For any STL container that I'm using, if I declare an iterator (of this particular container type) using the iterator's default constructor, what will the iterator be initialised to?
For example, I have:
std::list<void*> address_list;
std::list<void*>::iterator iter;
What will iter be initialised to?
By convention a "NULL iterator" for containers, which is used to indicate no result, compares equal to the result of container.end().
std::vector<X>::iterator iter = std::find(my_vec.begin(), my_vec.end(), x);
if (iter == my_vec.end()) {
//no result found; iter points to "nothing"
}
However, since a default-constructed container iterator is not associated with any particular container, there is no good value it could take. Therefore it is just an uninitialized variable and the only legal operation to do with it is to assign a valid iterator to it.
std::vector<X>::iterator iter; //no particular value
iter = some_vector.begin(); //iter is now usable
For other kinds of iterators this might not be true. E.g in case of istream_iterator, a default-constructed iterator represents (compares equal to) an istream_iterator which has reached the EOF of an input stream.
The default constructor initializes an iterator to a singular value:
Iterators can also have singular values that are not associated with any sequence. [ Example: After the declaration of an uninitialized pointer x (as with int* x;), x must always be assumed to have a singular value of a pointer. —end example ] Results of most expressions are undefined for singular values [24.2.1 §5]
The iterator is not initialized, just as int x; declares an integer which isn't initialized. It does not have a properly defined value.
来源：https://stackoverflow.com/questions/3395180/what-is-an-iterators-default-value