通过索引访问矢量迭代器？

Question

Recently I came across this code in my codebase (Simplified for here, of course)

auto toDelete = std::make_shared<std::string>("FooBar");
std::vector<decltype(toDelete)> myVec{toDelete};
auto iter = std::find_if(std::begin(myVec), std::end(myVec), 
   [](const decltype(toDelete) _next)
   {
      return *_next == "FooBar";
   });

if (iter != std::end(myVec))
{
   std::shared_ptr<std::string> deletedString = iter[0];
   std::cout << *deletedString;
   myVec.erase(iter);
}

Online Example

Now, I noticed that here we are accessing an iterator by indexing!

std::shared_ptr<std::string> deletedString = iter[0];

I've never seen anyone access an iterator by indexing before, so all I can guess at is that the iterator gets treated like a pointer, and then we access the first element pointed to at the pointer. So is that code actually equivalent to:

std::shared_ptr<std::string> deletedString = *iter;

Or is it Undefined Behavior?

Answer 1

From the cppreference documentation for RandomAccessIterator:

Expression: i[n]

Operational semantics: *(i+n)

Since a std::vector 's iterators meet the requirements of RandomAccessIterator, indexing them is equivalent to addition and dereferencing, like an ordinary pointer. iter[0] is equivalent to *(iter+0) , or *iter .

Answer 2

This is Standard conforming behavior

24.2.7 Random access iterators [random.access.iterators]

1 A class or pointer type X satisfies the requirements of a random access iterator if, in addition to satisfying the requirements for bidirectional iterators, the following expressions are valid as shown in Table 118.

a[n] convertible to reference: *(a + n)

Note that it is not required that the particular iterator is implemented as a pointer. Any iterator class with an overloaded operator[] , operator* and operator+ with the above semantics will work. For std::vector , the iterator category is random access iterator, and it it required to work.