C printf compiler warning - c

Question

#include <stdio.h>
#include <stdlib.h>
#include <fcntl.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/mman.h>
#include <sys/stat.h>
#include <errno.h>
int main(int argc, char *argv[])
{
int fd, offset;
char *data;
struct stat sbuf;
int counter;
if (argc != 2) {
fprintf(stderr, "usage: mmapdemo offset\n");
exit(1);
}
if ((fd = open("mmapdemo.c", O_RDONLY)) == -1) {
perror("open");
exit(1);
}
if (stat("mmapdemo.c", &sbuf) == -1) {
perror("stat");
exit(1);
}
offset = atoi(argv[1]);
if (offset < 0 || offset > sbuf.st_size-1) {
fprintf(stderr, "mmapdemo: offset must be in the range 0-%ld\n",sbuf.st_size-1);
exit(1);
}
data = mmap((caddr_t)0, sbuf.st_size, PROT_READ, MAP_SHARED, fd, 0);
if (data == (caddr_t)(-1)) {
perror("mmap");
exit(1);
}
// print the while file byte by byte
while(counter<=sbuf.st_size)
printf("%c", data++);
return 0;
}
This gives me error as follows:
gcc mmapdemo.c -o mmapdemo
mmapdemo.c: In function 'main':
mmapdemo.c:48: warning: format '%c' expects type 'int', but argument 2 has type 'char *'
Please help me to solve the problem.

Answer 1

printf("%c", *data++);
data is a char *. The %c format specifier tells printf to expect a char. To get a char from a char *, you need to dereference the pointer using the * operator.
That said, your program still won't work properly because you're not incrementing counter in your print loop, nor have you initialized it. I would go with:
for (size_t i = 0; i < sbuf.st_size; ++i) {
printf("%c", data[i]);
}
instead. I haven't inspected the rest of your program, but given that there were three serious errors in three lines that I looked at, I doubt that the rest is bug-free.

Answer 2

to print it out byte by byte, need to use
printf("%c ", *data++)
or to print out the hex values:
printf("%02X", *data++);