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Darina [25.2K]
3 years ago
6

What is the output of the following code? stackList stack; int x, y; x = 2; y = 3; stack.push(8); stack.push(x); stack.push(x +

5); y = stack.top(); stack.pop(); stack.push(x + y); x = stack.top(); stack.pop(); cout << "x = " << x << endl;
Computers and Technology
1 answer:
Sunny_sXe [5.5K]3 years ago
3 0

Stack is LIFO data structure (Last In First Out) where the last element entered in stack will be the last one to be out of stack. It has three operations: push() : used to insert an element in stack, pop() : used to delete an element from the stack, top() : used to return the top of the stack i.e. the newest member of the stack. All these operations will take place at the top.

<u>Explanation:</u>

Now, looking at the program, x and y are initialized the values of 2 and 3 respectively. The stack pushes 8 onto the stack making it the first member of the stack. Then the value of x which is 2 is pushed onto the stack. Next, (x+5) = (2+5) = 7 is pushed onto the stack.

Pop() is used to delete hence 7 is popped out from the stack. top() is assigned to y which is 2 in this case and again 2 is popped out from the stack. Now, (x+y) = (2+2) = 4 is pushed onto the stack. And the top() is assigned to x which is 4. 4 is again popped out from the stack. Hence the value of x is 4.

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Following are the solution to this question:

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As TCP ports (4 tuples) and UDP ports (2 tuples) are using demux keys to define socks, however, demux keys in UDP are unique from others in TCP for both the following reasons:

TCP uses a separate socket for every contact between both the server and client whereas UDP utilizes its same sockets to process all clients accessed by UDP servers with all communications, it maintains TCP is much more dependable and hence has to hold a different port.

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