why does while(1) does not work?

Hello everybody,

I'm a C programming beginner
I was asking myself why this code (it compiles properly) prints only zeros

Code:

---

#include <stdio.h>

int main()
{

  int number = 2;

  while (1)
  {
        printf("&#37;d\n", number);
        number *= 2;
  }

  return 0;

}

---

Thank you in advance!!

What do you mean "doesn't work".

If you put a "getchar()" inside your loop, perhaps you'll see it print 2, 4, 8, 16, 32, etc. But eventually the single 1-bit [all these numbers have only one bit set to one in them, it moves to the right when you multiply by 2] will fall off the left edge of the number [after 31 iterations of the loop if you are using a reasonably modern OS and compiler], and it will have no bits set to 0 - which means that the entire number is zero -> 0 * 2 = 0 so you'll forever see zero.

--
Mats

Well the first few lines do what you expect.
Then arithmetic overflow takes over and you end up with zero.

doh it was so logic! :(

the fact is that I've seen a version of a friend of mine that differ only for while(foo)

Code:

---

--- my_friend_version.c        2007-09-26 12:14:41.000000000 +0200
+++ my_version.c        2007-09-26 12:14:25.000000000 +0200
@@ -3,11 +3,12 @@
 int main()
 {
  int number = 2;
-
-  while (number)
+ 
+  while (1)
          {
                printf("%d\n", number);
                number *= 2;
          }
  return 0;
 }

---

and the version with while(number) stops after overflow. So I thaught my one should do the same; but control flow exits while when buffer overflows.

I have undestand

thank you again

Yes, your friends version stops when it overflows. But beware that not all types of overflow will be easily detected this way. If you multiply by 3, 5, 7 or some other other number, you may end up with digits in the lower part of the number, and thus the number will always have some non-zero value.

Let's make this an 8 bit number (as it would require quite a lot more typing for a 32-bit number, but the principle is the same. Number in() is the binary form of the result), number in [] is ACTUAL result before overflow is removed

Code:

---

  2 * 7 =  14 [  14] (00001110)
 14 * 7 =  98 [  98] (01100010)
 98 * 7 = 174 [ 686] (10101110)
174 * 7 =  98 [1218] (01100010)

---

As you can see, it gets back to 98 after a few loops, and thus will NEVER exit the loop.

--
Mats

Another way to check for overflow is to use

Code:

---

#include <stdio.h>
#include <limits.h>

int main()
{

  int number = 2;

  while (INT_MAX / number >= 2)
  {
        printf("&#37;d\n", number);
        number *= 2;
  }

  return 0;

}

---

INT_MAX is the maximum value that can be stored into an int. It's located in <limits.h>. There's also UINT_MAX (maximum value of an unsigned int), INT_MIN, LONG_MAX, and many other values in that header file.

Of course, I haven't tested that code, but it should work. :)

well, since "we" are programmers (it would be better to say "you...") :)

Code:

---


#include <stdio.h>

int main()
{
 
  int counter = 1;

  while (counter++ <= 1000)
    printf("thank you!\n");

  return 0;

}

---

bye, Luca