http://www.perlmonks.org?node_id=1057242

If you do nothing an infinite number of times, in the end do you end up having done anything?

Yesterday on #perl6 I had a quick talk with TimToady and others about:

loop {}

It was not the first time I saw this construct, but for the first time something bugged me about it.

At first I thought it was because of the idea that Perl 6 is supposed to be parallelism-friendly, but then I realized this is not the issue here.

I couldn't quite grasp what was the problem until I asked myself: "How long is the execution of this instruction supposed to take?"

I suppose a no-op is not instantaneous. Well, for a computer I mean. I guess it takes at least a clock cycle or something. But at least conceptually, in a way an optimizer should be aware of, such an operation is supposed to be instantaneous. Thus its execution time is zero. Thus if you execute it once, twice, ten or a thousand times, it should still be zero.

So, what if you do it an infinite number of times? Is it still zero? Infinite? In other words, what is 0 * Inf?

In maths, zero times infinity is an undefined concept. Perl 6 already knows that:

$ perl6 -e 'say 0 * Inf' NaN

Indeed we get NaN (Not A Number).

So, what happens when you ask a computer to loop a no-op? Well, the computer could actually run some code to perform the loop and just the loop, but a compiler should not request something like that, should it?

I mean if you put a no-op inside a program, like this for instance:

say 'hello';  ; say 'good bye'

The space between the two semi-colons can be considered as a no-op, right? Yet I very much doubt the compiler spends a clock cycle to execute it. More probably it just ignores it and goes directly from say 'hello' to say 'good bye'. Same if you run this:

;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;

Obviously such a program would not take twice as much time as:

;;;;;;;;;;;;;;;;;;;;;;;;;;

Because a finite sequence of no-op can reasonably be replaced by a single no-op, right?

To some extend, 'loop {}' is a way of writing an infinite sequence of semi-colons. So, why should it take seriously the no-op inside a loop? I think it shouldn't. It should just ignore it.

However, a compiler can only ignore a finite number of successive no-op. Otherwise it would face an undefined concept, just as in 0 * Inf.

That's why, if you ask me, I'd tell you that I think loop {} should die with a "undefined behavior" message. (or "singular behavior", "degenerated control structure", "infinite empty loop", ... you chose)