newbie question

Discussion in 'Perl Misc' started by kaptain kernel, Sep 26, 2003.

  1. sorry , i'm new to perl.

    i'm trying to decipher some perl code at work, and i was wondering what does
    this mean:

    $myvariable=100*


    doe this mean multiple myvariable by 100?
     
    kaptain kernel, Sep 26, 2003
    #1
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  2. kaptain kernel

    Anno Siegel Guest

    kaptain kernel <> wrote in comp.lang.perl.misc:
    > sorry , i'm new to perl.
    >
    > i'm trying to decipher some perl code at work, and i was wondering what does
    > this mean:
    >
    > $myvariable=100*
    >
    >
    > doe this mean multiple myvariable by 100?


    No. It's not Perl. Copy and paste the actual code.

    Anno
     
    Anno Siegel, Sep 26, 2003
    #2
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  3. Anno Siegel wrote:

    > kaptain kernel <> wrote in comp.lang.perl.misc:
    >> sorry , i'm new to perl.
    >>
    >> i'm trying to decipher some perl code at work, and i was wondering what
    >> does this mean:
    >>
    >> $myvariable=100*
    >>
    >>
    >> doe this mean multiple myvariable by 100?

    >
    > No. It's not Perl. Copy and paste the actual code.
    >
    > Anno


    my bad - the code was poorly formatted.

    it looked like this:

    $myvariable=100*


    $anothervariable




    when it should be

    $myvariable=100*$anothervariable
     
    kaptain kernel, Sep 26, 2003
    #3
  4. kaptain kernel

    Anno Siegel Guest

    kaptain kernel <> wrote in comp.lang.perl.misc:
    > Anno Siegel wrote:
    >
    > > kaptain kernel <> wrote in comp.lang.perl.misc:
    > >> sorry , i'm new to perl.
    > >>
    > >> i'm trying to decipher some perl code at work, and i was wondering what
    > >> does this mean:
    > >>
    > >> $myvariable=100*
    > >>
    > >>
    > >> doe this mean multiple myvariable by 100?

    > >
    > > No. It's not Perl. Copy and paste the actual code.
    > >
    > > Anno

    >
    > my bad - the code was poorly formatted.
    >
    > it looked like this:
    >
    > $myvariable=100*
    >
    >
    > $anothervariable
    >
    >
    >
    >
    > when it should be
    >
    > $myvariable=100*$anothervariable


    To answer your question, that multiplies the value in $anothervariable
    with 100 and assigns the result to $myvariable. The statement is still
    missing a semicolon (";") at the end, which would be needed if more
    statements were to follow.

    However, this isn't going to take you anywhere. You are not going
    to learn Perl by presenting tiny snippets of Perl code to the group
    asking for an explanation. The group isn't going to go along much
    either.

    Get a decent tutorial, like _Learning Perl_ from O'Reilly. If that
    raises specific questions, you can ask them here.

    Anno
     
    Anno Siegel, Sep 26, 2003
    #4
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