为什么std :: stack默认使用std :: deque？

Question

Since the only operations required for a container to be used in a stack are:

• back()
• push_back()
• pop_back()

Why is the default container for it a deque instead of a vector?

Don't deque reallocations give a buffer of elements before front() so that push_front() is an efficient operation? Aren't these elements wasted since they will never ever be used in the context of a stack?

If there is no overhead for using a deque this way instead of a vector, why is the default for priority_queue a vector not a deque also? (priority_queue requires front(), push_back(), and pop_back() - essentially the same as for stack)

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Updated based on the Answers below:

It appears that the way deque is usually implemented is a variable size array of fixed size arrays. This makes growing faster than a vector (which requires reallocation and copying), so for something like a stack which is all about adding and removing elements, deque is likely a better choice.

priority_queue requires indexing heavily, as every removal and insertion requires you to run pop_heap() or push_heap(). This probably makes vector a better choice there since adding an element is still amortized constant anyways.

Answer 1

As the container grows, a reallocation for a vector requires copying all the elements into the new block of memory. Growing a deque allocates a new block and links it to the list of blocks - no copies are required.

Of course you can specify that a different backing container be used if you like. So if you have a stack that you know is not going to grow much, tell it to use a vector instead of a deque if that's your preference.

Answer 2

See Herb Sutter's Guru of the Week 54 for the relative merits of vector and deque where either would do.

I imagine the inconsistency between priority_queue and queue is simply that different people implemented them.