number of calls to hash function in unordered_set

[Frank Tetzel <s1445051 at mail dot zih dot tu-dresden dot de>]

Hi,

can anybody give me a reasonable explanation why the hash function for
a simple unordered_set is called twice as much as I would expect.

Example:


#include <iostream>
#include <unordered_set>

struct MyHash{
	static int numberOfCalls;

	size_t operator()(int i) const noexcept{
		++numberOfCalls;
		//return std::hash<int>{}(i);
		return i;
	}
};
int MyHash::numberOfCalls = 0;

int main(){
	std::unordered_set<int,MyHash> set;

	std::cout << MyHash::numberOfCalls << '\n';
	set.reserve(1024);
	std::cout << MyHash::numberOfCalls << '\n';
	for(int i=0; i<1024; ++i){
		set.emplace(i);
	}
	std::cout << MyHash::numberOfCalls << '\n';

	return 0;
}




output:
0
0
2047

It seems to always be (inserted_elements * 2) - 1.
When adding 2 elements, it hashes 3 times. When adding 1 element, it
only hashes once. What is happening here? Why is the hash calculated
twice? What is so special about the last or first element?

Btw, I'm running gcc 8.1.1.

Best regards,
Frank