Before the update oiskuu says in the node above that $InfStr "is not the while condition here. (emphasis supplied)"

Strictly speaking that's true; in fact $InfStr isn't a condition at all, but that misses the point.

But that's not what my preceeding node says (omitting the detail that the condition is while(InfStr() =~ /(a)/g){).

The condition is InfStr() =~ /(a) and it IS always true because the sub always returns a value in which the test is true (My update: And that's pretty much the explanation espoused in oiskuu's "Update" above. And, oh yes, I added the initial phrase in this node to make the thread more readable.)

In reply to Re^3: Loop Quandary by ww
in thread Loop Quandary by joshpk105

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