bin2chr("01110011") # = 15 function ?

gcmartijn at gmail.com gcmartijn at gmail.com
Mon Feb 25 21:10:27 CET 2008


H!

I'm searching for a simple
bin2chr("01110011") function that can convert all my 8bit data to a
chr so I can use something like this:

def bin(i):
    l = ['0000', '0001', '0010', '0011', '0100', '0101', '0110',
'0111',
         '1000', '1001', '1010', '1011', '1100', '1101', '1110',
'1111']
    s = ''.join(map(lambda x, l=l: l[int(x, 16)], hex(i)[2:]))
    if s[0] == '1' and i > 0:
        s = '0000' + s
    return s

print ord("s") # = 115
print bin(ord("s")) # = 01110011

test= open('output.ext,'wb')
test.write(bin2chr("01110011"))
test.write(bin2chr("01111011"))
test.write(bin2chr("01110111"))
test.close()

I think this is very easy but I'm not very good in python.
In a other program I use I do something like this

Function bin2int(b$)
    blen=Len(b)
    For f=1 To blen
        n=n Shl 1 + (Mid(b,f,1)="1")
    Next
    Return n
End Function



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