Cast int to void*

Question

I'm trying to create a void* from an int. But gcc always give me a warning, that i cannot cast an int to a void*. Here is my code:

#define M_TABLE_SIZE 64*1024
static const void* M_OFFSET = M_TABLE_SIZE;

I already tried (void*)M_TABLE_SIZE but then I get an error that I cannot use the * operator.

What is the correct method to cast an int to a void*?

May be helpful for you. [What does it mean to convert int to void* or vice versa?](http://stackoverflow.com/questions/8618637/what-does-it-mean-to-convert-int-to-void-or-vice-versa) — ani627, Oct 14 '14 at 19:01

score 6 · Accepted Answer · answered Oct 14 '14 at 19:07

6

This

static const void* M_OFFSET = (void*)M_TABLE_SIZE;

expands to

static const void* M_OFFSET = (void*)64*1024;

Since gcc compiles that you are performing arithmetic between void* and an int (1024). Put your define inside a bracket:

#define M_TABLE_SIZE (64*1024)

Now, you can do:

static const void* M_OFFSET = (void*) M_TABLE_SIZE;

without a problem. It's always a good practice to put your #define's in brackets to avoid such surprise.

Unless you have a valid address at that value, you are going to invoke undefined behaviour when try to use that pointer. So make sure you understand what you are doing!

answered Oct 14 '14 at 19:07

P.P

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I guess the other important fact is that the cast operator has higher precedence that the multiplication operator – David Heffernan Oct 14 '14 at 19:11
@BlueMoon Thanks a lot! I assumed that gcc makes a 65536 out of my define, but I was wrong. – kaetzacoatl Oct 14 '14 at 19:15

Grzegorz Szpetkowski · Answer 2 · 2014-10-14T19:28:37.843

For a fairly recent compiler (that supports C99) you should not store or represent address as plain int value. If you really need such trickery, then consider one of dedicated types, which are intptr_t and uintptr_t. Referring to N1570 7.20.1.4/p1 (Integer types capable of holding object pointers):

The following type designates a signed integer type with the property that any valid pointer to void can be converted to this type, then converted back to pointer to void, and the result will compare equal to the original pointer:

intptr_t

The following type designates an unsigned integer type with the property that any valid pointer to void can be converted to this type, then converted back to pointer to void, and the result will compare equal to the original pointer:

uintptr_t

These types are optional.

In your case it might be:

#include <stdint.h>

#define M_TABLE_SIZE ((intptr_t) (64*1024))
static const void* M_OFFSET = (void *) M_TABLE_SIZE;

fhsilva · Answer 3 · 2014-10-14T19:22:14.147

0

First you are using a define, which is not a variable. The preprocessor will replace your code by this:

static const void* M_OFFSET = 64*1024;

This is unlikely what you are trying to do.

You may declare a const int variable and cast its reference to the void*.

edited Oct 14 '14 at 19:22

answered Oct 14 '14 at 19:00

fhsilva

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The preprocesor absolutely will not perform arithmetic – David Heffernan Oct 14 '14 at 19:12
You are right! Thanks for pointing that out. Just edited. However, I believe that even if the define was for the "65536", it would not be what @kaetzacoatl wanted. – fhsilva Oct 14 '14 at 19:24

Cast int to void*

3 Answers3