Why isn't my std::set sorted? - c++

Question

I have a class to store data that looks like this:
class DataLine
{
public:
std::string name;
boost::posix_time::time_duration time;
double x, y, z;
DataLine(std::string _name, boost::posix_time::time_duration _time, double _x,
double _y, double _z); //assign all these, not going to do it here
bool operator < (DataLine* dataLine) { return time < dataLine->time; }
}
Then I read in a bunch of data and .insert it into a std::set of the objects:
std::set<DataLine*> data;
data.insert( new DataLine(newname, newtime, newx, newy, newz) );
//...insert all data - IS OUT OF ORDER HERE
Then I run through my data and do stuff with it while appending new elements to the set.
boost::posix_time::time_duration machineTime(0,0,0);
for(std::set<DataLine*>::reverse_iterator it = data.rbegin(); it != data.rend(); ++it)
{
if(machineTime < (*it)->time)
{
machineTime = (*it)->time;
}
machineTime += processDataLine(*it); //do stuff with data, might add to append list below
for(std::vector<AppendList*>::iterator iter = appendList.begin(); iter != appendList.end(); ++iter)
{
data.insert( new DataLine( (*iter)->name, machineTime,
(*iter)->x, (*iter)->y, (*iter)->z); );
}
}
When I try to loop through the set of data both before and after inserting the elements all my data is out of order! Here are some times outputted when looped using
for(std::set<DataLine*>::iterator it = data.begin(); it != data.end(); ++it)
{
std::cout << std::endl << (*it)->time;
}
14:39:55.003001
14:39:55.003002
14:39:55.001000
14:39:59.122000
14:39:58.697000
14:39:57.576000
14:39:56.980000
Why aren't these times sorted in order?

Answer 1

It is sorted. It's sorted based on the data type you're storing in the set, which is a pointer to a DataLine. In other words, it'll sort according to the location in memory of your objects which is probably creation order (but may not be, depending on how the memory allocation functions work in your implementation).
If you want to sort based on the DataLine type itself, don't use a pointer. Store the objects themselves.
You can see a similar effect from the following code which creates two sets. The first is a set of integer pointers, the second a set of actual integers:
#include <iostream>
#include <iomanip>
#include <set>
using namespace std;
int main (void) {
set<int*> ipset;
set<int> iset;
cout << "inserting: ";
for (int i = 0; i < 10; i++) {
int val = (i * 7) % 13;
cout << ' ' << setw(2) << val;
ipset.insert (new int (val));
iset.insert (val);
}
cout << '\n';
cout << "integer pointer set:";
for (set<int*>::iterator it = ipset.begin(); it != ipset.end(); ++it)
cout << ' ' << setw(2) << **it;
cout << '\n';
cout << "integer set: ";
for (set<int>::iterator it = iset.begin(); it != iset.end(); ++it)
cout << ' ' << setw(2) << *it;
cout << '\n';
cout << "integer pointer set pointers:\n";
for (set<int*>::iterator it = ipset.begin(); it != ipset.end(); ++it)
cout << " " << *it << '\n';
cout << '\n';
return 0;
}
When you run that code, you see something like:
inserting: 0 7 1 8 2 9 3 10 4 11
integer pointer set: 0 7 1 8 2 9 3 10 4 11
integer set: 0 1 2 3 4 7 8 9 10 11
integer pointer set pointers:
0x907c020
0x907c060
0x907c0a0
0x907c0e0
0x907c120
0x907c160
0x907c1a0
0x907c1e0
0x907c220
0x907c260
You can see the unordered way in which values are added to the two sets (first line) and the way the pointer set in this case matches the order of input (second line). That's because the addresses are what's being used for ordering as you can see by the fact that the final section shows the ordered addresses.
Although, as mentioned, it may not necessarily match the input order, since the memory arena may be somewhat fragmented (as one example).
The set containing the actual integers (as opposed to pointers to integers) is clearly sorted by the integer value itself (third line).

Answer 2

You need to define member operator < like below, and save objects in std::set instead of raw pointers. Because for raw pointers, the default comparision criteria is based on the pointer value itself.
bool operator < (const DataLine &dataLine) const
{
return time < dataLine.time;
}
...
std::set<DataLine> data;