Exemplo n.º 1
0
Arquivo: impl.cpp Projeto: Lev-T/tasks
 bool uTest( UnitTest *utest_p)
 {
     DList<int> list;
     list.push_back(5);
     list.push_back(10);
     Unit *u = list.first();
     
     UTEST_CHECK( utest_p, u);
     UTEST_CHECK( utest_p, list.size() == 2);
     
     int& f = u->val();
     f = 15;
     
     int sum = 0;
     for ( Unit* e = list.first (); e!= 0; e = e->next() )
     {
         sum += e->val();
     }
     
     UTEST_CHECK( utest_p, sum == 25);
     
     u = list.first();
     
     UTEST_CHECK( utest_p, u->next() == list.last() );
     UTEST_CHECK( utest_p, u->val() == 15);
     
     Unit * last = list.erase( u);
     
     UTEST_CHECK( utest_p, last->next() == 0);
     UTEST_CHECK( utest_p, last->prev() == 0);
     UTEST_CHECK( utest_p, last == list.first());
     UTEST_CHECK( utest_p, last == list.last());
     
     // Test clear
     list.clear();
     
     UTEST_CHECK( utest_p, list.size() == 0);
     UTEST_CHECK( utest_p, list.empty() );
     UTEST_CHECK( utest_p, list.first() == 0);
     UTEST_CHECK( utest_p, list.last() == 0);
     
     //--- Test insert and reverse
     list.push_front( 10);// list: 10
     list.push_front( 20);// list: 20 10
     list.push_front( 40);// list: 40 20 10
     
     u= list.first()->next();
     list.insert( u, 30); // list: 40 30 20 10
     
     list.reverse(); // list: 10 20 30 40
     
     UTEST_CHECK( utest_p, list.first()->val() == 10);
     UTEST_CHECK( utest_p, list.first()->next()->val() == 20);
     UTEST_CHECK( utest_p, list.first()->next()->next()->val() == 30);
     
     UTEST_CHECK( utest_p, list.last()->val() == 40);
     
     
     return utest_p->result();
 }
Exemplo n.º 2
0
int main(int argc, char* argv[]) {
	DList dl;

	dl.reverse();

	dl.push_back(200);
	dl.push_back(300);

	std::cout << dl;

	dl.push_front(100);

	std::cout << dl;

	dl.reverse();

	std::cout << dl;

	// Let compiler figure out the type for dl2 based on expression
	const decltype(dl) dl2 {10, 20, 30, 40, 50};
	
	std::cout << dl2;
	//dl2.push_front(0);

	decltype(dl) dl3 = {1000, 2000, 3000, 4000, 5000};
	dl3.reverse();

	std::cout << dl3;

	return 0;
}