C- printf hex values giving unexpected output

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!

Keith Thompson · Accepted Answer · 2013-12-30 20:59:27Z

2

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.)

answered Dec 30, 2013 at 20:59

Keith Thompson

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TypeIA · Accepted Answer · 2013-12-30 21:07:44Z

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]);

edited Dec 30, 2013 at 21:07

answered Dec 30, 2013 at 20:59

TypeIA

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1 Comment

TypeIA Over a year ago

@dasblinkenlight You're right, I didn't think through the - 1. Will remove that comment, thanks.

Elliott Frisch · Accepted Answer · 2013-12-30 21:01:21Z

1

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

answered Dec 30, 2013 at 21:01

Elliott Frisch

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Comments

Floris · Accepted Answer · 2013-12-30 21:07:52Z

1

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]);

edited Dec 30, 2013 at 21:07

answered Dec 30, 2013 at 20:57

Floris

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4 Comments

Keith Thompson Over a year ago

Your first printf has undefined behavior. How do you know that the first sizeof (int) bytes of s are correctly aligned? What about byte order? %x requires an unsigned int, not an int. %d requires an int; strlen() returns a size_t result. (Yes, I downvoted.)

Floris Over a year ago

@KeithThompson I appreciate your inputs. I have updated my answer.

Keith Thompson Over a year ago

I'm saying no compiler, regardless of how modern it is, is required to cope with badly aligned pointers. (x86 happens to handle misaligned addresses in hardware; other platforms require strict alignment). The point is that your first printf has undefined behavior, and IMHO is poor style as well. And I see you deleted it from your answer as I was writing this, so I'll withdraw the downvote. But you're computing strlen(s) on every iteration of the loop, which is inefficient.

Floris Over a year ago

@KeithThompson - I guess I have never used a platform that had trouble with misaligned pointers, but I see your point. And I've eliminated the "gross inefficiency" in the loop.

Vorsprung · Accepted Answer · 2013-12-30 21:07:18Z

0

#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));
}

answered Dec 30, 2013 at 21:07

Vorsprung

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Comments

sujin · Accepted Answer · 2013-12-30 21:08:03Z

0

print all the array index one by one.

printf("%x ", *s);
printf("%x ", *(s+1));
....
....

edited Dec 30, 2013 at 21:08

answered Dec 30, 2013 at 21:00

sujin

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