Bloat.py Python Code Obfuscation picoCTF 2022

[@rasmusmose8059]

There are two sys.exit(0) in the program, so only two ways, the program can terminate. The first sys.exit(0) is where you found the password. Here I added 1==1 or

[@SanjinDedic]

sorry can you elaborate a little bit, I can see the 2 different sys.exit() points on line 11 and 40 but I don't yet understand how they can be used to crack this challenge

[@rasmusmose8059]

@@SanjinDedic You know, that if you write the password wrong, then the program will terminate. This can only happen in two places; at line 11 or at the end of the program. At the end of the program it's supposed to stop, so what you have to do is to avoid the program to stop at line 11. This can be done by making the first part of the if-statement true. E.g. change line 5 to if True or arg432 == a[71]+a[64]+a[79]+a[79]+a[88]+a[66]+a[71]+a[64]+a[77]+a[66]+a[68]: