Map two integers to another integer [closed] - c++

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I have two integers and I want to create a hash map which maps the two integers to another integer.
In order to so I created the following program, however it is giving me errors:
#include <iostream>
#include <unordered_map>
using namespace std;
int main()
{
std::pair <int, int> var1;
var1=std::make_pair(10,20);
cout<<"\n var1.f="<<var1.first<<"\t 2."<<var1.second<<"\n";
std::unordered_map <std::pair <int,int>, int> yeah;
return 0;
}
Is there some way to get rid of the errors?
Is there some way to do the same in a scalable manner?
Error:
n file included from /usr/local/gcc-4.8.1/include/c++/4.8.1/bits/hashtable.h:35:0,
from /usr/local/gcc-4.8.1/include/c++/4.8.1/unordered_map:47,
from main.cpp:2:
/usr/local/gcc-4.8.1/include/c++/4.8.1/bits/hashtable_policy.h: In instantiation of ‘struct std::__detail::_Hash_code_base<std::pair<int, int>, std::pair<const std::pair<int, int>, int>, std::__detail::_Select1st, std::hash<std::pair<int, int> >, std::__detail::_Mod_range_hashing, std::__detail::_Default_ranged_hash, true>’:
/usr/local/gcc-4.8.1/include/c++/4.8.1/bits/hashtable_policy.h:1402:10: required from ‘struct std::__detail::_Hashtable_base<std::pair<int, int>, std::pair<const std::pair<int, int>, int>, std::__detail::_Select1st, std::equal_to<std::pair<int, int> >, std::hash<std::pair<int, int> >, std::__detail::_Mod_range_hashing, std::__detail::_Default_ranged_hash, std::__detail::_Hashtable_traits<true, false, true> >’
/usr/local/gcc-4.8.1/include/c++/4.8.1/bits/hashtable.h:174:11: required from ‘class std::_Hashtable<std::pair<int, int>, std::pair<const std::pair<int, int>, int>, std::allocator<std::pair<const std::pair<int, int>, int> >, std::__detail::_Select1st, std::equal_to<std::pair<int, int> >, std::hash<std::pair<int, int> >, std::__detail::_Mod_range_hashing, std::__detail::_Default_ranged_hash, std::__detail::_Prime_rehash_policy, std::__detail::_Hashtable_traits<true, false, true> >’
/usr/local/gcc-4.8.1/include/c++/4.8.1/bits/unordered_map.h:100:18: required from ‘class std::unordered_map<std::pair<int, int>, int>’
main.cpp:12:50: required from here
/usr/local/gcc-4.8.1/include/c++/4.8.1/bits/hashtable_policy.h:1070:12: error: invalid use of incomplete type ‘struct std::hash<std::pair<int, int> >’
struct _Hash_code_base<_Key, _Value, _ExtractKey, _H1, _H2,
^
In file included from /usr/local/gcc-4.8.1/include/c++/4.8.1/bits/basic_string.h:3033:0,
from /usr/local/gcc-4.8.1/include/c++/4.8.1/string:52,
from /usr/local/gcc-4.8.1/include/c++/4.8.1/bits/locale_classes.h:40,
from /usr/local/gcc-4.8.1/include/c++/4.8.1/bits/ios_base.h:41,
from /usr/local/gcc-4.8.1/include/c++/4.8.1/ios:42,
from /usr/local/gcc-4.8.1/include/c++/4.8.1/ostream:38,
from /usr/local/gcc-4.8.1/include/c++/4.8.1/iostream:39,
from main.cpp:1:
/usr/local/gcc-4.8.1/include/c++/4.8.1/bits/functional_hash.h:58:12: error: declaration of ‘struct std::hash<std::pair<int, int> >’
struct hash;
^
In file included from /usr/local/gcc-4.8.1/include/c++/4.8.1/bits/hashtable.h:35:0,
from /usr/local/gcc-4.8.1/include/c++/4.8.1/unordered_map:47,
from main.cpp:2:

Answer 1

I assume the error is something like "The C++ Standard doesn't provide a hash for this type"? The problem is that anything you use as the key in an unordered map must be hashable, and there is no built-in hash function for pair<int, int>. A couple possibilities are:
-Write your own hasher for pair<int, int> and use it as the third template parameter to the unordered_map. This post may be able to help with that.
-Use std::unordered_map<int, std::unordered_map<int, int>> instead. Be careful not to accidentally create copies of the inner map when you're performing operations.
-Use map instead of unordered_map. This will give you logarithmic operations instead of constant-time, but unless the map is huge and you're doing tons of lookups per second, you won't notice the difference.
Edit in response to your edit: Yes, that's what those error messages amount to, though they don't say it in quite as friendly a fashion as Microsoft's compiler does. :)