C- printf hex values giving unexpected output - c

Question

I'm trying to learn some of the basics of C and I have been reading nonstop about gets/fgets/puts/scanf and can't seem to get this right...
My code is:
#include <string.h>
#include <stdio.h>
#define BUFF 256
void main()
{
char s[BUFF];
fgets(s, BUFF, stdin);
s[strlen(s)-1]='\0';
printf("%x %s %d", s, s, strlen(s));
}
I'm trying to get the hex format of the variable s to print, my input is AAAA and the output I want is 41414141 AAAA 4, the output I'm getting is 12fe80 AAAA 4.
I thought I needed to cast s (as a u int) for the hex interpretation, but that didn't work either.
I would really appreciate an explanation on this as well as help, I'm really trying to learn this.
Thank you!

Answer 1

void main()
This isn't your problem, but the correct definition is
int main(void)
void main() is useful mostly as a tool to detect C textbooks written by incompetent authors.
printf("%x %s %d", s, s, strlen(s));
This has undefined behavior:
"%x" requires an argument of type unsigned int; you're giving it a char* (the array expression s is converted to a pointer to its first element).
%d requires an argument of type int; you're giving it a size_t. Either convert it to int, or use %zu.
If you want to print a hexadecimal representation of the contents of the string s, there's no direct way to do that. You'll just have to loop over the characters contained in s and print the value of each one in hexadecimal. (You should cast each char value to unsigned int so it works properly with the "%02x" format.)

Answer 2

%x works on numbers. You're passing in a pointer to a string. So in this case printf() is interpreting the pointer (memory address) as a number and printing that address in hex format. Sounds like you just want to print the ASCII values, in hex, of each character in the input:
for (int i = 0; i < strlen(s); ++i)
printf("%02x", (unsigned) s[i]);

Answer 3

Try this
$ cat hello.c
#include <string.h>
#include <stdio.h>
#define BUFF 256
int main()
{
int i = 0; /* a counter */
char s[BUFF];
fgets(s, BUFF, stdin);
s[strlen(s)-1]='\0';
while (i < strlen(s)) { /* to get the character(s) in s */
printf("%x", s[i]); /* as hex */
i++;
}
printf(" %s %d\n", s, strlen(s)); /* as before */
}
$ gcc hello.c -o hello && echo "AAAA" | ./hello
41414141 AAAA 4

Answer 4

The hex value you are printing is the address of the string, not the contents (since s is a char*).
To see the full contents of the string as hex bytes, you will have to do something like this:
int n = strlen(s);
for(int ii=0; ii<n; ii++) printf("%02x", (unsigned int) s[ii]);

Answer 5

#include <stdio.h>
#include <string.h>
main() {
char a[]="AAAA";
printf("%x%x%x%x %s %d", a[0],a[1],a[2],a[3],a, (int)strlen(a));
}

Answer 6

print all the array index one by one.
printf("%x ", *s);
printf("%x ", *(s+1));
....
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