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Python hexadecimal comparison

Asked 2009-12-11T13:35:41.030
13

I got a problem I was hoping someone could help me figure out!

I have a string with a hexadecimal number = '0x00000000' which means:

0x01000000 = apple  
0x00010000 = orange  
0x00000100 = banana

All combinations with those are possible. i.e., 0x01010000 = apple & orange

How can I from my string determine what fruit it is? I made a dictionary with all the combinations and then comparing to that, and it works! But I am wondering about a nicer way of doing it.

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2 Answers

2

You could first of all convert your string to an integer:

s = "0x01010000"
i = int(s, 16) #i = 269484032

then, you could set up a list for the fruits:

fruits = [(0x01000000, "apple"), (0x00010000, "orange"), (0x00000100, "banana")]

for determing what fruits you have that is enough:

s = "0x01010000"
i = int(s, 16)
for fid,fname in fruits:
    if i&fid>0:
        print "The fruit '%s' is contained in '%s'" % (fname, s)

The output here is:

The fruit 'apple' is contained in '0x01010000'
The fruit 'orange' is contained in '0x01010000'
answered 2009-12-11T13:40:53.297
0
def WhichFruit(n):
    if n & int('0x01000000',16):
        print 'apple'
    if n & int('0x00010000',16):
        print 'orange'
    if n & int('0x00000100',16):
        print 'banana'

WhichFruit(int('0x01010000',16))
answered 2009-12-11T13:43:16.210

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