dereference a pointer and then take the address of dereference

Question

I read STL and a usage of pointer puzzles me.

destroy(&*first);

first is a pointer, then "&*first" is equal to first, why not use first directly? destroy is declared as follow:

void destroy(T* pointer)

T is a template parameter.

score 7 · Answer 1 · edited May 23 '17 at 12:29

7

This is most likely due to operator overloading. first is the name typically given to iterators, which overload operator* to return a reference to the pointed to element of the container then operator& is used to get the address of the variable the iterator was pointing to. You can read more about iterators here.

However, if first is a pointer and not a user defined iterator, then yes you can just use first directly. Note that pointers can be iterators (specifically, RandomAccessIterators).

edited May 23 '17 at 12:29

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answered Aug 10 '14 at 02:30

Rapptz

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Strictly speaking, pointers are iterators, so the last part isn't that true. – chris Aug 10 '14 at 02:31
@chris Yeah, true. Edited to take this into account. – Rapptz Aug 10 '14 at 02:33

score 3 · Answer 2 · answered Aug 10 '14 at 02:47

If the argument is already a raw pointer, then &* combination does nothing, as you already noted. However, if the argument is a class type with overloaded unary * operator, then &* is no longer a no-op.

The classic usage of &* pattern is to convert a "generic" pointer into a raw pointer. For example, if it is an iterator associated with a vector element, then &*it will give you an ordinary raw pointer to that vector element.

dereference a pointer and then take the address of dereference

2 Answers2