When a string is "" or a vector has no elements, is `begin()` equal to `end()`?

Question

When a std::string is "" or a vector has no elements, is begin() equal to end()?

If so, what the value of begin()/end()?

How about compiling and checking it? about you second question, you don't care. — Elazar, May 13 '13 at 23:53

score 6 · Accepted Answer · answered May 13 '13 at 23:55

6

When a std::string is "" or a vector has no elements, is begin() equal to end()?

Yes, for any empty standard library container, including std::string and std::vector, begin() will return the same iterator as end().

If so, what the value of begin()/end()?

It will be an iterator which is unique to that container, but should not be dereferenced. Doing so would cause undefined behavior.

answered May 13 '13 at 23:55

Drew Dormann

Why is it undefined instead of throwing an exception? efficiency? – Elazar May 13 '13 at 23:58
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@Elazar, Because "you pay for what you need" is a common sight in C++. – chris May 13 '13 at 23:58
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@Elezar exceptions aren't for programmer mistakes. – Pubby May 13 '13 at 23:58
thanks, chris @Pubby (it's "elazar" BTW). so conceptually it should have been an `assert`, except the library is tuned to release mode and hence doesn't have asserts in it. Am I right? Or are there other gotcha's here? – Elazar May 14 '13 at 00:09
@Elazar There are never `assert`s either. If you are iterating through any container it is the responsibility of *your code* to check whether you have reached the `end()` iterator. C++ will not check all your iterator dereferences to see if they're invalid. It's assumed that *you're* checking. – Drew Dormann May 14 '13 at 14:22

score 4 · Answer 2 · answered May 13 '13 at 23:54

4

Yes. Any STL container type with no elements will have begin() == end(), as the range of elements is the half-open range [begin(), end()).

Hope this helps!

answered May 13 '13 at 23:54

templatetypedef

2 Answers2