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by ryao
402 days ago
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If they prove a NULL check is always false, it means you have dead code. For example: if (p == NULL) return;
if (p == NULL) doSomething();
It is safe to delete the second one. Even if it is not deleted, it will never be executed.What is problematic is when they remove something like memset() right before a free operation, when the memset() is needed to sanitize sensitive data like encryption keys. There are ways of forcing compilers to retain the memset(), such as using functions designed not to be optimized out, such as explicit_bzero(). You can see how we took care of this problem in OpenZFS here: https://github.com/openzfs/zfs/pull/14544 |
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Unfortunately, we cannot have nice things because of optimizing compilers and the holy C standard.
The compiler "knows" that signed integer overflow is undefined. In practice, it just assumes that integer overflow cannot ever happen and uses this "fact" to "optimize" this program. Since signed integers "cannot" overflow, it "proves" that the condition always evaluates to false. This leads it to conclude that both the condition and the consequent are dead code.
Then it just deletes the safety check and introduces potential security vulnerabilities into the software.
They had to add literal compiler builtins to let people detect overflow conditions and make the compiler actually generate the code they want it to generate.
Fighting the compiler's assumptions and axioms gets annoying at some point and people eventually discover the mercy of compiler flags such as -fwrapv and -fno-strict-aliasing. Anyone doing systems programming with strict aliasing enabled is probably doing it wrong. Can't even cast pointers without the compiler screwing things up.