bitwise operators

This is a discussion on bitwise operators within the C Programming forums, part of the General Programming Boards category; Hi, I need to write a function that will return 1 if x is > 0 otherwise return 0 using ...

  1. #1
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    bitwise operators

    Hi,

    I need to write a function that will return 1 if x is > 0 otherwise return 0 using only the following operations and no loops:
    ! ~ & ^ | + << >>


    My code works for positive and negative numbers, but does not work for 0. How can I evaluate 0 as being not greater then 0?

    Here is my code:
    Code:
    int isPositive(int x) {
    
    x = x >> 31;
    x = x & 1;
    x = !x;
    
    
      return x;
    }

    I'm probably missing something obvious, but no matter which way I try it, 0 always comes back as 1.

    Any ideas?

  2. #2
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    Hello,

    This is becuase the line: x = !x changes your value before returning the answer. Let's take a look step by step as x = 0:

    "0 = 0 >> 31;" = 0
    "0 = 0 & 1;" = 0
    "0 = !0;" = 1
    "return 1;"


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    Segmentation Fault: I am an error in which a running program attempts to access memory not allocated to it and core dumps with a segmentation violation error. This is often caused by improper usage of pointers, attempts to access a non-existent or read-only physical memory address, re-use of memory if freed within the same scope, de-referencing a null pointer, or (in C) inadvertently using a non-pointer variable as a pointer.

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    Right, 0 should be 0, not 1. But if I shift 0 to the right it is going to have the same affect as a positive number.

  4. #4
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    It's supposed to do that, stack overflow otherwise it would return the wrong answer for any input. It only returns the wrong answer for zero.

  5. #5
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    You used the '=' operator. Doesn't that violate the conditions?
    If you understand what you're doing, you're not learning anything.

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    I can use = I can't use ==

  7. #7
    Just Lurking Dave_Sinkula's Avatar
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    Code:
       return !!x & !((x >> 31) & 1);
    The !!x will be 0 for zero and 1 for any nonzero value of x.
    Last edited by Dave_Sinkula; 02-17-2005 at 03:08 PM.
    7. It is easier to write an incorrect program than understand a correct one.
    40. There are two ways to write error-free programs; only the third one works.*

  8. #8
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    Code:
    #include <stdio.h>
    
    int ispositive(int x)
    {
      return !((x >> 31) & 1) & !!x;
    }
    
    int main(void)
    {
      int i;
    
      for(i = -3;i < 4;++i)
        printf("%d - %d\n", i, ispositive(i));
    
      return 0;
    }
    My output:
    Code:
    -3 - 0
    -2 - 0
    -1 - 0
    0 - 0
    1 - 1
    2 - 1
    3 - 1
    EDIT: Bah! Dave_Sinkula beat me to it
    If you understand what you're doing, you're not learning anything.

  9. #9
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    Code:
    int isPositive(int x) 
    {
       return ((x + 0x7FFFFFFF) >> 31) & 1;
    }
    hehe i cheated.
    Last edited by Brian; 02-17-2005 at 03:12 PM.

  10. #10
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    You used '=' too, Brian!
    If you understand what you're doing, you're not learning anything.

  11. #11
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    he said we're allowed.

  12. #12
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    So he did. I missed that post. Sorry.
    If you understand what you're doing, you're not learning anything.

  13. #13
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    Quote Originally Posted by itsme86
    So he did. I missed that post. Sorry.
    there I fixed it :P

  14. #14
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    = yes, == no.

    she not he

  15. #15
    C Programmer Stack Overflow's Avatar
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    Heh,

    How about these:
    Code:
    #define isPositive(x) ((x + 0x7FFFFFFF) >> 31 & 1)
    Code:
    #define isPositive(x) !!x & !((x >> 31) & 1)
    Made to make itsme86 happy.


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    Segmentation Fault: I am an error in which a running program attempts to access memory not allocated to it and core dumps with a segmentation violation error. This is often caused by improper usage of pointers, attempts to access a non-existent or read-only physical memory address, re-use of memory if freed within the same scope, de-referencing a null pointer, or (in C) inadvertently using a non-pointer variable as a pointer.

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