Re: lun - Lucky Number

On 15/03/2025 19:44, Marcel Logen wrote:

Richard Heathfield in sci.crypt:
 

<snip>

   for(i = 0; i < n; i++)
   {
     for(j = 7; j >= 0; j--)
     {
       fprintf(stdout, "%d", !!(b[i] & (1 << j)));

                               ^^
What does this do? Is it necessary?
(As you see, I'm not a C programmer. ;-)

Okay. In C, ! is the `not' operator.
0 is false, 1 is true, non-zero is true, but not true is false. So:
x = 6;
printf("%d", x); prints 6.
printf("%d", !x); prints 0 because not non-zero is false.
printf("%d", !!x); prints 1 because not not non-zero is true.
So !!x maps 0 to 0 and everything else to 1. If you're printing bits, it's just the job.

 

     }
     putchar('\n');
   }
   printf("--------\n");
}
>
int main(void)
{
   unsigned char t[8] = {31, 41, 59, 26, 53, 58, 97, 32};
   unsigned char r[8] = {0, 0, 0, 0, 0, 0, 0, 0};
   unsigned char l[8] = {0, 0, 0, 0, 0, 0, 0, 0};
>
   puts("t first");
   dump(t, 8);
   puts("now turn it right 90 deg");
   tuwer(r, t);
   dump(r, 8);
   puts("and back again");
   tuwel(l, r);
   dump(t, 8);

 "dump(l, 8);" (?)

Oh. Er... yeah, good spot. Oops. Fortunately, it amounts to the same thing in this case.

 

   return 0;
}

[...]

#define BIT_QRY(x,i) ((x[(i)>>3] & (1<<((i)&7)))!=0)
#define BIT_SET(x,i) (x)[(i)>>3]|=(1<<((i)&7))
#define BIT_CLR(x,i) (x)[(i)>>3]&=(1<<((i)&7))^0xFF

 Can you explain this a little bit?

Three extraordinarily useful macros.
x is a pointer to an array of unsigned char.
i is a bit index, so to speak.
Say i is 77, which is
01001101
i >> 3 gives us 00001001, or 9.
i & 7 gives us 00000101, or 5.
So these macros respectively query, set, or clear bit 5 of byte 9 (counting from x).
A somewhat hurried explanation, so feel free to ask further.
--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
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