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?
Asked
Active
Viewed 1.7k times
1 Answers
55
#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);
});
}
-
9Or, write a make_unordered_hash templated function for type deduction. There is also actually a C++ proposal to more generally address this problem http://isocpp.org/files/papers/n3602.html – mmocny Apr 09 '13 at 14:05