How to remove base element from the stack and return it in the same order without using push, pop, any method

Question

I am programming a method called popButtom () in C ++ using stacks.

The method must do the following: ¨Eliminate the element of the base and leave the stack in the same order but without the element elminado¨, I can not use pop or push.

For example:

ini stack:

A
B
C
D

end stack:

A
B
C

I have programmed the following, but I do not know that I can have bad:

void popFull()
{
    struct node *A, *B;

    top1 = top;

    while (top1 != NULL)
    {
        B = top1->ptr;        
        A = top1;                     

        B->ptr = A;

        top1 = B;
    }
}

Regards Mariam

score 1 · Accepted Answer · edited May 23 '17 at 10:32

So, I'll see what I can do to answer this, though it would be very helpful if you could include a more complete version of your code, because I'm not entirely sure what type of data structure some of your variables are because there are no declarations included. As well, could you clarify what you mean by "but I do not know that I can have bad:"? I think these changes would make your question easier to answer.

In any case, I'll try to answer your question by interpreting it as "How do I eliminate the element at the base and leave the stack in the same order, not using pop or push." (I assume this is some sort of assignment?)

To that end I'll propose several options. C++11 has another function that isn't push() or pop() which you can use by doing stack.emplace() which just adds an item to the top of the stack. It is functionally the same as stack.push but it might be a nice hack. It's obviously a bit of a technicality, and there actually is a difference (it's very nuanced though, here's a link if you're interested: C++: Stack's push() vs emplace()) but you might be able to get away with it.

Next, I'll say that if you cannot use stack.pop() or stack.push() this next option is a possibility, but only if you initialize stack with a container class of vector, because otherwise the items are not contiguous in memory and there is no guarantee that it will work. I'm referring to, of course, pointer arithmetic. Here: Copy std::stack into an std::vector is another answer that deals with this, but I'll give a brief overview of what they did. If you initialize your stack using a std::vector as in this example in the documention, you can then copy your stack to a vector, and then operate freely on that vector, then copy back to a stack.

Here's what I mean (keep in mind this only works if the container class is vector because it seem like you're just designing a function to take in an argument and not initialize your own).

//this is how it will have to have been initailized
//for this to be guarenteed to work    
std::stack<int, std::vector<int>> myStack; 

int* begin = &stack.top()+1;
int* end = being+stack.size();
std::vector stackContents(begin,end);

And hurray, smooth sailing from here, now you can remove the item freely using your method of choice on the vector. Then, when you've modified the vector, you can create another stack to return by doing the opposite:

std::stack<int, std::vector<int>> newStack (stackContents);
return newStack;

Obviously this is a major workaround, and in the real world pop() and push() are useful functions and are included for a reason. This might actually be a good time to touch on the idea that stacks are designed to be accessed from either end. That's why it's been categorized as Last In First Out, because the idea of order matters and trying to circumvent that order means that a stack wasn't the proper data structure to use in the first place. Either way, that's my two cents, and I hope this helps.

How to remove base element from the stack and return it in the same order without using push, pop, any method

1 Answers1