C/C++ What does the space above 'E' represent in a C++ deque diagram?

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The discussion clarifies misconceptions regarding the representation of a C++ deque in a diagram. It emphasizes that deques do not have a "capacity" but rather a "size," which indicates the number of elements currently in the container. The conversation highlights errors in the diagram, specifically the incorrect number of empty slots after pop operations and the misrepresentation of the order of elements. The final output of the deque after the operations is confirmed to be 7, 6, 5, 42, 42, 1, 2, 3, 4.

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GeorgeCostanz
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I have a minor question about this diagram I'm hoping someone can clear up for me.

The deque has initial capacity 4. The following operations are performed.

push_back(1)
push_back(2)
push_back(3)

pop_front()
pop_front()

push_back(4)

push_front(5)
push_front(6)
push_front(7)

final answer:
OPmbBeP.png


I'm just wondering what the space above the letter 'E' is supposed to signify. I thought there were only 2 empty slots in the deque for the 2 pop_front operations at initial capacity 4.
 
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There are a number of things wrong with this question and the diagram. One is the word "capacity". Vectors have a capacity. Deques don't. Vectors and deques have a size, the number of elements currently in the container. Capacity is something different. I'm assuming that size was meant here: There are four items in the deque prior to those operations.

The second thing wrong is those three holes in the middle of the diagram. As you noted, that should be two holes. Those two pop_front() calls will remove the first two items from the deque, and since all operations up until then were on the back, the removed items are from that initial content of four items.

The third thing that's wrong: What happened to the 1 and the 2?

Finally, the order of the first three items is incorrect. It should be 7,6,5 rather than 5,6,7.
Code:
#include <deque>
#include <iostream>

int main () {
   std::deque<int> d(4,42);
   d.push_back(1);
   d.push_back(2);
   d.push_back(3);

   d.pop_front();
   d.pop_front();

   d.push_back(4);

   d.push_front(5);
   d.push_front(6);
   d.push_front(7);

   for (std::deque<int>::iterator iter = d.begin(); iter < d.end(); ++iter) {
      std::cout << *iter << " ";
   }
   std::cout << "\n";
}

The above prints 7 6 5 42 42 1 2 3 4.
 
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okay thanks. i found it on a random HW solution set on the internet.
 
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