bin2chr("01110011") # = 15 function ?
gcmartijn at gmail.com
gcmartijn at gmail.com
Mon Feb 25 15:10:27 EST 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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