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Your response confuses me so I must be misunderstanding it. Code is code, all I need is an indirect jump to a pointer somewhere in memory to run the arbitrary c
by tinktank 8y ago
Your response confuses me so I must be misunderstanding it. Code is code, all I need is an indirect jump to a pointer somewhere in memory to run the arbitrary code I've installed to compromise the system, so why are unikernels not susceptible to this attack?
- nutjob2 8y ago"arbitrary code I've installed" How do you do that if there is no facility to load and run code? That's what he's getting at, the binary installed has fixed functionality and has no facilities to do anything but what it's programmed to do. It may have no file system nor almost any other feature that an OS has. It likely supports network ports, but they are unlikely to allow arbitrary code execution becuase again, they will have a fixed function. The only way to do what you're talking about is replacing or modifying the binary, which means you have physical access or there is a serious breakdown in security which nothing could withstand.
- tinktank 8y agoSurely the moment I have a vulnerability that allows me to execute code (via a buffer overflow or similar) I have the ability to run the code that adds that feature? In my understanding even if I have a piece of code that just loops printing "hello world" and nothing else if it's got a vulnerability that allows me to execute code all bets are off.
- close04 8y agoI guess he means that the OS cannot do more than it was intended to do. A software with a very limited set of features and functionality that you can't expand even if you wanted to. Which would mean an attacker would have to get by with only that instead of the "endless possibilities" an OS usually provides.