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

In reply to Re^2: $var == 1 fails when $var = 1 (lost post) by syphilis
in thread $var == 1 fails when $var = 1 by mjlush

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