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C++26: Trivial infinite loops are no longer undefined behaviour

173 points · 290 comments · ibobev

  1. JoshTriplett · · 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. ameliaquining · · focus · HN ↗
      I'm curious, what exactly do you imagine going wrong here?
      1. yk · · focus · HN ↗
        I would expect an infinite loop

            while(true) std::this_thread::yield(); 
        
        to be designed to play nice with the scheduler, while I would assume a infinite loop

            while(true);
        
        to not play nice with the scheduler. Now, I can't really imagine where this matters except for horrible hacky attempts at faking a real time scheduler on windows, but breaking horrible hacky attempts at faking a real time scheduler sounds like the kind of bug you hear about in the evening news.
        1. nicebyte · · focus · HN ↗
          under what conditions would the infinite loop be scheduled over something else after it has run out of its time slice?
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