in reply to printing options with elsif statement

You're confusing equality (==) with assignment (=) - an easy mistake to make even for experienced programmers.

if ($foo == 3) checks if $foo is equal to 3.

if ($foo = 3) gives $foo a value of 3. This assignment then itself evaluates to 3, which is "true", so the if statement is always true.

dave hj~

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Re: Re: printing options with elsif statement
by splatohara (Novice) on Feb 18, 2002 at 17:34 UTC
    ok I have changed it to check whether the variable $modtyp is x, 
    $modtyp = 9;
if ($modtyp == 9)
{
print " response1";
}
elsif ($modtyp == 10)
{
print " response2\n";
}

the reponses are not being typed even if the value of $modtyp
can be seen to be set to a value being checked!!!

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[d/l]
      Well, that code snippet works... I just tried it.

      dash2@cherokee:~ > perl


$modtyp = 9;
if ($modtyp == 9)
{
print " response1";
}
elsif ($modtyp == 10)
{
print " response2\n";
}
 response1dash2@cherokee:~ >

      [download]
      So, as Sherlock Holmes said, once you have eliminated the impossible, whatever remains, however improbable, must be true. Check elsewhere in your code. Check you are printing newlines, cos sometimes perl doesn't print out until you hit a newline. Check you are printing to where you think you are - did you select some other filehandle? Check that $modtyp is exactly what you think it is by doing print "modtyp is '$modtyp'". Debug. Fix. Good luck.

      dave hj~

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