如何在unordered_map中使用lambda函数作为哈希函数?

I wonder if it is possible to use lambda function as custom hash function for unordered_map in C++11? If so, what is the syntax?

我想知道是否可以在c++ 11中使用lambda函数作为unordered_map的自定义哈希函数?如果是，语法是什么?

1 个解决方案

#1

#include<unordered_map>
#include<string>

int main() {
    auto my_hash = [](std::string const& foo) {
        return std::hash<std::string>()(foo);
    };

    std::unordered_map<std::string, int, decltype(my_hash)> my_map(10, my_hash); 
}

You need to pass lambda object to unordered_map constructor, since lambda types are not default constructible.

您需要将lambda对象传递给unordered_map构造函数，因为lambda类型不是默认的可构造的。

As @mmocny suggested in comment, it's also possible to define make function to enable type deduction if you really want to get rid of decltype:

正如@mmocny在评论中建议的，如果你真的想要摆脱decltype，也可以定义make函数来启用类型演绎:

#include<unordered_map>
#include<string>

template<
        class Key,
        class T,
        class Hash = std::hash<Key>
        // skipped EqualTo and Allocator for simplicity
>
std::unordered_map<Key, T, Hash> make_unordered_map(
        typename std::unordered_map<Key, T, Hash>::size_type bucket_count = 10,
        const Hash& hash = Hash()) {
    return std::unordered_map<Key, T, Hash>(bucket_count, hash);
}

int main() {
    auto my_map = make_unordered_map<std::string, int>(10,
            [](std::string const& foo) {
                return std::hash<std::string>()(foo);
            });
}

#1

#include<unordered_map>
#include<string>

int main() {
    auto my_hash = [](std::string const& foo) {
        return std::hash<std::string>()(foo);
    };

    std::unordered_map<std::string, int, decltype(my_hash)> my_map(10, my_hash); 
}