Converting int to byte streams in python

Question

I am new to python so please bear with if it sounds a novice question. I looked it up but different sources tell me different things so I decided to ask it here. And doc is not very clear.

I have an int that I want to convert to bytes.

#1024 should give me 0000010000000000
print bytes([1024])
[1024]

If I use bytes() I assume I get a list since when I print I end up getting [1024]. However if I do

print bytes([1024])[0] 
[

I get [ back. Why is this? So it does not return a list?

Is there a way to get the byte streams back given an integer? Ideally I want something as follows:

x = tobyte(1024)
print x[0] #should give me  0  00000000
print x[1] #should give me  8  00000100

I need to be able to use x[0] elsewhere in the code i.e. to be passed to base64 encode where I expect my data to be 64 bits

It's depend how you want to code your bytes (little endian or big). You want the string representation or the bytes representation ? — Pierre Turpin
– Pierre Turpin, Commented Jun 3, 2014 at 19:53
Note that bytes is just another name for the str type, and str([1024]) just produces a string representing the list: '[1024]'. — chepner
– chepner, Commented Jun 3, 2014 at 20:09

chepner · Accepted Answer · 2014-06-03 19:53:25Z

2

To get the individual bytes from an integer, use struct:

individual_bytes = struct.unpack("BB", struct.pack("<H", 1024))

First, "<I" packs the integer as a 16-bit value, using little-endian ordering. Then "BB" unpacks the byte string into two separate 8-bit values.

answered Jun 3, 2014 at 19:53

chepner

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3 Comments

Sven Marnach Over a year ago

You certainly did the better job decrypting this question. :)

as3rdaccount Over a year ago

Thanks chepner. Since my data is supposed to be 64 bits is there a way to unpack it without specifying 8 Bs (for cleaniliness reasons)? i.e. struct.unpack("BBBBBBBB", struct.pack("<L", 1024))

Steve Barnes Over a year ago

struct.unpack("8B", struct.pack("<L", 1024)) will unpack into an 8 byte array.

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