Why are there different types of pointers for different data types in C?

Question

If we have to hold an address of any data type then we require a pointer of that data type.

But a pointer is simply an address, and an address is always int

Answer 1

The following example can help to understand the differences between pointers of different types:

#include <stdio.h>

int main()
{
    // Pointer to char
    char * cp = "Abcdefghijk";
    // Pointer to int
    int * ip = (int *)cp; // To the same address

    // Try address arithmetic
    printf("Test of char*:\n");
    printf("address %p contains data %c\n", cp, *cp);
    printf("address %p contains data %c\n", (cp+1), *(cp+1));
    printf("Test of int*:\n");
    printf("address %p contains data %c\n", ip, *ip);
    printf("address %p contains data %c\n", (ip + 1), *(ip + 1));

    return 0;
}

The output is:

It is important to understand that address+1 expression gives different result depending on address type, i.e. +1 means sizeof(addressed data), like sizeof(*address).

So, if in your system (for your compiler) sizeof(int) and sizeof(char) are different (e.g., 4 and 1), results of cp+1 and ip+1 is also different. In my system it is:

E05859(hex) - E05858(hex) = 14702684(dec) - 14702681(dec) = 1 byte for char
E0585C(hex) - E05858(hex) = 14702684(dec) - 14702680(dec) = 4 bytes for int

Note: specific address values are not important in this case. The only difference is the variable type the pointers hold, which clearly is important.

Update:

By the way, address (pointer) arithmetic is not limited by +1 or ++, so many examples can be made, like:

int arr[] = { 1, 2, 3, 4, 5, 6 };
int *p1 = &arr[1];
int *p4 = &arr[4];
printf("Distance between %d and %d is %d\n", *p1, *p4, p4 - p1);
printf("But addresses are %p and %p have absolute difference in %d\n", p1, p4, int(p4) - int(p1));

With output:

So, for better understanding, read the tutorial.