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How could I use bit masking to make all the bits in a number 1 if they were all 0, and all 0 if they were not?

Using an unsigned variable:

So, if I have 0000-0000, I would like that to become 1111-1111. If I have 0101-0110 (or 0000-0001, or 1111-1111, etc), I would like that to become 0000-0000.

Is this possible to do without using any conditional?

MirroredFate
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  • What language? Also, "0000-0000" looks like a string, not an unsigned integer.. In any case, look at the documentation in your language for bit complement (~ in C/C++). – Matt Aug 23 '13 at 16:42
  • I put 0000-0000 for readability... – MirroredFate Aug 23 '13 at 16:54
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    So basically you want this code without the "if"?: `if value == 0: result = 255; else: result = 0` – user9876 Aug 23 '13 at 16:55

3 Answers3

3

Sure, it's possible:

int y = 0xff;
y = ~(y & 1 | y>>1 & 1 | y>>2 & 1 | ...) - 1

But unless this is an academic exercise, you really shouldn't. If you're concerned about performance, y = y != 0 is almost certainly faster.

Explanation:

y & 1 takes the first bit of the number. y >> k shifts the number right by k bits, allowing us to get that bit by y >> k & 1. We simply | them together, which results in one if any bit is set or zero if not. Subtracting 1 gives us 0 if any bit was set, and -1 if not. The binary representation of -1 is 1111...

Shift:

1010 - y
1010 - y >> 0
 101 - y >> 1
  10 - y >> 2
   1 - y >> 3

Take the first bit:

   0 - y >> 0 & 1
   1 - y >> 1 & 1
   0 - y >> 3 & 1
   1 - y >> 4 & 1

Or them: 

   1 - 0 | 1 | 0 | 1

Negate:

0000 - 1-1
Kevin
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2

Not in an efficient way probably.

If you really want you can maybe:

int mask = 0;
int result = 0;


for(int i = 0; i < sizeof(number) * 8; i++)
{
    mask |= number & 1 << i;
}


for(int i = 0; i < sizeof(number) * 8; i++)
{
    result |= mask & 1 << i;
}

~result is your answer.

TomF
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0

How about this:

def check_for_zero(value):
    # Same as "if value == 0: return 0; else: return 1"
    # value must be an 8-bit number.

    # Need to check all 8 bits of value.  But we can compress it...
    x = value | (value >> 4)
    # Now just need to check the low 4 bits of x.  Compress again...
    x = x | (x >> 2)
    # Now just need to check the low 2 bits of x.  Compress again...
    x = x | (x >> 1)
    # Now just need to check the low 1 bit of x.  Success!
    return x & 1

def expand_to_8bit(bit):
    # Same as "if bit == 0: return 0; else: return 255"
    # Must pass bit == 0 or bit == 1

    # bit is a 1-bit value.  Expand it...
    x = bit | (bit << 1)
    # x is a 2-bit value.  Expand it...
    x = x | (x << 2)
    # x is a 4-bit value.  Expand it...
    x = x | (x << 4)
    # x is a 8-bit value.  Done!
    return x

def foo(value):
    x = check_for_zero(value)
    x = x ^ 1  # Flips the bit
    return expand_to_8bit(x)
user9876
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  • Sorry. I may have been misleading when I replied that "just no if statement" was correct. I don't want any conditional whatsoever. There should be no check on the value of the number. – MirroredFate Aug 23 '13 at 17:05
  • The only checks are the "assert" statements, which you can take out. They're just there so if you pass 1000000 to this code, it blows up obviously rather than doing the wrong thing. – user9876 Aug 23 '13 at 17:07
  • OK, I removed the asserts for clarity. – user9876 Aug 23 '13 at 17:09