in reply to Re: $var == 1 fails when $var = 1 (lost post)
in thread $var == 1 fails when $var = 1

Well once again

Why ?

Does that link actually answer either of the 2 issues raised by the OP ?
Those 2 issues can be paraphrased as:

1) Why does perl assert that $x is 1 when $x != 1 ?
2) How is it that $x+$y+$z != $x+$z+$y ?

Cheers,
Rob
  • Comment on Re^2: $var == 1 fails when $var = 1 (lost post)

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Re^3: $var == 1 fails when $var = 1 (lost post)
by LanX (Saint) on Sep 21, 2018 at 15:41 UTC
    primarily b/c I was worried about the lost post.

    Cheers Rolf
    (addicted to the Perl Programming Language :)
    Wikisyntax for the Monastery FootballPerl is like chess, only without the dice

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