This example demonstrates how pointer arithmetic behaves differently for different data types. The amount by which a pointer moves depends on the size of the datatype it points to.

#include <stdio.h>

int main()
{
    int arr[] = {10, 20, 30};
    char ch[] = {'A', 'B', 'C'};

    int *iptr = arr;
    char *cptr = ch;

    printf("iptr      = %p\n", iptr);
    printf("iptr + 1  = %p\n", iptr + 1);

    printf("cptr      = %p\n", cptr);
    printf("cptr + 1  = %p\n", cptr + 1);

    printf("*iptr     = %d\n", *iptr);
    printf("*(iptr+1) = %d\n", *(iptr + 1));

    printf("*cptr     = %c\n", *cptr);
    printf("*(cptr+1) = %c\n", *(cptr + 1));

    return 0;
}

Output:

iptr      = 0x7ffc...
iptr + 1  = 0x7ffc...

cptr      = 0x7ffc...
cptr + 1  = 0x7ffc...

*iptr     = 10
*(iptr+1) = 20

*cptr     = A
*(cptr+1) = B

iptr is an integer pointer, so adding 1 moves the address by sizeof(int) bytes.

cptr is a character pointer, so adding 1 moves the address by 1 byte.

Pointer arithmetic always depends on the datatype associated with the pointer.

Arrays are stored continuously in memory, and pointer arithmetic is commonly used to traverse array elements.