Converting double to void* in C

Question

I'm writing an interpreter and I'd like to be able to store whatever value a function returns into a void pointer. I've had no problem storing ints and various pointers as void pointers but I get an error when trying to cast a double as a void pointer. I understand that doubles are stored differently than integers and pointers at the bit level, but I don't understand why I can't place whatever bits I want into the pointer (assuming it has enough memory allocated) and then take them out later, casting them as a double.

Is it possible to cast a double to a void pointer using syntax I'm not aware of or am I misunderstanding how void pointers work?

It sounds like you are doing something horribly wrong. Please look at David's answer. — pmr, Jul 04 '11 at 19:31
check my answer in this other question http://stackoverflow.com/questions/6539023/storing-values-in-union-in-c/6539185#6539185 — Vinicius Kamakura, Jul 04 '11 at 19:32

David Heffernan · Accepted Answer · 2011-07-04T19:35:11.427

16

On many systems a double is 8 bytes wide and a pointer is 4 bytes wide. The former, therefore, would not fit into the latter.

You would appear to be abusing void*. Your solution is going to involve allocating storage space at least as big as the largest type you need to store in some variant-like structure, e.g. a union.

edited Jul 04 '11 at 19:35

answered Jul 04 '11 at 19:25

David Heffernan

601,492
42
1,072
1,490

2

I can't believe I've never heard of a union before. This is exactly what I was looking for. – Jack Jul 04 '11 at 19:41

score 3 · Answer 2 · answered Jul 04 '11 at 22:10

3

Of course it's possible to cast it. Void pointers is what makes polymorphism possible in C. You need to know ahead of time what you're passing to your function.

void *p_v ;
double *p_d ;
p_d = malloc( sizeof( double ) ) ;
p_v = ( void * ) p_d ;

answered Jul 04 '11 at 22:10

James

31
1

2

Your example appears to work because `p_d` is a pointer. The op was asking about a non-pointer to void* conversion. – Mar 03 '16 at 20:06

cpuwolf · Answer 3 · 2021-12-28T07:14:30.117

-1

continue with Fabio. and make sure you are running on x64

int main()
{
    double d = 1.00e+00; // 0x3ff0000000000000
    double* pd = &d;
    void** p = (void**)pd;
    void* dp = *p;
    printf("%f %p %p %p \n", d, pd, p, dp);

    printf("%d %f\n", dp, *(double*)p);

    printf("%f\n", *(double*)(void **)&d);
}

output

1.000000 0000000353F1FA30 0000000353F1FA30 3FF0000000000000
0 1.000000
1.000000

edited Dec 28 '21 at 07:14

answered Dec 28 '21 at 07:02

cpuwolf

27
3

score -3 · Answer 4 · answered Mar 06 '16 at 08:51

-3

Here is it

int main ( ) {
    double d = 1.00e+00 ; // 0x3ff0000000000000
    double * pd = & d ;
    void * * p = ( void * * ) pd ;
    void * dp = * p ;
    printf ( "%f %p %p %p \n" , d , pd , p , dp ) ;
    return 0 ;
} ;

output

1.000000 0x7fff89a7de80 0x7fff89a7de80 0x3ff0000000000000

2nd and 3rd addresses could be different. A shortcut

void * dp = * ( void * * ) & d ;

Cheers

answered Mar 06 '16 at 08:51

Fabio

211
1
8

This is undefined behaviour (strict aliasing violation) – M.M Oct 05 '17 at 04:21
@M.M it is not. I'm casting the pointer, I know what I'm doing, it's not c++, it compiles and gives the right output. I would not suggest to use a such technique in a long code, but I'm right. – Fabio Oct 05 '17 at 12:09
The expression `*p` has type `void *` but is accessing memory that is actually of type `double *` . Those could even be different sizes – M.M Oct 05 '17 at 20:44
you're right, I assumed that everything was 64 bit aligned... otherwise you need iterate a read with pchar of a given length – Fabio Oct 06 '17 at 10:39
1

`* ( void * * ) & d` is undefined behavior, and it does violate strict aliasing. – S.S. Anne Sep 12 '19 at 20:19

Converting double to void* in C

4 Answers4

Linked

Related