std :: hash是否保证等于“相等”的哈希值？浮点数字？ [英] Does std::hash guarantee equal hashes for "equal" floating point numbers?

问题描述

是浮点特化的 std :: hash （比如，对于 double s或 float s）关于 几乎平等 ？也就是说，如果两个值（例如（1./std::sqrt（5。）/ std :: sqrt（5。））和 .2 ）应该比较相等，但不会对 == 运算符这样做， std :: hash behave？

那么，我可以依靠 double 作为 std :: unordered_map 键可按预期工作？

---

散列浮点值，但要求提高;我在问C ++ 11的保证。

解决方案

std :: hash 对所有可以实例化
的类型都有同样的保证：如果两个对象相等，它们的哈希码将是
相等。否则，他们
不会有很大的可能性。因此，您可以依靠 double 作为
unordered_map 中的键来按预期工作：if two双打不是
等于（如 == 所定义），它们可能会有不同的
散列（即使它们不是，它们因为
unordered_map 也检查是否相等）。

显然，如果你的值是不准确的
计算的结果，它们不适用于 unordered_map
（也可能不适用于任何地图）的适当键。

Is the floating point specialisation of std::hash (say, for doubles or floats) reliable regarding almost-equality? That is, if two values (such as (1./std::sqrt(5.)/std::sqrt(5.)) and .2) should compare equal but will not do so with the == operator, how will std::hash behave?

So, can I rely on a double as an std::unordered_map key to work as expected?

---

I have seen "Hashing floating point values" but that asks about boost; I'm asking about the C++11 guarantees.

解决方案

std::hash has same guarantees for all types over which it can be instantiated: if two objects are equal, their hash codes will be equal. Otherwise, there's a very large probability that they won't. So you can rely on a double as a key in an unordered_map to work as expected: if two doubles are not equal (as defined by ==), they will probably have a different hash (and even if they don't, they're different keys, because unordered_map also checks for equality).

Obviously, if your values are the results of inexact calculations, they aren't appropriate keys for unordered_map (nor perhaps for any map).

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