Transform bits into byte series

Question

Given a Python integer which is within the size of 4 bits, how does one transform it – with bitwise arithmetic instead of string processing – into an integer within the size of 4 bytes, for which each bit in the original corresponds to a byte which is the bit repeated 8 times?

For example: 0b1011 should become 0b11111111000000001111111111111111

Scott Griffiths · Accepted Answer · 2011-02-21 09:59:31Z

4

With apologies to ncoghlan:

expanded_bits = [
    0b00000000000000000000000000000000,
    0b00000000000000000000000011111111,
    0b00000000000000001111111100000000,
    0b00000000000000001111111111111111,
    0b00000000111111110000000000000000,
    0b00000000111111110000000011111111,
    0b00000000111111111111111100000000,
    0b00000000111111111111111111111111,
    0b11111111000000000000000000000000,
    0b11111111000000000000000011111111,
    0b11111111000000001111111100000000,
    0b11111111000000001111111111111111,
    0b11111111111111110000000000000000,
    0b11111111111111110000000011111111,
    0b11111111111111111111111100000000,
    0b11111111111111111111111111111111,
    ]

Then just index this list with the nibble you want to transform:

>>> bin(expanded_bits[0b1011])
"0b11111111000000001111111111111111"

edited Feb 21, 2011 at 9:59

Scott Griffiths

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answered Feb 21, 2011 at 3:37

senderle

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6 Comments

senderle Over a year ago

@John Machin -- OK, sure. But just so I understand, why?

Keith Over a year ago

@senderle table lookups are much faster.

senderle Over a year ago

@Keith, was what I had before John Machin's edit not a lookup table?

Keith Over a year ago

@senderle I don't know what you had before. Perhaps I'm jumping into the middle of something here? Now that I take a closer look I think I answered to wrong question.

John Machin Over a year ago

@senderle: Less whitespace, looks better. You can view previous versions by clicking on the edited 1 hour ago link.

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Raph Levien · Accepted Answer · 2011-02-21 02:38:31Z

1

I'd just do a loop:

x = 0b1011
y = 0
for i in range(4):
    if x & (1 << i):
        y |= (255 << (i * 8))
print "%x" % y

answered Feb 21, 2011 at 2:38

Raph Levien

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1 Comment

ncoghlan Over a year ago

And, having done that once, I would cache the result in a 16 entry lookup table.

Adeel Zafar Soomro · Accepted Answer · 2011-02-21 06:56:03Z

0

The following recursive solution uses only addition, left/right shift operators and bitwise & operator with integers:

def xform_rec(n):
    if n == 0:
        return 0
    else:
        if 0 == n & 0b1:
            return xform_rec(n >> 1) << 8
        else:
            return 0b11111111 + (xform_rec(n >> 1) << 8)

Or, as a one-liner:

def xform_rec(n):
    return 0 if n == 0 else (0 if 0 == n & 0b1 else 0b11111111) + (xform_rec(n >> 1) << 8)

Examples:

>>> print bin(xform_rec(0b1011))
0b11111111000000001111111111111111
>>> print bin(xform_rec(0b0000))
0b0
>>> print bin(xform_rec(0b1111))
0b11111111111111111111111111111111)

edited Feb 21, 2011 at 6:56

answered Feb 21, 2011 at 3:38

Adeel Zafar Soomro

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