Affecting an int to a char array , why does printf shows this?

Question

With this code:

#include <stdio.h>
#include <unistd.h>

int main()
{

  char buffer[64];
  int check;
  int i = 2;

  buffer[-2] = i;
  printf("%x\n",buffer[-2]);

  i=25;
  buffer[-2] = i;
  printf("%x\n",buffer[-2]);

  i=255; 
  buffer[-2] = i;
  printf("%x\n",buffer[-2]);   

  return 0;
}

I get this output (compilation with gcc and -m32 option).

2
19
ffffffff

So, I can understand the two first values (2 and 19), but I really don't understand the 3rd as 255 is equal to ff and not ffffffff.

Do you have any idea why it's displayed like this ?

Thank you

p.s: Please don't suggest to affect "chars" to a "char" array, this code is just for training, thank you.

p.s (bis): please don't say it's not valid, because if it is, why are the 1rst two output of the program in accordance with the code ??

I compiled this code with gcc -m32 -o code code.c

valid indices for buffer are `0` through `63`. Your program causes undefined behaviour, therefore any output is possible — M.M, Jan 26 '17 at 09:15
@M.M Please try youself, it work just find, see the 2 first outpu of buffer[-2], they are valid — gfd78Fgi, Jan 26 '17 at 09:20
Apart from the undefined behavior: `char` is a *signed* integer on your platform, and printing it with `%x` does a sign extension to `-1` which is then printed in hex as `ffffffff`. So you would get the same output when using `buffer[+2]`. — Martin R, Jan 26 '17 at 09:22
@gfd78Fgi, nops, a negative index is only allowed in cases like: `char buffer[64]; char *ptr = buffer + 2; ptr[-2] = i;` — David Ranieri, Jan 26 '17 at 09:23
please see [what is undefined behaviour](http://stackoverflow.com/questions/2397984/undefined-unspecified-and-implementation-defined-behavior) before you attract even more downvotes — M.M, Jan 26 '17 at 09:25
@MartinR using wrong format specifier is also undefined behaviour. On OP's system the "sign extension" is likely to come from default argument promotions, not from `%x` — M.M, Jan 26 '17 at 09:26
@M.M: Well, that is what I meant (perhaps expressed badly): `char` is promoted to `int` with sign extension in the variable argument list, and that is printed with `%x` as `ffffffff`. — Martin R, Jan 26 '17 at 09:28

score 2 · Answer 1 · answered Jan 26 '17 at 09:15

2

Do you have any idea why it's displayed like this ?

No, because your program has undefined behavior:

buffer[-2] = i; 
//     ^^

You're accessing buffer out of bounds, causing UB - therefore anything can happen.

answered Jan 26 '17 at 09:15

Vittorio Romeo

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Please try youself before telling this, it perfectly works ! – gfd78Fgi Jan 26 '17 at 09:17
1

@gfd78Fgi: sure thing. Here's [UndefinedBehaviorSanitizer](http://melpon.org/wandbox/permlink/et7sZYKEBTVwYwhw) telling you that you're wrong. – Vittorio Romeo Jan 26 '17 at 09:18

score 2 · Answer 2 · edited May 23 '17 at 12:33

2

 buffer[-2] = i; //invalid; out of range

It is not valid, because you would be accessing memory outside the bounds of the array. Please see this stack overflow answer.

edited May 23 '17 at 12:33

Community

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answered Jan 26 '17 at 09:17

msc

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Mats · Answer 3 · 2017-01-26T09:32:00.453

You seem to be asking about the behaviour of this code:

int i = 255;
char c = i;
printf("%x\n", c);

The reason it prints ffffffff is because printf is a vararg function so all integer-types smaller than int are promoted to int. So you could have written:

int i = 255;
char c = i;
printf("%x\n", (int)c);

Then the question comes, why is (int)c the same as ffffffff. It is because c is assigned ff, which is beyond its limits of -128 to 127. It is wrapped to -1. It is then converted to the int value -1, and the int value -1 is coded as ffffffff.

Note: As many others have written, your program has undefined behaviour. This means that your program may print other things, or crash, when run on a different compiler, a different computer, or even a different version of your compiler. It is considered bad practise to include undefined behaviour in a program.

For completeness it should be pointed out that on a system with unsigned `char` there will be no sign extension and the output will be `ff`. — Klas Lindbäck, Jan 26 '17 at 09:31

Affecting an int to a char array , why does printf shows this?

3 Answers3