#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.
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
:
#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);
});
}
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