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I have some function
void print_elem(const std::vector<int>::iterator it, const std::vector<int> &vec) {/*.....*/}
If I leave out that the vector is a reference to the original object, I get copies of the vector. Why doesn't the same hold true for the iterator? Why doesn't the iterator need to be a reference also?
For instance if I wanted to iterate through the vector, print each element and wanted to stop when I hit the end of the vector, unless I pass a reference to the vector the iteration just continuously iterates through the first vector copy. But if I pass through a reference the iteration goes through the original vector object. But why does the iterator not get copied the way the vector without reference does?
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When we pass an array to a function, a pointer is actually passed. However, to pass a vector there are two ways to do so: Pass By value. Pass By Reference.
In general: If you pass a non- const reference, the caller doesn't know if the iterator is being modified. You could pass a const reference, but usually iterators are small enough that it gives no advantage over passing by value.
A vector<int> is not same as int[] (to the compiler). vector<int> is non-array, non-reference, and non-pointer - it is being passed by value, and hence it will call copy-constructor. So, you must use vector<int>& (preferably with const , if function isn't modifying it) to pass it as a reference.
The iterator models a pointer, and it most likely either is one, or contains one which points to the vector or its contents. When you copy it, the copy is in fact a different iterator, but it stores the same value, so it still points to the same thing, just like a pointer would.
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