std::set - sorting order specified?

[Gernot Frisch]

Hi,

is the sorting order of std::set / std::map defined? i.e. is the
*begin() of std::set<int> always smaller than *(--end()) ?

 

[Rolf Magnus]

Hi,

is the sorting order of std::set / std::map defined?

i.e. is the *begin() of std::set<int> always smaller than *(--end()) ?

No. --end() shoudn't work at all.

 

[Ivan Vecerina]

: Gernot Frisch wrote:
: > i.e. is the *begin() of std::set<int> always smaller than *(--end()) ?
:
: No. --end() shoudn't work at all.

But it will, for UDT with a defined prefix operator--.
Just not for raw pointers...

Ivan

 

[Neelesh Bodas]

No. --end() shoudn't work at all.

This code compiles and runs well :

#include <iostream>
#include <set>
using namespace std;

int main()
{
set<int> a;
a.insert(1);
a.insert(2);
a.insert(3);
cout << (*(--a.end())) << endl;
}

Am I missing something?

 

[Gernot Frisch]

This code compiles and runs well :

#include <iostream>
#include <set>
using namespace std;

int main()
{
set<int> a;
a.insert(1);
a.insert(2);
a.insert(3);
cout << (*(--a.end())) << endl;
}

Why shouldn't it work? end() is an iterator, which has an -- operator,
that returns an iterator, which has an * operator returning a
reference to the item.