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c++pointers
Kangyabing
Kangyabing

Reputation: 61

dereference a pointer and then take the address of dereference

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.

Upvotes: 6

Views: 805

Answers (2)

AnT stands with Russia
AnT stands with Russia

Reputation: 320739

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.

Upvotes: 3

Rapptz
Rapptz

Reputation: 21317

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).

Upvotes: 7

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