spacerose

вместо dataout из 1 лабы (для решения 2 лабы)

May 14th, 2020
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  1. for (auto obj : pointers){
  2. cout<<" "<<obj->name;
  3. for (auto obj2:obj->pointers){
  4. cout<<" "<<obj2->name;
  5. for (auto obj3:obj2->pointers){
  6. cout<<" "<<obj3->name;
  7. for (auto obj4:obj3->pointers){
  8. cout<" "<<obj4->name;
  9. for (auto obj5:obj4->pointers){
  10. cout<<" "<<obj5->name;
  11. for(auto obj6:obj5->pointers){
  12. cout<<" "<<obj6->name;
  13. }
  14. }
  15. }
  16. }
  17. }
  18. }
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