3 ms·
Treating otherwise unsafe code as safe because nontermination makes it dead doesn't strike me as particularly useful. Is there any other case where knowing the
by moonchild 6y ago
Treating otherwise unsafe code as safe because nontermination makes it dead doesn't strike me as particularly useful. Is there any other case where knowing the termination status of a function is helpful in proving safety?
- adwn 6y ago> Treating otherwise unsafe code as safe because nontermination makes it dead doesn't strike me as particularly useful. It doesn't matter whether it's useful or not. What matters is that the compiler can't determine whether it's safe or not.