Thread

1. JoshTriplett · 1789746575 · focus · HN ↗

   > When both conditions are met, the loop body is replaced with a call to std::this_thread::yield().

   Insert screaming here.

   An infinite loop, with no library calls whatsoever, gets a system call inserted. That's a horrible surprise waiting to happen.

   The entire concept of the "forward progress guarantee" is broken. An infinite loop should compile to an infinite loop. Nothing more, nothing less.

   1. marcosdumay · 1789751761 · focus · HN ↗

      I don't understand your problem. Did you expect your C++ program to get uninterrupted access to the computer? What progression do you think isn't happening there?

      I think you are misinterpreting that. That phrase unambiguously says the loop is preserved on the final binary.

      1. JoshTriplett · 1789751975 · focus · HN ↗

         I expect an infinite loop to be compiled into, for instance, a jump instruction jumping to itself. The OS, if there is any, is welcome to interrupt and context switch. I don't expect code that has no function calls at all to have a system call inserted into it.

         1. leni536 · 1789756432 · focus · HN ↗

            A call to a standard library function is still subject to the as if rule. It doesn't have to manifest into a call instruction to a standard library function. Much like memcpy in source code doesn't have to manifest to a call instruction.

            1. fc417fc802 · 1789768479 · focus · HN ↗

               Me: I don't expect to be stabbed.

               You: But you only might be stabbed. It isn't required to happen only permitted.

               1. leni536 · 1789832456 · focus · HN ↗

                  A bad but conforming implementation of the standard can screw you over on every line of your program.

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