How do I find whether a number is positive, negative or zero without using if or for?

Question

I want to implement the sign and zero flag setting in microprocessor. So I need to write a function to find whether the number is positive, negative or zero without using if or for loops and also only Boolean and bitwise operators are allowed. I did the following. But how to I implement it for zero condition ?

int status (int x) {
    int sign = (x >> 31);
    return sign;
}

Any suggestions ?

Also related: http://stackoverflow.com/questions/3912375/check-if-a-number-x-is-positive-x0-by-only-using-bitwise-operators-in-c — jogojapan, May 24 '13 at 05:28
@jogojapan...I know this Q. is similar to your suggestion, but he want to use only boolean operator. — Navnath Godse, May 24 '13 at 05:31
@Navnath The question says boolean _and_ bitwise operators are allowed, or am I reading this wrongly? — jogojapan, May 24 '13 at 05:32
@noufal ok although I love devnull answer, but let me think using bitwise op — Grijesh Chauhan, May 24 '13 at 05:33
a book called "hacker's delight",has a lot of example about these trick including yours. — MYMNeo, May 24 '13 at 06:09
I got it, It was quite simple It can be easly done by `return ((!!x) | (x >> 31));` — noufal, May 25 '13 at 15:55

score 7 · Answer 1 · answered May 24 '13 at 05:16

7

The following will return -1 for negative values, 0 for zero, 1 for positive values of x.

int status (int x) {
  int sign = (x > 0) - (x < 0); 
  return sign;
}

answered May 24 '13 at 05:16

devnull

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+ nice suggestion, but OP need via Boolean, do you have any suggestion? – Grijesh Chauhan May 24 '13 at 05:23
devnull I can't use `greater than (>)` or `less than (<)` operators.. – noufal May 24 '13 at 05:26

score 2 · Answer 2 · answered May 24 '13 at 06:30

2

Is this enough for your purpose?

int status(int val)
{
    int minus = (val&0x80000000)&&TRUE;
    int pos = (val&0x7FFFFFFF)&&(!(val&0x80000000));
    int r = 0 - minus + pos;
    return r;
}

answered May 24 '13 at 06:30

raj raj

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score 0 · Answer 3 · answered May 24 '13 at 07:02

0

Try this

int status (unsigned no) {
  int sign = 0;

   // If Zero             // If -Ve = -1 OR If +Ve = -2
   (sign = ( no | 0 )) && ( sign = (~( ( (no << 1) >> 1 ) == no)) );

   // If +Ve
   (sign == -2) && (sign = 1);

  return sign;
}

answered May 24 '13 at 07:02

Navnath Godse

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I got it, It was quite simple It can be easly done by `return ((!!x) | (x >> 31));` – noufal May 25 '13 at 15:57
@noufal...Yes your code is working correct but it is not portable. It will only work for 32 bit. – Navnath Godse May 28 '13 at 05:33
@Navanth yes... that's correct.. but here its ok for me.. do you know how to do otherwise ? – noufal May 28 '13 at 14:45
@noufal...You can use `sizeof` operator as -> `return ((!!x) | (x >> (sizeof(int)*8) - 1));` but you need to compile for respective platform to get correct result. – Navnath Godse May 29 '13 at 04:16

How do I find whether a number is positive, negative or zero without using if or for?

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