Arithmetic with void pointers in C++

Question

I need to access an object in a buffer, pointed by a void pointer. The object is located at a certain offset but since arithmetic on a void pointer is prohibited how can I access the object?

Arithmetic on void* is not just prohibited, but meaningless. There is no sizeof(void). — Lightness Races in Orbit
– Lightness Races in Orbit, Commented Feb 20, 2011 at 19:34
Interestingly(?), gcc allows arithmetic operations on void*, while g++ seems not to allow it. — Ise Wisteria
– Ise Wisteria, Commented Feb 20, 2011 at 22:55
@IseWisteria: As of version 4.7.2, g++ permits it (with a warning) if you specify -fpermissive. — Keith Thompson
– Keith Thompson, Commented Jul 18, 2014 at 21:41

Matěj Zábský · Accepted Answer · 2014-07-18 21:32:05Z

11

You can cast the pointer to char* (+1 on such pointer is offset by one byte) or any other pointer type if that suits your needs better.

However, this approach is grossly error prone! You better check your design, something smells here! void* are in 99% of cases unnecessary in C++, designs that use them are usually more "C" than "C++". Remember, templates and inheritance should be the way to do these things.

edited Jul 18, 2014 at 21:32

answered Feb 20, 2011 at 19:27

Matěj Zábský

17.3k17 gold badges72 silver badges115 bronze badges

Sign up to request clarification or add additional context in comments.

7 Comments

David Heffernan Over a year ago

You might have a valid reason for doing this for interop across a language boundary.

Lightness Races in Orbit Over a year ago

+1 Absolutely correct. This is a complete hack and smells of a very poor design. Alternatively, of the misfortune of having to deal with C APIs. :)

BatchyX Over a year ago

sorry, but i don't see your point. You cannot escape using void* pointers when doing binary IO. Even boost's memory map api returns a void* pointer. avoiding void* is a good thing only when you can.

Matěj Zábský Over a year ago

@BatchyX Yes, there are specific cases where using void* can't be avoided - which is why I wrote only 99%, but I think it is much more likely that OP is just abusing void* to get around the strongly typed nature of C++.

TonyK Over a year ago

@MatějZábský: That is rather condescending of you.

|

Collectives™ on Stack Overflow

Arithmetic with void pointers in C++

1 Answer 1

7 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

7 Comments

Your Answer

Sign up or log in

Post as a guest

Related