generate random long in python OR convert str to long

Question

I need to generate random variable type=long, so it should looks like:x=0xabcdef1234567890

EDIT:

>>> x='0xa123456789abcdef'
>>> type(x)
<type 'str'>
>>> long(x, 16)
11611200575284956655L         
>>> x=0xa123456789abcdef       
>>> type(x)
<type 'long'>

You're looking at a hexadecimal number (represented by prefixing with `0x...`): https://en.wikipedia.org/wiki/Hexadecimal — Reut Sharabani, Sep 09 '15 at 20:34
possible duplicate of [Convert hex string to int in Python](http://stackoverflow.com/questions/209513/convert-hex-string-to-int-in-python) — 301_Moved_Permanently, Sep 09 '15 at 20:35
`x=0xabcdef1234567890` is creating a long ... what does the string have to do with anything ... — Joran Beasley, Sep 09 '15 at 20:36
the question is: How to generate randomly variable with type=long ? — bay, Sep 09 '15 at 20:47
Since long has unlimited precision, you cannot generate a random value with uniform distribution. — stark, Sep 09 '15 at 20:50

taesu · Answer 1 · 2015-09-09T20:53:39.813

0

x = 0xabcdef1234567890
type(x) #long
x = 51924361L
type(x) #long
x = 0xDEFABCECBDAECBFBAEl
type(x) #long
x = -4721885298529L
type(x) #long

x = long(1)
type(x) #long


def generate_random_long():
    import random
    return long(random.choice(range(0,10000000)))

def generate_random_long_from_string():
    import random
    return long('0x'+''.join(random.choice('0123456789abcdef') for n in xrange(12)),16)

edited Sep 09 '15 at 20:53

answered Sep 09 '15 at 20:38

taesu

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how to get value (ex: x = 0xabcdef1234567890) randomly? – bay Sep 09 '15 at 20:43
1

Note that `long()` does not exists anymore in python3. Use `int()` instead. – 301_Moved_Permanently Sep 09 '15 at 20:45

generate random long in python OR convert str to long

1 Answers1