std::advance behavior when advancing beyond end of container [duplicate]

Question

What is the behavior of std::advance when you have say:

std::vector<int> foo(10,10); auto i = foo.begin(); std::advance(i, 20); 

What is the value of i? Is it foo.end()?

Answer 1

The standard defines std::advance() in terms of the types of iterator it's being used on (24.3.4 "Iterator operations"):

These function templates use + and - for random access iterators (and are, therefore, constant time for them); for input, forward and bidirectional iterators they use ++ to provide linear time implementations.

The requirements for these operations on various iterator types are also outlined in the standard (in Tables 72, 74, 75 and 76):

• For an input or forward iterator

  ++r precondition: r is dereferenceable 

• for a bidirectional iterator:

  --r precondition: there exists s such that r == ++s 

• For random access iterators, the +, +=, -, and -= operations are defined in terms of the bidirectional & forward iterator prefix ++ and -- operations, so the same preconditions hold.

So advancing an iterator beyond the 'past-the-end' value (as might be returned by the end() function on containers) or advancing before the first dereferenceable element of an iterator's valid range (as might be returned by begin() on a container) is undefined behavior since you're violating the preconditions of the ++ or -- operation.

Since it's undefined behavior you can't 'expect' anything in particular. But you'll likely crash at some point (hopefully sooner rather than later, so you can fix the bug).

Answer 2

According to the C++ Standard §24.3.4 std::advance(i, 20) has the same effect as for ( int n=0; n < 20; ++n ) ++i; for positive n. From the other side (§24.1.3) if i is past-the-end, then ++i operation is undefined. So the result of std::advance(i, 20) is undefined.