Re^2: Loop Quandary

$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
[download]

Returning a reference from the sub can work around the problem:

  my $x = "ababa";
  sub foo { \$x; }
  print join " ", @-, @+ while ${foo()} =~ m/a/g;
[download]

I hope this explains things a little bit better.

Comment on Re^2: Loop Quandary Select or Download Code

Replies are listed 'Best First'.
Re^3: Loop Quandary by ww (Archbishop) on Nov 22, 2013 at 17:03 UTC
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.)	[reply] [d/l] [select]

Replies are listed 'Best First'.

Re^3: Loop Quandary
by ww (Archbishop) on Nov 22, 2013 at 17:03 UTC

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.)

[reply]
[d/l]
[select]