Get nth byte of integer

Question

I have the following integer:

target = 0xd386d209
print hex(target)

How can I print the nth byte of this integer? For example, expected output for the first byte would be:

0x09

Answer 1

You can do this with the help of bit manipulation. Create a bit mask for an entire byte, then bitshift that mask the number of bytes you'd like. Mask out the byte using binary AND and finally bitshift back the result to the first position:

target = 0xd386d209
bit_index = 0
mask = 0xFF << (8 * bit_index)
print hex((target & mask) >> (8 * bit_index))

You can simplify it a little bit by shifting the input number first. Then you don't need to bitshift the mask value at all:

target = 0xd386d209
bit_index = 0
mask = 0xFF
print hex((target >> (8 * bit_index)) & mask)

Answer 2

def byte(number, i):
    return (number & (0xff << (i * 8))) >> (i * 8)

Answer 3

>>> def print_n_byte(target, n):
...     return hex((target&(0xFF<<(8*n)))>>(8*n))
... 
>>> print_n_byte(0xd386d209, 0)
'0x9L'
>>> print_n_byte(0xd386d209, 1)
'0xd2L'
>>> print_n_byte(0xd386d209, 2)
'0x86L'