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How does one edit void pointers? Obviously, this is possible in C, because there are many standard library functions that do this.

I thought I'd implement a function swapping two array elements (I know there is a standard function for it, it's just an exercise), so ideally I'd want to do this:

void swap(void *arr, int a, int b, int elSize) {
    void arrEl = arr[a];
    arr[a] = arr[b];
    arr[b] = arrEl;
}

Of course, I can't dereference a void pointer, nor can I have a variable of type void, so I have to work around that somehow, but I discovered I don't know how to edit a void type array.
So how should I implement above pseudo-C?

I saw this thread, but I am not sure how to apply that and think seeing the correct version of above pseudo-C will make it a bit more clear.

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  • Why you declared arrEl as void? Commented Nov 17, 2013 at 14:51
  • @haccks How do you mean? I know it is invalid C, but please read the last paragraph. Commented Nov 17, 2013 at 14:52
  • @haccks And what if I'd want to pass in a struct to this function? If this were a 'real' function I'd have to write another function like this, only with different types. I just wanted to learn about void pointers and then came across this. Commented Nov 17, 2013 at 14:58
  • You can pass struct (its address) to your function as arr is of void * type. Commented Nov 17, 2013 at 15:11

3 Answers 3

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Technically you can't have a void array or void values. You can have a void pointer that points to an array of some type.

To use it you have to cast it to a non-void pointer, like e.g.

int value0 = ((int *) arr)[0];

And the simplest solution to not have to do this is to declare the pointer arr as a pointer to the type it actually is.

If you can't do that, then if you have the element size, you can declare temporary variable-length arrays to handle the swapping, like

char tmp[elSize];

memcpy(tmp, (char *) arr + elSize * a, elSize);
memcpy((char *) arr + elSize * a, (char *) arr + elSize * b, elSize);
memcpy((char *) arr + elSize * b, tmp, elSize);

The above swaps between the values at "index" a and b.

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4 Comments

But say elSize > sizeof(int), wouldn't I swap only part of a possibly large struct, thus corrupting the passed in array?
Thank you for the updates. One last question: is this temporary array trick considered 'standard'?
@11684 It's standard in C.
I'd use char* or unsigned char*, not uint8_t*. Presumably elSize is intended to be the size in bytes, which by definition is the size of a char. (It's possible to have CHAR_BIT > 8; in that case, uint8_t won't exist.)
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Joachim was a bit faster, but I'll add my answer anyway.

#include <string.h>

void swap(void *arr, size_t a, size_t b, size_t elSize) {
    char temp[elSize];

    memcpy(temp, (char*)arr+a*elSize, elSize);
    memcpy((char*)arr+a*elSize, (char*)arr+b*elSize, elSize);
    memcpy((char*)arr+b*elSize, temp, elSize);
}

#include <stdio.h>
int main(void)
{

int array[3] = {1,13,42};

char text[] = "This is text\n" ;

swap(text, 0, 2, 4);

printf("%s", text );

swap(array, 0, 2, sizeof array[0] );

printf("{%d %d %d}\n", array[0], array[1], array[2] );

return 0;
}
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2 Comments

Is there any difference between using char and uint8_t at all (except the first is obviously easier to type)?
There will probably be a difference on machines with CHAR_BIT unequal to 8. (these are rare today for general purpose machines, but some DSP's have 16 or 32 bit chars, IIRC) BTW: I don't expect these machines to have uint8_t defined. In any case: the smallest adressable unit of memory is char. (and therefore sizeof (char) is 1)
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By assigning arr[a] to arrEl, in fact you are dereferencing a void type pointer. And for this you must have to cast it before doing any operation to it. As; 

int arrEl = ((int *)arr)[a];

or 

int arrEl = *((int *)arr + a);
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1 Comment

If you knew int was the right type, you wouldn't need to use void* in the first place, and you wouldn't need to pass elSize. memcpy is almost certainly the right solution.

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