Clarification regarding signed and unsigned integers in C language

Question

I have tried following program in C language and the result of the program is:

Output : 0 is printed at last when the i is unsigned int, when the value of i is 0, then the control should not go inside the while loop since I have given i>0 condition, but 0 is printed, I didn't understand how it is possible.

When the i is signed int, then 0 is not printed.

Please let me know what is happening in the program.

#include <stdio.h>
#include <string.h>

main()
{
   unsigned int i=1;
   int c=1;
   while(i>0)
   {
     i=i*2;         
     c++;
     printf("%d\n",i);          
   }

   printf("%d c value is",c) ;
}

Read this http://stackoverflow.com/questions/14695118/2147483648-0-returns-true-in-c/14695202#14695202 — ani627, Sep 12 '14 at 04:57
Vamsi - **always enable warnings when you compile your code** For example, at a minimum use: `gcc -Wall -Wextra -o youroutput yourcode.c` That will show you the problem areas every time :) — David C. Rankin, Sep 12 '14 at 05:59

score 5 · Answer 1 · answered Sep 12 '14 at 05:42

Using %d with unsigned int causes undefined behaviour. To fix your program, change %d to %u and then the results will make more sense.

When i gets to 2³¹ (assuming you have 32-bit ints) and you do i * 2 then the result , which would be 2³² actually wraps around to 0, which is why your loop ends. Unsigned integers wrap around like this (technical definition is arithmetic modulo UINT_MAX+1).

If i is signed int (and you use %d) then you cause undefined behaviour when i * 2 would exceed INT_MAX, so anything can happen.

0xF1 · Answer 2 · 2014-09-12T06:05:57.260

2

You have printed i inside while after computing i = i*2; hence when i is unsigned and has value 2^31, the while condition is true and it is again multiplied by 2, where it becomes 0 due to integer overflow, which is printed and loop gets over the next time while(i > 0) condition is checked.

For signed int, the behavior is undefined as pointed out by @Anonymous.

However here it seems that (you cannot rely on this behavior!!) condition of while fails when i equals 2^31 because it is treated as negative number now (sign bit is 1), hence it is not multiplied by 2 and doesn't become 0.

If you do not want to see 0 output, you can try doing printf before i = i*2 statement.

edited Sep 12 '14 at 06:05

answered Sep 12 '14 at 05:10

0xF1

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The behavior for signed int is undefined. – Paul Hankin Sep 12 '14 at 05:32
Hmm.. right, but here it seems to be happening that way only. Thanks for mentioning that anyways... I have included that in my answer. – 0xF1 Sep 12 '14 at 06:03

score 2 · Answer 3 · edited Sep 12 '14 at 07:27

In case of unsigned int:

supposing int takes 4 bytes, the value of i becomes "1073741824" for which the binary notation is

 "01000000 00000000 00000000 00000000"

Multiplying it by two again makes the decimal value "2147483648" which the binary notation is

"10000000 00000000 00000000 00000000".

Again multiplying it by two makes the binary notation as follows

"00000001 00000000 00000000 00000000 00000000".

But as int occupies only four bytes, 4 least significant bytes are all filled zero, so the value of i becomes zero.

In case of signed int:

Most significant bit is used for indication of negative or positive values. 1 for negative numbers. so it prints negative number further multiplying i by 2. Which makes the while condition fails. as negative number is not greater than zero.

Nice explanation. Please format it properly, and read @Matt McNabb's answer to know about "undefined behavior" with `signed int`. — 0xF1, Sep 12 '14 at 06:10

R Sahu · Answer 4 · 2014-09-12T06:02:20.407

1

Change the while loop so that printf is called first. Also, use %u for unsigned int instead of %d.

while(i>0)
{
  printf("%u\n",i);          
  i=i*2;         
  c++;
}

With this change you will be printing only the value of i that pass the conditional of the while statement.

edited Sep 12 '14 at 06:02

answered Sep 12 '14 at 05:10

R Sahu

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Clarification regarding signed and unsigned integers in C language

4 Answers4