C++ 11 将自定义 hash function 传递到 unordered_map

Question

I am not sure I understand why this compiles.

#include <iostream>
#include <iterator>
#include <string>
#include <unordered_map>
#include <utility>
#include <vector>
#include <unordered_set>
#include <algorithm>
#include <queue>
using namespace std;



void thing() {
    auto hash = [](const std::pair<int, int> &pair) {
        return std::hash<string>()(to_string(pair.first));
    };
    auto map = std::unordered_map<std::pair<int, int>, int, decltype(hash)>();
}


int main() {

}

I haven't passed the hash function into the constructor -- how does it know the implementation to hash a pair?

Answer 1

It may compile because the lambda expression's type is an anonymous class - which can be instantiated (but see note below). The code block is that class' operator()() method. This class is instantiated as a default parameter value by the constructor std::unordered_map - using the lambda expression's type, which is a template argument:

unordered_map( InputIt first, InputIt last,
               size_type bucket_count = /*implementation-defined*/,
               const Hash& hash = Hash(), /* <<<< INSTANTIATION HAPPENS HERE */
               const key_equal& equal = key_equal(),
               const Allocator& alloc = Allocator() );

For more on the nature of lambda expressions, see this SO question:

What is a lambda expression in C++11?

---

Note: As commenters note, your code doesn't actually compile with C++11 , because the implicit default constructor for the anonymous class defined by a lambda expression is deleted in C++11. This has been changed in C++20: Non-capturing lambdas get a default constructor, so now the code does compile.