C - 空指针和偏移量

Question

Say I have a void pointer (more like; array), and I want to get the items inside it. So, I know that pointer[i] won't work since it's void and I don't know the type; I tried using the offset technique:

void function(void* p, int eltSize){
  int offset = 3;
  for(i = 0; i<offset; i++){
   memcpy(p+(i*eltsize), otherPointer, eltSize);//OtherPointer has same type.
  } 
  //End function
}

This function works good and everything, but the only problem is that at the end of main(..) I get segmentation fault. I know it's because of the pointer and how I accessed the items of it, but I don't know how to correct the problem and avoid segmentation fault.

Answer 1

As pointed out by @sunqingyao and @flutter, you can not use arithmetic with void pointers in Standard C; instead, use a char * (a chunk of bytes a la qsort ):

#include <stdio.h>
#include <string.h>

void function(void *ptr, size_t eltSize, void *otherPointer, size_t offset)
{
    char *p = ptr;

    for (size_t i = 0; i < offset; i++) {
        memcpy(p + (i * eltSize), otherPointer, eltSize);
    }
}

int main(void)
{
    int arr[] = {1, 2, 3};
    int otherValue = 4;

    function(arr, sizeof *arr, &otherValue, sizeof arr / sizeof *arr);
    for (int i = 0; i < 3; i++) {
        printf("%d\n", arr[i]);
    }
    return 0;
}

Answer 2

Quoted from N1570 6.5.6 Additive operators(emphasis mine):

2 For addition, either both operands shall have arithmetic type, or one operand shall be a pointer to a complete object type and the other shall have integer type. (Incrementing is equivalent to adding 1.)

Obviously, void isn't a complete object type. Thus, applying + operator on void * invokes undefined behaviour, which may result in segmentation fault or anything else.

One approach to solve your problem would be declaring parameter p as a char * .