Re: set::find with custom equality operator==

From:
tohava <tohava@gmail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Wed, 3 Feb 2010 08:49:18 CST
Message-ID:
<1d0e376f-f987-4101-af7b-3f1947cf5578@q4g2000yqm.googlegroups.com>
On Feb 3, 1:12 am, dragoncoder <pktiw...@gmail.com> wrote:

Hi all, I have the following code.

// db_operators.h

#ifndef INCLUDED_DB_OPERATORS
#define INCLUDED_DB_OPERATORS

#include <cstring>

struct ecodb2{
   unsigned long long id;
   char code[21];

};

inline bool operator==(const ecodb2& lhs,
                        const ecodb2& rhs)
{
     std::cout << "Some text" << std::endl;
     return ((lhs.id == rhs.id) || !std::strcmp(lhs.code, rhs.code));

}

struct Compare
{
     bool operator()(const ecodb2& lhs,
                     const ecodb2& rhs) const
     {
         bool retval = false;
         if(lhs.id < rhs.id)
             retval = true;
         else if(lhs.id == rhs.id)
             retval = std::strcmp(lhs.code, rhs.code);
         return retval;
     }

};

#endif

// main.cpp

#include <iostream>
#include <cstring>
#include <set>

#include <db_operators.h>

int main()
{
     std::set<ecodb2, Compare> industry_codes;

     ecodb2 val;
     val.id = 1;

     std::strncpy(val.code, "GOV", sizeof(val.code));
     industry_codes.insert(val);

     std::strncpy(val.code, "ABC", sizeof(val.code));

     std::set<ecodb2>::iterator it = industry_codes.find(val);
     if(it != industry_codes.end())
         std::cout << "Found in set" << std::endl;
     else
         std::cout << "Not found in set" << std::endl;

}

While running this code, I expect to see Found in set (as either id or
code match meaning a perfect match) and also "Some text" being
printed, becasue set::find should call my custom operator==() function
to find a match but the output I get is "Not found in set" and also
"Some text" is not getting printed.


I don't think it should be possible for 2 variables
of type ecodb2 to be satisfy both x < y and x == y.
This can happen with your code (in the case
where x and y share an id but not a code). My guess
would be that find checks at some point for some x and y
whether x < y, and if this is the case, it assumes that x != y,
which in your example is NOT the case.

I would recommend implementing this using two maps,
one with id as key, and one with code.

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