Quick Q: Why doesn't std::queue::pop return value?
Quick A: To guarantee exception safety and for performance.
Some time ago on SO:
Why doesn't std::queue::pop return value?
So, whats the difference, pop function could have done the same thing.It could indeed have done the same thing. The reason it didn't, is because a pop that returned the popped element is unsafe in the presence of exceptions (having to return by value and thus creating a copy).
Consider this scenario (with a naive/made up pop implementation, to ilustrate my point):
template<class T> class queue { T* elements; std::size_t top_position; // stuff here T pop() { auto x = elements[top_position]; // TODO: call destructor for elements[top_position] here --top_position; // alter queue state here return x; // calls T(const T&) which may throw }If the copy constructor of T throws on return, you have already altered the state of the queue (
top_positionin my naive implementation) and the element is removed from the queue (and not returned). For all intents and purposes (no matter how you catch the exception in client code) the element at the top of the queue is lost.This implementation is also inefficient in the case when you do not need the popped value (i.e. it creates a copy of the element that nobody will use).
This can be implemented safely and efficiently, with two separate operations (
void popandconst T& front()).
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