Compare two std multimaps for same element

Posted on 2008-09-30
Last Modified: 2010-04-21
Is there a quick and dirty way to compare two std::multimap to find an identical element?
Question by:steenpat
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LVL 53

Expert Comment

ID: 22609807
>> to find an identical element?

Do you mean any identical element ? Or a specific one ?

Author Comment

ID: 22610432
I'm storing a map of directories and the files underneath them.
So if I have a map that contains [0] - "Files", "readme.txt"
and I have another map that contains the same item, I want to be able to find this out, so I know if the src directory and destination directory, files, etc are going to be the same and alert the user before copying files over.

Author Comment

ID: 22610443
But yeah any identical element will do.
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LVL 53

Accepted Solution

Infinity08 earned 250 total points
ID: 22612434
>> But yeah any identical element will do.

From your explanation, it seems you need all identical items, not just one (any).

To answer your question : I wouldn't use a multimap for this. Something like this might be more appropriate :

        typedef std::string Directory;
        typedef std::string File;
        typedef std::map<Directory, std::set<File> > FileList;

Below is some example code of how you could do it (compile and run the code).

You can also do it with multimaps if you want, but it'll be slightly more complicated.
#include <iostream>

#include <string>

#include <map>

#include <set>

#include <algorithm>

typedef std::string Directory;

typedef std::string File;

typedef std::map<Directory, std::set<File> > FileList;

size_t intersection(FileList &res, const FileList &fl1, const FileList &fl2) {

  size_t cnt = 0;


  FileList::const_iterator it, it2;

  for (it = fl1.begin(); it != fl1.end(); ++it) {

    if ((it2 = fl2.find(it->first)) != fl2.end()) {

      std::set<File>::const_iterator it3;

      for (it3 = it->second.begin(); it3 != it->second.end(); ++it3) {

        if (it2->second.find(*it3) != it2->second.end()) {







  return cnt;


int main(void) {

  FileList flist1, flist2;

















  FileList duplicates;

  size_t cnt = intersection(duplicates, flist1, flist2);


  std::cout << cnt << " duplicates were found :" << std::endl;

  FileList::const_iterator it;

  for (it = duplicates.begin(); it != duplicates.end(); ++it) {

    std::cout << "  -> directory " << it->first << " :" << std::endl;

    std::set<File>::const_iterator it2;

    for (it2 = it->second.begin(); it2 != it->second.end(); ++it2) {

      std::cout << "     * file " << *it2 << std::endl;



  return 0;


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LVL 39

Assisted Solution

itsmeandnobodyelse earned 250 total points
ID: 22614335
>>>> but it'll be slightly more complicated

You didn't tend to under-estimating before. Why now?   ;-)

steenpat,  the std::multimap is one of the less comfortable classes (you may also call them 'poor'). In your case it especially worse as you may have a big count of files mapped to one directory. For that a multi-map isn't suitable.

Directories and their files best were stored in a tree container. Microsoft STL has a _Tree class which is the basic data structure for map and set class. AFAIK it is not recommended to use it. The FileList Infinity posted above is a good alternative, though not very comfortable (and not fast) both for traversing or finding duplicates.

Maybe the following is an alternative:

typedef std::string Path;
typedef std::string Directory;
typedef std::string Name;

std::set<Path> allPaths;
std::map<Name, std::vector<Directory> > index;

You would store any full path in allPaths and build-up the index same time

    while (getNextFileOrFolder(parentPath, name))
         allPaths.insert(parentPath + "/" + name);

To check whether a given name has duplicates then is

   std::map<Name, std::vector<Directory> >::iterator f;

    // check if name was used more than once
   if ((f  = index.find(name)) != index.end() && f->size() > 1)
          std::cout << "folders containing " << name << ":" << std::endl;
          std::vector<Directory> v& = *f;
          for (int i = 0; i < (int)v.size(); ++i)
               std::cout << v[i] << std::endl;


Author Closing Comment

ID: 31501750
Both options influenced my final decision, so I split the points.

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