in reply to while loops


$num=20; $while($num<30){ print "num is $num\n"; $num=$num+1;\n ; }
should be

$num=20;<BR> while($num<30){ print "num is $num\n"; $num+=$num;\n; }

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Re: Re: while loops
by rob_au (Abbot) on Jan 11, 2002 at 18:41 UTC
    No, not at all - Your code will only pass through a single iteration of the loop as on the first pass, the value of $num is added to $num. The result is that $num equals 40 and the while loop is exited. I think you may have been meaning, ignoring formatting syntax errors ...

    $num = 20; while ($num < 30) { print "num is $num\n"; ++$num; }

    Although, good pick up on the syntax error with the line - $while($num<30){ - Although this had been addressed previously within this thread.


    perl -e 's&&[@.]/&&s&.com.&_&&&print'

      yes , thanks

      in this case would $num++ also work

Re^2: while loops
by lvanhout (Curate) on Jun 15, 2004 at 17:42 UTC
    Still not there.

    $num=20; while($num<30) { print "num is $num\n"; <b><u>$num+=$num;\n</b></u> } #I think you ment: $num += 1; #Though these should work also: $num++; ++$num;
    There is also an extra "\n;" in the code in the reply.