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mitxelatoday at 12:30 AM1 replyview on HN

I thought it was generally so the compiler can merge two computation loops without proving if one of them runs forever .


Replies

murderfstoday at 1:56 AM

Correct. See N1528: "Why undefined behavior for infinite loops?" https://www.open-std.org/jtc1/sc22/wg14/www/docs/n1528.htm