这个循环不断给我细分错误的任何解决方法？

Question

The purpose of the loop is to go through two stacks that are being compared at the same time, and placed into temporary stacks(c and d) while the rest is being popped, however this gives me segmentation faults. I am still learning c++ and segfaults have been my biggest problem.

    while((a.top()==b.top())&&(!a.empty())&&(!b.empty())){
        e=a.top();
        f=b.top();
        a.pop();
        b.pop();
        c.push(e);
        d.push(f);
    }

EDIT: a and b are two stacks that have been previously defined and are to be compared to each other.

c and d are the temporary stacks that will hold the data of a and b as they are popped out,

e and f are value placeholders for the item inside of a and b to be placed in c and d.

Answer 1

while(!(a.empty() || b.empty())
      && (a.top() == b.top()))
{
        e=a.top();
        f=b.top();
        a.pop();
        b.pop();
        c.push(e);
        d.push(f);
}

note the order of your tests. your code will exhibit undefined behaviour if either container is empty. The order of the tests matter because the evaluation of the second operand of && will not be evaluated if the first evaluates to false.

edit:

I was a little bothered by the number of copies going in in the loop (yes, I too am guilty of looking for ways to prematurely optimise a program :-) ).

Depending on how many matching items are on your stacks, you might get better performance like this (c++11 or better):

while(!(a.empty() || b.empty())
      && (a.top() == b.top()))
{
    c.push(std::move(a.top()));  // move the top item of a to the top of c
    a.pop();                     // destroy the shrivelled husk of what used to be 
                                 // at the top of a.

    d.push(std::move(b.top()));
    b.pop();
}

Answer 2

Hard to say without mode code but assuming these are std::stack objects, you might want to check they are not empty before calling top on them Change

 while((a.top()==b.top())&&(!a.empty())&&(!b.empty())){

to

while((!a.empty())&&(!b.empty())&&(a.top()==b.top())){

top returns a reference but if there is nothing on the stack then what would that be a reference to?? This would lead to undefined behaviour. There is no point checking for emptiness after calling top . By changing the order in this way, you are certain to not call top on either a or b if one (or both) are empty.

Assuming that the information in this link http://www.cplusplus.com/reference/stack/stack/top/ is correct, then it suggests that top calls back on the underlying container and this will certainly result in undefined behaviour for some (if not all) containers when they are empty.