How to convert a hexstring to shellcode format?

I have a bytes key define as:

KEY = b’\xb5\x89\xd5\x03\x03\x96`\x9dq\xa7\x81\xed\xb2gYR’

I want this to be formatted like shellcode, i.e two hexa characters like: \x41\x42\x43…

So I tried to do it like this:

KEY = b'\xb5\x89\xd5\x03\x03\x96`\x9dq\xa7\x81\xed\xb2gYR'
hexkey = KEY.hex()
l = []

for i in range(0, len(hexkey) - 2, 2):
    b = '\\x' + hexkey[i] + hexkey[i+1]
    l.append(b)

but this isn’t escaping the backslash for some reason, I get this output:

[‘\\xb5’, ‘\\x89’, ‘\\xd5’, ‘\\x03’, ‘\\x03’, ‘\\x96’, ‘\\x60’,
‘\\x9d’, ‘\\x71’, ‘\\xa7’, ‘\\x81’, ‘\\xed’, ‘\\xb2’, ‘\\x67’,
‘\\x59’]

what am i doing wrong? Is there a better way to do this?

>Solution :

.join() and print them. You can iterate over the bytes directly as well:

KEY = b'\xb5\x89\xd5\x03\x03\x96`\x9dq\xa7\x81\xed\xb2gYR'
print(''.join(f'\\x{b:02x}' for b in KEY))

Output:

\xb5\x89\xd5\x03\x03\x96\x60\x9d\x71\xa7\x81\xed\xb2\x67\x59\x52