How to get just first element of set in c++

Question

I am new to c++, I have declared set of sets:

std::set< std::set<int> > return_moves;

and also pushed some values in it. I want to access the first element of this set so that I can count a number of elements in that inner set. I am trying to get it by

return_moves.begin().size()

I am getting this error:

set_diff.cpp: In function ‘int main()’:
set_diff.cpp:62:47: error: ‘std::set<std::set<int> >::iterator {aka struct std::_Rb_tree_const_iterator<std::set<int> >}’ has no member named ‘size’

Please help me to rectify my syntax.

Answer 1

Elaborating on Ed's comment, a set in C++ is by default weakly ordered, so there is some guarantee to the iteration ordering, but as a mathematical concept there is no "order" in a set, and hence implementations do not usually allow obtaining a "first" element.

If you want just to obtain any element, without iterating you can use

auto someElementIterator = myset.begin()

which will return an iterator to some element. To "pop" it you can follow it with

myset.erase(someElementIterator)

and you can get the element using *someElementIterator.

Regarding your error:

An iterator is in some some sense like a pointer, you need -> not . To access the element size (rather than the iterator size, which doesn't exist). so:

someElementIterator->size()

and not someElementIterator.size().

Answer 2

I'm late sorry, i just passed and found the solution for next developpers.
To get the first element of the std::set you could just use :

std::set< std::set<int> > return_moves;
auto oneMove = *(return_moves.begin()); // will return the first set<int>

oneMove.size(); // will return the size of the first set<int>

Because return_moves.begin() return the iterator of the first element of the set, and by adding the * we will recive the value of the first element.