How to advance the array base address using a function? - c

Question

i am trying writing a function that forwards the pointer by a specified size in bytes.
void
increment_ptr(void **_ptr, int size){
*_ptr = (char *)(*_ptr) + size;
}
int
main(int argc, char **argv){
int arr[] = {11,20};
printf("%d\n", arr[0]);
increment_ptr((void **)&arr, sizeof(int));
printf("%d\n", arr[0]);
return 0;
}
output i am getting is :
vm#vm:~$ ./a.out
11
15
I am expecting the output to be 11, and 20.
Can anybody pls help ?

Answer 1

arr is not an ordinary variable that you can change. It is not a pointer - its type is int[2].
Try this code:
#include <stdio.h>
void
increment_ptr(void **_ptr, int size){
*_ptr = (char *)(*_ptr) + size;
}
int
main(int argc, char **argv){
int arr[] = {11,20};
int* arr1 = arr;
printf("arr is %p\n", (void*)arr);
printf("arr1 is %p\n", (void*)arr1);
printf("arr gives %d\n", arr[0]);
printf("arr1 gives %d\n", arr1[0]);
increment_ptr((void **)&arr, sizeof(int));
increment_ptr((void **)&arr1, sizeof(int));
printf("arr is %p\n", (void*)arr);
printf("arr1 is %p\n", (void*)arr1);
printf("arr gives %d\n", arr[0]);
printf("arr1 gives %d\n", arr1[0]);
return 0;
}
Possible output:
arr is 0xfff1d1b8
arr1 is 0xfff1d1b8
arr gives 11
arr1 gives 11
arr is 0xfff1d1b8 // Notice arr isn't changed
arr1 is 0xfff1d1bc // arr1 works as expected
arr gives 15
arr1 gives 20
In other words: You can't advance the base address of such an array
You might also find this code interesting:
#include <stdio.h>
int main(int argc, char **argv){
int arr[2] = {11,20};
printf("arr is %p\n", (void*)arr);
printf("arr is %p\n", (void*)&arr);
return 0;
}
Possible output:
arr is 0061FED8
&arr is 0061FED8
Notice how printing arr as a pointer gives the same as printing &arr
So what happens in your code is that the number 11 (aka arr[0]) is read as a pointer (casted to char*), then add 4 (so you have 15) which overwrites the current 11. Note: This result indicates the pointers and ints are both 4 bytes on your system.

Answer 2

The code
int arr[2];
declares a memory space. So, when you acquire its address it returns you the base address of this memory space, which is identical to arr.
This is different to the case when you declare a pointer. A pointer is a variable that saves an address, but an array is just the memory space.

Answer 3

When you declare a variable (array or not), it has an address that is unchanged for the rest of that variable's lifetime. It is not possible to move a variable to a new address.
Your code just causes undefined behaviour (strict aliasing violation - *_ptr uses an lvalue of type void * to access memory declared as int).