Using void pointer to an array

Question

I was just trying to use a void pointer to an integer array ,I tried to see if i can print the array back by casting it back into int. But it is giving me some random value. Can you tell me where i am going wrong?

#include<stdio.h>
#include<stdlib.h>

int main(){
    int a[5];
    int x;
    int j;

    a[0]=1;
    a[1]=2;
    a[2]=3;
    a[3]=4;

    void *arr=a;

    for(j=0;j<4;j++){
        x = *(int *)(arr+j);
        printf("%d",x);
    }
    return 0;
}

Output is this:

133554432131072512

Why is it not pinting elements of array a[] i.e 1,2,3,4 ?

You can improve the output by putting a newline into the format string that prints the numbers: printf("%d\n", x);. — Jonathan Leffler
– Jonathan Leffler, Commented Jan 11, 2012 at 0:58
Yupp.I was not concentrating on the formatting because i wrote this to help myself test this functionality for another program. — Anusha Pachunuri
– Anusha Pachunuri, Commented Jan 11, 2012 at 1:27

ruakh · Accepted Answer · 2012-01-11 00:45:27Z

16

You need to cast arr before adding j. Here is a minimal fix:

x = *(((int *)arr)+j);

but I think it's clearer to write:

x = ((int *)arr)[j];

answered Jan 11, 2012 at 0:45

ruakh

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2 Comments

Anusha Pachunuri Over a year ago

I was wondering why *(int *)(arr+j) is wrong? Can you please explain me?

dreamlax Over a year ago

@AnushaPachunuri: Please see my answer, but to sum up, it's wrong because you can't add numbers to void *, you need to convert the void * first and then add numbers to it.

ouah · Accepted Answer · 2012-01-11 00:50:21Z

6

You are doing pointer arithmetic on void * which is not valid in C.

In GNU C (C with gcc extensions), it is actually permitted and the sizeof (void) is considered to be 1.

http://gcc.gnu.org/onlinedocs/gcc/Pointer-Arith.html

"addition and subtraction operations are supported on pointers to void and on pointers to functions. This is done by treating the size of a void or of a function as 1."

answered Jan 11, 2012 at 0:50

ouah

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1 Comment

Jonathan Leffler Over a year ago

+1: for pointing out that arithmetic on void * is a GCC extension.

dreamlax · Accepted Answer · 2012-01-11 00:50:45Z

2

The C standard does not define behaviour for arithmetic of void *, so you need to cast your void * to another pointer type first before doing arithmetic with it.

Some compilers [as an extension] treat pointer arithmetic of void * the same as char *, so each ‘+1’ will only increase the address by 1, rather than by the size of the pointed-to object. This is not standardised though so you can't rely on this behaviour.

answered Jan 11, 2012 at 0:50

dreamlax

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5 Comments

Anusha Pachunuri Over a year ago

oh so u mean pointer arithmetic is not applicable for void * pointers. Well i see the point. Thanks for a great explanation.

Jonathan Leffler Over a year ago

ISO/IEC 9899:1999. §6.2.5 ¶19 The void type comprises an empty set of values; it is an incomplete type that cannot be completed. and then in ¶20 it adds A pointer type may be derived from a function type, an object type, or an incomplete type ... where an incomplete type is clearly not an object type, so you cannot treat void * as a pointer to an object type. §6.5.2 The additive operators says For addition, either both operands shall have arithmetic type, or one operand shall be a pointer to an object type and the other shall have integer type. So you cannot do arithmetic on void *.

dreamlax Over a year ago

@JonathanLeffler: The behaviour is not defined by the C standard, but it is not explicitly undefined either, and it's usually these areas where C implementations put in such behaviour and call it an extension.

ouah Over a year ago

@dreamlax as mentioned by @JonathanLeffler addition of an integer and an operand of void * type is a clear violation of the constraints of the additive operator (6.5.6p3). So it requires a diagnostic and the implementation is free to not translate the program.

dreamlax Over a year ago

@ouah: I'm not arguing anything here, if anything I'm agreeing with you; I said that some compilers treat it as an extension. What I'm saying is that it is not undefined behaviour to use void * in arithmetic, I'm saying it is simply not defined at all (i.e. the standard does not define any behaviour for arithmetic involving void *). There is a massive difference between undefined behaviour and something that is not defined by the C standard.

asaelr · Accepted Answer · 2012-01-11 00:50:02Z

1

you should not add numbers to void pointers. cast it before. (x = *((int *)arr+j);)

When you add number to a pointer, the compiler multiply this number with the size of the type that is pointed, so if you add number to a pointer to wrong type, you will get wrong result.

if I remember correct, add to void* is illegal, but some compilers adds the exact number in bytes (like it is char*). `

answered Jan 11, 2012 at 0:50

asaelr

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