behaviour of binary predicate for stl map and multimap.... - c++

Question

I have the following code:
#include <iostream>
#include <stdio.h>
#include <cmath>
#include <map>
using namespace std;
struct vals
{
int cods[5];
int sz;
};
struct myComp
{
bool operator()(vals A, vals B) const
{
int i=0;
while(A.cods[i]==B.cods[i] && i<A.sz)
i++;
if(i==A.sz)
return false; //<-----this is the value im changing..
else
return A.cods[i] > B.cods[i];
}
};
map< vals, int, myComp> Mp;
int main()
{
vals g, h;
g.sz=h.sz=3;
g.cods[0] = 12;
g.cods[1] = 22;
g.cods[2] = 32;
Mp.insert(pair< vals, int >(g,4));
Mp.insert(pair< vals, int >(g,7));
cout<<Mp.count(g)<<endl;
cout<<Mp.size()<<endl;
return 0;
}
Now, when declare Mp as map and put false in the binary predicate..
The output is:
1
1
Mp => map && binary predicate:true ==> output: 0 2
Mp => multimap && binary predicate:true ===> output: 0 2
Mp => multimap && binary predicate:false ===> output: 2 2
I thought that the return value of predicate just tells the stl whether to put an element either infront of it or behind it. But I don't get how does this effect the size of the map itself..
Please throw some light on this. Thank you.

Answer 1

Your comparison must implement a strict weak ordering. This requirement is not met when you use
if(i==A.sz)
return true;
in your comparator. In this case, all the elements in the arrays the same. The functor cannot return true if both its arguments are equal. The map cannot function correctly if you do not have a strict weak ordering comparison.
You can greatly simplify your functor by using std::lexicographical_compare:
#include <algorithm> // for std::lexicographical_compare
#include <functional> // for std::greater
...
bool operator()(vals A, vals B) const
{
return std::lexicographical_compare(A, A+A.sz, B, B+B.sz); // less-than
//return std::lexicographical_compare(A, A+A.sz, B, B+B.sz, std::greater<int>()); // gt
}