How would shallow generators compose with lambda?

Brendan Eich brendan at mozilla.com
Thu May 14 15:42:12 PDT 2009


On May 14, 2009, at 2:50 PM, Mark S. Miller wrote:

>> If so, did you have a different way of reasoning about the reasons  
>> today why
>> finally might not run that I mentioned (iloop detection or other  
>> hard stop)?
>
> Those hard stops kill all further activity within that event loop.
> Once a universe has been destroyed, no further bad things can happen
> in that universe.

There's always the next universe (new event starts another control  
flow). Life goes on, in the JS serial multiverse, and those finally  
clauses failed to run even though control abruptly left the lambda  
under the hypothesis.


> Infinite loops don't kill their universe, so this case is similar but
> different. As the halting problem teaches us, an infinite loop is
> generally indistinguishable from a loop that hasn't terminated yet.
> Since control flow has not yet escaped the loop, it hasn't yet
> bypassed the finally, and so no invariants have yet been violated.

True enough, but see above.

This isn't entirely academic, since information leaks include  
termination channels.

/be


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