My guess is in the first loop the match happens on the same string each time, so you'll find each a in turn then exit the loop

In the second example, you call InfStr on every cycle of the loop, which makes it a regex against a new string, so it'll find the first a in the string each time and never progress

Nice explanation. Interestingly, adding the empty prototype to the sub definition and removing return changes the behaviour, as Perl optimizes the sub away:

sub InfStr () { 'ababa' }
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which is equivalent to

use constant INFSTR => 'ababa';
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لսႽ† ᥲᥒ⚪⟊Ⴙᘓᖇ Ꮅᘓᖇ⎱ Ⴙᥲ𝇋ƙᘓᖇ

"while" repeats the statement *while* the condition is true.

$InfStr() is not the while condition here.

Update:
The reason first loop ends is because $str stores the necessary magic for m//g.
Second loop is infinite since InfStr() returns a fresh value each iteration.

$ perl -MDevel::Peek -e'(my $x = "ababa") =~ m/a/g; print Dump($x);'
SV = PVMG(0x7b65a0) at 0x786680
  REFCNT = 1
  FLAGS = (PADMY,SMG,POK,pPOK)
  IV = 0
  NV = 0
  PV = 0x778780 "ababa"\0
  CUR = 5
  LEN = 8
  MAGIC = 0x780ee0
    MG_VIRTUAL = &PL_vtbl_mglob
    MG_TYPE = PERL_MAGIC_regex_global(g)
    MG_LEN = 1
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Returning a reference from the sub can work around the problem:

  my $x = "ababa";
  sub foo { \$x; }
  print join " ", @-, @+ while ${foo()} =~ m/a/g;
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I hope this explains things a little bit better.

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.)