python - If statement 'not working' despite conditions being true -

currently going through bin file full of hex data process, match i'm using try , find group of 3 hex bytes in file aren't working correctly. values identical not printing string i've got set confirm match, @ present i'm trying match first 3 bytes know works etc. code follows:

match1 = "\\x00\\n\\x00" print ("match1 ="+match1)  if bytedata == match1:     print ("byte string 030791 found!") elif bytedata == match1:     print ("byte string 050791 found!") exit() 

the value of bytedata '\x00\n\x00' script ignores , moves exit statement. file being opened follows :

file = open('file.bin', 'rb') while true: byte = file.read(3) 

when printing value of byte reports "\x00\n\x00" have ideas why match isn't working properly?

match1 not contain 3 bytes. contains 10:

>>> match1 = "\\x00\\n\\x00" >>> len(match1) 10 

you escaped escape sequences, \\x00 four bytes, \ backslash, letter x followed 2 0 digits.

remove backslash escapes:

match1 = "\x00\n\x00" 

don't try print directly; terminals won't make nulls visible, newline. use repr() function produce debug output looks python string can reproduce value in code or interactive interpreter:

print ("match1 =", repr(match1)) 

this how interactive interpreter shows expression results (unless produced none):

>>> match1 = "\x00\n\x00" >>> len(match1) 3 >>> match1 '\x00\n\x00' >>> print("match1 =", repr(match1)) match1 = '\x00\n\x00' 

next, if using python 3, you'll still won't have match because opened file in binary mode , getting bytes objects, match1 variable str text sequence. if want 2 types match you'll either have convert (encode text or decode bytes), or make match1 bytes object start with:

match1 = b'\x00\n\x00' 

the b prefix makes bytes literal.