On 1/18/2011 8:08 AM, Christian Witts wrote:
On 18/01/2011 14:45, Tom Lin wrote:
Hi guys,
Please help me with this:
Convert an IP address from binary string to decimal format.There are
some preconditions:
1.IP address is in the form of '000010010....001100'.32 bits with no dot.
2.int(string, base) is not allowed, You have to implement the conversion .
Christian: note rule 2!
3.Performance should be considered.
For example an IP address like '11111111111111111111111111111111 '
would be converted to '255.255.255.255'
That's how I implement it. But I think it looks ugly
Tom:
What exactly do you mean by "ugly"?
and I wonder if there is a better way to do this.
What do you mean by "better"?
The algorithm you programmed is one good way to accomplish your result.
The only problems I see are using re, ** and int() which are less
efficient than other ways.
The best might be - use slicing to get the 8 "bit" substrings, == to
test for "1" and bit shifting.
Start with 0, add1 if current bit == "1" and shift left.
Then try * 2 instead of shift left.
Run timing tests on the various approaches. Given that this is a very
fast algorithm timing tests might not be useful.
import re
import sys
def convert(bin_ip):
patt = re.compile(r'\d{8}')
bin_list = patt.findall(str(bin_ip))
dec_list = []
for bin in bin_list:
sum = 0
i = 7
for n in bin:
if int(n):
sum = sum + 2**i
i = i - 1
dec_list.append(str(sum))
dec_ip = '.'.join(dec_list)
print dec_ip
if __name__ == '__main__':
bin_ip = sys.argv[1:]
convert(bin_ip)
If I knew the input to be perfect it would be a simple matter
'.'.join((str(int(input_ip[x:x+8], 2)) for x in range(4)))
--
Bob Gailer
919-636-4239
Chapel Hill NC
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