Python 2.7 concat byte

Question

I've two integer of one byte, so each integer can be in the range 0-255. Suppose

    a = 247
    b = 8

Now the binary representation of each is:

   a = 11110111
   b = 00001000

What I need are the following operations:

1) Concat those two binary sequence, so following the example, "a" concat "b" would result in:

 concat = 1111011100001000 -> that is 63240

2) Consider only the first n most significative bits, suppose the first 10 bits, this would resoult in:

 msb = 1111011100000000

3) Revert the bits order, this would result in:

 reverse = 0001000011101111

For the question number 2 I know it's just a shift operation, for the 3 I've guessed to convert in hex, then swap and then convert to integer, but I guess also that there are more elegant way.

For the number 1, I've wrote this solution, but I'm looking for something more elegant and performant:

 a = 247
 b = 8

 first = hex(a)[2:].zfill(2)
 second = hex(b)[2:].zfill(2)

 concat = int("0x"+first+second,0)

Thank you

Answer 1

Those operations are basic bit manipulations, with the exception of the last step:

1. 247, left shift by 8 bits (= move to higher byte)
2. keep 8 as it is (= lower byte)
3. add those results
4. bit-and with a mask that zeroes out the last 6 bits
5. reverse the bit order (not a bit operation; more methods here)

In Python 2.7 code:

a = 247 << 8
print "a = {0: >5} - {1:0>16}".format(a, bin(a)[2:])

b = 8
print "b = {0: >5} - {1:0>16}".format(b, bin(b)[2:])

c = a + b
print "c = {0: >5} - {1:0>16}".format(c, bin(c)[2:])

d = c & ~63  # 63 = 111111, ~63 is the inverse
print "d = {0: >5} - {1:0>16}".format(d, bin(d)[2:])

e = int('{:08b}'.format(d)[::-1], 2)
print "e = {0: >5} - {1:0>16}".format(e, bin(e)[2:])

Output

a = 63232 - 1111011100000000
b =     8 - 0000000000001000
c = 63240 - 1111011100001000
d = 63232 - 1111011100000000
e =   239 - 0000000011101111