Convert iterator to pointer?

Question

I have a std::vector with n elements. Now I need to pass a pointer to a vector that has the last n-1 elements to a function.

For example, my vector<int> foo contains (5,2,6,87,251). A function takes vector<int>* and I want to pass it a pointer to (2,6,87,251).

Can I just (safely) take the iterator ++foo.begin(), convert it to a pointer and pass that to the function? Or use &foo[1]?

UPDATE: People suggest that I change my function to take an iterator rather than a pointer. That seems not possible in my situation, since the function I mentioned is the find function of unordered_set<std::vector*>. So in that case, is copying the n-1 elements from foo into a new vector and calling find with a pointer to that the only option? Very inefficient! It's like Shlemiel the painter, especially since i have to query many subsets: the last n-1, then n-2, etc. elements and see if they are in the unordered_set.

Solution 1

That seems not possible in my situation, since the function I mentioned is the find function of unordered_set<std::vector*>.

Are you using custom hash/predicate function objects? If not, then you must pass unordered_set<std::vector<int>*>::find() the pointer to the exact vector that you want to find. A pointer to another vector with the same contents will not work. This is not very useful for lookups, to say the least.

Using unordered_set<std::vector<int> > would be better, because then you could perform lookups by value. I think that would also require a custom hash function object because hash does not to my knowledge have a specialization for vector<int>.

Either way, a pointer into the middle of a vector is not itself a vector, as others have explained. You cannot convert an iterator into a pointer to vector without copying its contents.

Solution 2

here it is, obtaining a reference to the coresponding pointer of an iterator use :

example:

string my_str= "hello world";

string::iterator it(my_str.begin());

char* pointer_inside_buffer=&(*it); //<--

[notice operator * returns a reference so doing & on a reference will give you the address].

Solution 3

If you can, a better choice may be to change the function to take either an iterator to an element or a brand new vector (if it does not modify).

While you can do this sort of things with arrays since you know how they are stored, it's probably a bad idea to do the same with vectors. &foo[1] does not have the type vector<int>*.

Also, while the STL implementation is available online, it's usually risky to try and rely on the internal structure of an abstraction.

Solution 4

A vector is a container with full ownership of it's elements. One vector cannot hold a partial view of another, even a const-view. That's the root cause here.

If you need that, make your own container that has views with weak_ptr's to the data, or look at ranges. Pair of iterators (even pointers work well as iterators into a vector) or, even better, boost::iterator_range that work pretty seamlessly.

It depends on the templatability of your code. Use std::pair if you need to hide the code in a cpp.

Solution 5

Your function shouldn't take vector<int>*; it should take vector<int>::iterator or vector<int>::const_iterator as appropriate. Then, just pass in foo.begin() + 1.

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