Saturday, October 01, 2005

why does std::stack::pop() returns void?

I have atleast 2 good explanations for this apparently counter intuitive way of defining the interface.

1. SGI explanation: http://www.sgi.com/tech/stl/stack.html
One might wonder why pop() returns void, instead of value_type. That is, why must one use top() and pop() to examine and remove the top element, instead of combining the two in a single member function? In fact, there is a good reason for this design. If pop() returned the top element, it would have to return by value rather than by reference: return by reference would create a dangling pointer. Return by value, however, is inefficient: it involves at least one redundant copy constructor call. Since it is impossible for pop() to return a value in such a way as to be both efficient and correct, it is more sensible for it to return no value at all and to require clients to use top() to inspect the value at the top of the stack.

2. std::stack < T > is a template. If pop() returned the top element, it would have to return by value rather than by reference as per the of above explanation. That means, at the caller side it must be copied in an another T type of object. That involves a copy constructor or copy assignment operator call. What if this type T is sophisticated enough and it throws an exception during copy construction or copy assignment? In that case, the rvalue, i.e. the stack top (returned by value) is simply lost and there is no other way to retrieve it from the stack as the stack's pop operation is successfully completed!
----
src: EXCEPTION HANDLING: A FALSE SENSE OF SECURITY by Tom Cargill

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mstone said...

Thank you for the post. You might want to update it a tiny bit now in order not to inadvertently confuse the reader: in C++11 "rvalue" means something else.