Why do I need to typecast the arguments to (void*) in C when using the %p format specifier in printf()?

I was experimenting with the following C code:

#include <stdio.h>
#include <stdlib.h>

int main(void) {
    int a = 5;
    int *ptr = &a;
    printf("Data stored: %d\n", *ptr); // Simply prints 5.
    printf("Address of variable a: %p\n", (void*) &a);
    printf("Address stored in ptr: %p\n", (void*) ptr);
    return EXIT_SUCCESS;
}

I compiled the program with:

gcc -Wall -Wextra -Wconversion -pedantic -std=c11 -fsanitize=address -fsanitize=undefined example-1.c -o example-1 

I understand that the %p format specifier expects an object address (using & or a valid pointer) but when I compile the code without typecasting to void* it generates a warning:

example-1.c: In function ‘main’:
example-1.c:8:37: warning: format ‘%p’ expects argument of type ‘void *’, but argument 2 has type ‘int *’ [-Wformat=]
     printf("Address of variable a: %p\n", &a);
                                    ~^     ~~
                                    %ls
example-1.c:9:34: warning: format ‘%p’ expects argument of type ‘void *’, but argument 2 has type ‘int *’ [-Wformat=]
     printf("Data stored in ptr: %p\n", ptr);
                                 ~^
                                 %ls

So, I naturally typecast the arguments to void* and then the program works fine, but I cannot make any logical inference about what is actually happening.