Comment by mitxela
10 hours ago
I thought it was generally so the compiler can merge two computation loops without proving if one of them runs forever .
10 hours ago
I thought it was generally so the compiler can merge two computation loops without proving if one of them runs forever .
Correct. See N1528: "Why undefined behavior for infinite loops?" https://www.open-std.org/jtc1/sc22/wg14/www/docs/n1528.htm