c ++ 11 std :: hash函数对象类的线程安全 [英] c++11 std::hash function object classes thread safety

问题描述

在c ++ 11中，在< functional> 线程中声明的哈希函数类对象是否安全？例如，从多个线程调用此函数是否安全？

In c++11 are the hash function class objects declared in <functional> thread safe? E.g., is it safe to call this function from multiple threads?

size_t hash1(const std::string& s) {
    std::hash<std::string> str_hash;
    return str_hash(s); 
}

或者，如果有一个全局对象 std： ：hash< std :: string> str_hash_global; ，那么从多个线程调用第二个函数是否安全？

or, if one has a global object std::hash<std::string> str_hash_global;, then is it safe to call this second function from multiple threads?

size_t hash2(const std::string& s) {
    return str_hash_global(s); 
}

推荐答案

标准库承诺您仅在标准库对象上调用 const 限定的成员函数，则标准库代码不会引起数据争用（参见[res。 on.data.races]）。

The standard library promises that if you only call const-qualified member functions on a standard library object, the standard library code does not cause a data race (cf. [res.on.data.races]).

标准模板 std :: hash 及其所有允许的模板专业化，以及任何满足 Hash 要求（[hash.requirements]）的用户提供的函子，都必须具有 const 合格的呼叫运营商，因此使用库提供的 std :: hash 专业化不应引起竞争。此外，由于[namespace.std]，程序提供的专业化必须满足相同的要求。

The standard template std::hash, as well as all its permissible specializations, as well as any user-provided functor that meets the Hash requirements ([hash.requirements]) must have a const-qualified call operator due to the requirements, and thus using the library-provided std::hash specializations should not cause a race. Moreover, due to [namespace.std], program-provided specializations must meet the same requirements.

最后，我想您通常会使用递归地吸引 const 调用：如果多个线程同时在地图中查找值，则它们必须使用地图的 const

Finally, I imagine that you would usually use the race-freeness guarantees by recursively appealing to const calls: If you multiple threads concurrently look up values in a map, they have to use the map's const interface to invoke the above library rule, but then the map only gets to use a constant value of the hasher (or a private copy), and so it can itself only perform race-free hash computations. Whether a racy non-const call operator exists is immeterial at that point.

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