Question

For example map1 is gaving values 1 to 10 with some address(begin to end).
i want to have values 10 to 1 with corresponding address in map2(begin to end)

map<long , int* > v;
map<long , int* > rv;

int i,a[10];
for(i=0; i<10; i++)
{
a[i] = i+1;
v.insert(pair<long, int *>(i+1,&a[i]));
}
itr = v.begin();
while(itr != v.end())
{
 cout << itr->first << " "<<itr->second;
 cout << endl;
 itr++;
}
rv.insert(v.rbegin(),v.rend());
cout << "copied array: "<<endl;
itr = rv.begin();
while(itr != rv.end())
{
cout << itr->first << " "<<itr->second;
cout << endl;
itr++;
 }

i tried above one but am getting values 1 to 10 only..my expected values 10 to 1.
please help me to find out….

Answer 1

STL map is an ordered container. The order of items that you get during the iteration is independent of the order in which you insert the items into the container.

The order of iteration is determined by two things:

• The value of the key, and
• The Compare class passed as the template parameter to the map

You can iterate the map in reverse order (your code snippet shows that you already know how it is done). The performance penalty for reverse-iterating a map, if any, is negligible. You can also provide a non-default Compare (std::greater<long> instead of the default std::less<long>) to have the default order of iteration altered.