CODEWITHSUNDEEP
javapythonpython-2.7
Satoshi Nakamoto
Satoshi Nakamoto

Reputation: 35

toByteArray in Python?

BigInteger i = new BigInteger("1000");
Byte[] arr = i.toByteArray();
Contents of arr: [3, -24]

Since byte range in java is: minimum -128 and a maximum value of 127 (inclusive).

The java.math.BigInteger.toByteArray() returns a byte array containing the two's-complement representation of this BigInteger. The byte array will be in big-endian byte-order: the most significant byte is in the zeroth element.

How can I do this in python 2.7? Since the byte range there is 0 <= byte <= 255 e.g.

x = 1000
list = doTheMagic(x)

and as a result I get:

list = [3, -24]

Upvotes: 1

Views: 675

Answers (3)

millerdev
millerdev

Reputation: 10341

user189's answer is almost correct. Sometimes BigInteger.toByteArray() adds an extra zero byte to its result. Here's a more correct solution:

def to_byte_array(num):
    bytea = []
    n = num
    while n:
        bytea.append(n % 256)
        n //= 256
    n_bytes = len(bytea)
    if 2 ** (n_bytes * 8 - 1) <= num < 2 ** (n_bytes * 8):
        bytea.append(0)
    return bytearray(reversed(bytea))

Upvotes: 0

user189
user189

Reputation: 604

If I understood correctly what you want, something like that could do the trick.

def doTheMagic(x):
    l = []
    while x:
        m = x%256
        l.append(m if m <=127 else m-256)
        x /= 256
    return l[::-1]

x = 1000
l = doTheMagic(x)

print l

Upvotes: 0

falsetru
falsetru

Reputation: 369164

Using struct.pack and struct.unpack:

>>> struct.unpack('2b', struct.pack('>h', 1000))
(3, -24)
>>> struct.unpack('4b', struct.pack('>I', 1000))
(0, 0, 3, -24)
>>> struct.unpack('8b', struct.pack('>Q', 1000))
(0, 0, 0, 0, 0, 0, 3, -24)

Unfortunately, you can't get abitrary length byte array using struct.pack/struct.unpack.

Upvotes: 3

Related Questions