[SOLVED] Void pointer arithmetic and dereference


This Content is from Stack Overflow. Question asked by Solruhama

I was looking at quick sort implementation for generic purpose and one of the parameters of the quick sort function was a void pointer array, and i saw the ff arithemetics on void pointers, so i was wondering what that actually does and if it’s even possible? thank you.

void _qsort(void* v, int size, int left, int right,int (*comp)(void*, void*));
void swap(void* v1, void* v2, int size)
    char buffer[size];
    memcpy(buffer, v1, size);
    memcpy(v1, v2, size);
    memcpy(v2, buffer, size);
//this part is body of the quick sort defined above
void* vl = (char*)(v + (left * size));
void* vr = (char*)(v + (mid * size));
swap(vl, vr, size);

so from above code,

  1. How is it possible to do arithmetic on void pointer v, like v+ (left*size)
  2. What does void* vl = (char*)(v + (left * size)); part mean? isn’t already casted to char pointer,if so why are we assinging it to a void pointer?
  3. In the swap part, what exactly is happening, like are the vl and vr changing their memory location, value or sth else?
    thank you in advance


You assume right: arithmetics on void pointers is not defined by the C Standard.

The program you are looking at uses a common compiler extension supported by gcc, clang, tcc and many others, that implements arithmetics on void pointers as if they were byte pointers. With this extension, v + (left * size) behaves as

    (void *)((char *)v + (left * size))

So the declaration void *vl = (char *)(v + (left * size)); is equivalent to:

    void *vl = (char *)((void *)((char *)v + (left * size)));

Note that the whole expression is cast implicitly to (void *) in C.

This declaration can be simplified as:

    void *vl = (char *)v + left * size;

This is probably what the programmer meant to write and their mistake went unreported because the compiler allows void * arithmetics with exactly the same effect.

Regarding your third question, the swap function exchanges the contents of the memory blocks pointed to by v1 and v2 using a local variable length array buffer of size bytes. qsort is usually called with a rather small element size, so this approach is OK, but calling qsort with an array of very long elements (more than a few megabytes) is allowed and could cause a stack overflow.

Here is a safer implementation:

void swap(void *v1, void *v2, size_t size) {
    unsigned char *p1 = v1;
    unsigned char *p2 = v2;
    while (size >= 8) {
        char buffer[8];
        memcpy(buffer, p1, sizeof buffer);
        memcpy(p1, p2, sizeof buffer);
        memcpy(p2, buffer, sizeof buffer);
        p1 += sizeof buffer;
        p2 += sizeof buffer;
        size -= sizeof buffer;
    if (size > 0) {
        if (size >= 4) {
            char buffer[4];
            memcpy(buffer, p1, sizeof buffer);
            memcpy(p1, p2, sizeof buffer);
            memcpy(p2, buffer, sizeof buffer);
            p1 += sizeof buffer;
            p2 += sizeof buffer;
            size -= sizeof buffer;
        while (size > 0) {
            unsigned char temp = *p1;
            *p1 = *p2;
            *p2 = temp;

Also note these remarks:

  • The standard library function qsort has a different prototype:

      void qsort(void *base, size_t nmemb, size_t size,
                 int (*compar)(const void *, const void *));
  • Using int for sizes and index values is not recommended. left * size might overflow for a large array on systems with 64-bit size_t and 32-bit int: The _qsort function assumes that left * size and right * size are within the range of type size_t and at most the size of the array pointed to by v, yet this size might exceed INT_MAX, causing the int multiplication to overflow with undefined behavior. The correct type is size_t and a sanity check at the beginning of the function can be used to detect invalid arguments.

This Question was asked in StackOverflow by Solruhama and Answered by chqrlie It is licensed under the terms of CC BY-SA 2.5. - CC BY-SA 3.0. - CC BY-SA 4.0.

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