elegant way of doing this?
The keys will in general be floating-point.
My apologies, it turns out that you can actually do it with a single decrement. I just placed it wrong in my program, that was the real error.
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AFAIK，lower/upper_bound 会将迭代器返回到该值的第一个元素，因此您可以减少它
AFAIK, lower/upper_bound will return iterator to FIRST element of this value, So you can decrease it
lower_bound指向大于或等于参数（或end）的最小元素。因此，将其递减一次即可得到所需的元素（如果存在）。lower_boundpoints to the smallest element greater than or equal to the argument (orend). Thus decrementing it once gives you the desired element (if it exists).