Increment void pointer by one byte? by two?

Question

I have a void pointer called ptr. I want to increment this value by a number of bytes. Is there a way to do this?

Please note that I want to do this in-place without creating any more variables.

Could I do something like ptr = (void *)(++((char *) ptr)); ?

score 83 · Accepted Answer · answered Jun 23 '11 at 06:04

83

You cannot perform arithmetic on a void pointer because pointer arithmetic is defined in terms of the size of the pointed-to object.

You can, however, cast the pointer to a char*, do arithmetic on that pointer, and then convert it back to a void*:

void* p = /* get a pointer somehow */;

// In C++:
p = static_cast<char*>(p) + 1;

// In C:
p = (char*)p + 1;

answered Jun 23 '11 at 06:04

James McNellis

338,529
73
897
968

3

shouldn't that be reinterpret_cast? – Game_Overture Dec 12 '12 at 21:18
5

@Balk: `static_cast` is sufficient. – James McNellis Dec 12 '12 at 22:42
I'm getting `invalid type conversion` error when converting `MyStruct*` to `char*` using the method in this answer. Is there some context missing? – Tomáš Zato - Reinstate Monica Sep 17 '19 at 12:23
@TomášZato-ReinstateMonica - yes: you are not casting a `void*` to any other pointer type (which can be done statically) but `MyStruct*` - see [here](https://stackoverflow.com/a/310489) – Beppo Jun 30 '20 at 09:24

Alok Save · Answer 2 · 2012-12-13T03:04:27.000

24

No arithmeatic operations can be done on void pointer.

The compiler doesn't know the size of the item(s) the void pointer is pointing to. You can cast the pointer to (char *) to do so.

In gcc there is an extension which treats the size of a void as 1. so one can use arithematic on a void* to add an offset in bytes, but using it would yield non-portable code.

edited Dec 13 '12 at 03:04

answered Jun 23 '11 at 06:03

Alok Save

196,531
48
417
525

2

In GCC it's possible because `sizeof(void)` yields `1`. – Blagovest Buyukliev Jun 23 '11 at 06:09
1

@Blagovest Buyukliev: It is extension in GCC which treats the size of a void as 1, Using it would yield non-portable code. – Alok Save Jun 23 '11 at 06:11
4

@Blag: As always, please compile with `-pedantic`. :) – Xeo Jun 23 '11 at 06:13

tdammers · Answer 3 · 2011-06-23T06:18:43.577

8

Just incrementing the void* does happen to work in gcc:

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

int main() {
    int i[] = { 23, 42 };
    void* a = &i;
    void* b = a + 4;
    printf("%i\n", *((int*)b));
    return 0;
}

It's conceptually (and officially) wrong though, so you want to make it explicit: cast it to char* and then back.

void* a = get_me_a_pointer();
void* b = (void*)((char*)a + some_number);

This makes it obvious that you're increasing by a number of bytes.

edited Jun 23 '11 at 06:18

answered Jun 23 '11 at 06:10

tdammers

19,940
1
35
54

1

As always, please compile with `-pedantic`. :) It's a non-conforming extension by GCC. – Xeo Jun 23 '11 at 06:13

score 1 · Answer 4 · answered May 23 '17 at 10:09

1

You can do:

++(*((char **)(&ptr)));

answered May 23 '17 at 10:09

EyalSh

29
3

11

Yeah, but why _would_ you? That's horrible obscure code that makes it needlessly difficult to parse what it's doing, for no good reason. Also, C-style casts are to be discouraged in C++, but that's the least of the problems here. – underscore_d May 23 '17 at 10:14
9

You can also shove a flaming stick into your eye. It would probably hurt less than looking at `++(*((char **)(&ptr)));`. – Andrew Henle May 23 '17 at 10:27
1

He wanted to do this in-place without creating any more variables, so I gave this solution. You can see it as an instructional way, but definitely not for using it in your code. – EyalSh Jan 22 '19 at 17:56

Increment void pointer by one byte? by two?

4 Answers4

Linked