Need help in my C code - Getting unexpected output

Question

#include <stdio.h>

int main()
{
int a = -1;
char *p = &a;
printf("%x, %x",*p, *(p+1));
return 0;
}

output is = ffffffff,ffffffff

I know signed integer is represents as ffffffff.

* Expected result should print only one character because it char *

How *(p+1) is ffffffff ?

Is there any difference in

char *p = &a;

and

char *p;
p = &a;

@rafix07, deleted my comment as soon as I posted it because I missed the fact that the OP is using a character pointer to point to an `int`. — daShier, Sep 07 '19 at 17:40
Note that `char *p = &a;` and `char *p; p = &a;` have the same effect. — Weather Vane, Sep 07 '19 at 17:40
Its char * so it should print only 1 byte right that is ff ? — Pankaj Suryawanshi, Sep 07 '19 at 17:41
@WeatherVane, the behavior is not undefined. Passing a byte to be parsed with `%x` is defined as being sign extended to an `int`, which is exactly what the OP is showing. — daShier, Sep 08 '19 at 18:37
@daShier uh yes: brain fart. With a variadic function, integer types smaller than `int` are promoted to `int`, floating point types smaller than `double` are promoted to `double`, although that is unrelated to the format specifier. For some reason I was thinking of `scanf`. — Weather Vane, Sep 08 '19 at 20:03

daShier · Accepted Answer · 2019-09-07T18:16:24.743

3

An int is more than one byte (how many depends on your system). So int a = -1; sets at least 2 bytes (or more) to FF. Now you create a byte pointer *p and point to the address of the first byte of your a integer. When you use the format x% on a byte, the byte gets padded to an int.

When you try to print the second byte of your int by using *(p+1), the %x will pad the output of the ff to a full int as well.

Update: The process is known as "sign extension" and when a smaller value is used in place of a longer value, the small value is "extended" out such that if the sign is positive, zeros are padded to the left, while if the sign is negative, ones are padded to the left so that the byte FF is seen as an int FFFFFFFF rather than 000000FF. This is considered a "feature" of C so that -1 is still -1.

edited Sep 07 '19 at 18:16

answered Sep 07 '19 at 17:46

daShier

2,056
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It means output is undefined behavior ? – Pankaj Suryawanshi Sep 07 '19 at 17:50
It's not undefined. It's called "sign extension" and is a "feature". – daShier Sep 07 '19 at 17:53
What is sign extension please add in your answer ? – Pankaj Suryawanshi Sep 07 '19 at 17:57
what is difference char *p = &a and char *p ; p = &a ; if same how ? – Pankaj Suryawanshi Sep 07 '19 at 17:58
Please add above question answer in your answer. – Pankaj Suryawanshi Sep 08 '19 at 12:59
@PankajSuryawanshi, your new question regarding initializing a character pointer is outside the scope of your original question. It is also easy to find the answer by doing a search here or the web in general. You can also try both methods yourself and look at the assembly code to see if they generate the same lines of code or not (even if you don't understand the assembly). – daShier Sep 08 '19 at 13:16

XBlueCode · Answer 2 · 2019-09-07T17:55:10.317

1

To cast the variable to unsigned char you can use the following format:

printf("%hhx, %hhx",*p, *(p+1));

And you can try the output of the following code, maybe you'll get better idea:

#include <stdio.h>

int main()
{
    int a = 0xaf0e3b;
    char *p = &a;
    printf("%hhx, %hhx, %hhx",*(p + 2), *(p+1), *p);
    return 0;
}

out:

af, 0e, 3b

edited Sep 07 '19 at 17:55

answered Sep 07 '19 at 17:49

XBlueCode

785
3
18

DevesH · Answer 3 · 2019-09-07T17:51:49.663

0

First of all - [Error] cannot convert 'int*' to 'char*' in initialization.

To print character use %c not %x.

%x means that printf will output its value in hexadecimal format.

Ran this code - (Getting space() and comma(,) as output)

int main() {
   int a = -1;
   char *p = (char *)&a;
   printf("%c, %c",*p, *(p+1));
   return 0;
}

edited Sep 07 '19 at 17:51

answered Sep 07 '19 at 17:46

DevesH

486
4
18

it's possible using `char *p = (char*)(&a);`if your compiler is not happy with that – Saif Faidi Sep 07 '19 at 17:47
formatting `0xff` value with `%c` is not really a good idea because it's not a printable char, so it will not give any significant information. – XBlueCode Sep 07 '19 at 18:01
it will copy only 1 byte, print will be based on endianness – Pankaj Suryawanshi Sep 07 '19 at 18:05

Need help in my C code - Getting unexpected output

3 Answers3