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