unordered_map not unordered

[morando]

#include <iostream>
#include <unordered_map>
#include <string>

typedef std::unordered_map<std::string, int> mm;

int main()
{
	mm m;

	m.insert( mm::value_type( "one",   1 ) );
	m.insert( mm::value_type( "two",   2 ) );
	m.insert( mm::value_type( "three", 3 ) );
	m.insert( mm::value_type( "four",  4 ) );

	const auto& end = m.end();
	auto it         = m.begin();
	for(; it != end; ++it)
	{
		std::cout << (*it).second << std::endl;
	}

	std::cin.ignore();
	return 0;
}

expected output:

1
2
3
4

actual output:

3
1
2
4

WTF???

[webJose]

What compiler/implementation of unordered_map?

[morando]

VS2010 default std lib

EDIT: Just tried it with g++ 4.6.2 (Mingw) , output is different:

2
4
3
1

But i need unique elements and not sorted, how to achive this?

[naraku9333]

I'm fairly certain the output is due to the fact unordered_map is a hashed container. Different hashing algorithms, different output.

[Peter87]

std::unordered_map is unordered so don't expect any particular order.

[cire]

But i need unique elements and not sorted, how to achive this?

You have unique elements, and they're not sorted. I think you have achieved that.

If you want the order of insertion preserved, use a vector. You'll have to manually check for duplicates if there's a possibility of them existing, but it'll get the job done.