We're going in circles here. You keep claiming that a compiler can parallelise an expresssion if EO is defined. Which is backwards to everyone else. You're going to need to go into more details in your explanation for anyone to buy in.

Using the func1/func2 example above, we define an order: left-to-right makes as much sense as anything else. This means that func1 is evaluated, then func2 is evaluated. That's a defined execution order. There may be others, but ltr and rtl are the only ways I can think of defining these things.

And this makes defined execution order serial.

[W]ith defined execution order, the compiler is not free to "do this is any order"

Is this an admission (finally) that defined EO prevents parallelism, and any optimisation that the compiler could elicit from such parallel computation? That is all I've been saying: defined EO == serial; undefined EO permits parallel. I think you've been grossly misinterpreting everything in this thread if you think anyone has been saying anything other than this, except you.

In reply to Re^33: Why is the execution order of subexpressions undefined? (magic ruts) by Tanktalus
in thread Why is the execution order of subexpressions undefined? by BrowserUk

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