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Home/ Questions/Q 868293
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T10:06:55+00:00 2026-05-15T10:06:55+00:00

I have a boost::unordered_map, but it appears to be in order, giving me an

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I have a boost::unordered_map, but it appears to be in order, giving me an overwhelming feeling of “You’re Doing It Wrong”. Why is the output to this in order? I would’ve expected the underlying hashing algorithm to have randomized this order:

#include <iostream>
#include <boost/unordered_map.hpp>

int main()
{
    boost::unordered_map<int, int> im;

    for(int i = 0; i < 50; ++i)
    {
        im.insert(std::make_pair(i, i));
    }

    boost::unordered_map<int, int>::const_iterator i;

    for(i = im.begin(); i != im.end(); ++i)
    {
        std::cout << i->first << ", " << i->second << std::endl;
    }

    return 0;
}

…gives me…

0, 0
1, 1
2, 2
...
47, 47
48, 48
49, 49

Upon examination of boost’s source code:

inline std::size_t hash_value(int v)
{
    return static_cast<std::size_t>(v);
}

…which would explain it. The answers below hold the higher level thinking, as well, which I found useful.

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  1. Editorial Team
    Editorial Team
    2026-05-15T10:06:56+00:00Added an answer on May 15, 2026 at 10:06 am

    While I can’t speak to the boost internals as I’m not a C++ guy, I can propose a few higher-level questions that may alleviate your concerns:

    1) What are the guarantees of an “unordered” map? Say you have an ordered map, and you want to create a map that does not guarantee ordering. An initial implementation may simply use the ordered map. It’s almost never a problem to provide stronger guarantees than you advertise.

    2) A hash function is something that hashes X -> int. If you already have an integer, you could use the identity function. While it may not be the most efficient in all cases, it could explain the behavior you’re seeing.

    Basically, seeing behavior like this is not necessarily a problem.

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