Re: Find 3: The Reckoning

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Just kidding....

Happy Halloween....
  ----- Original Message -----=20
  From: Liona Kerslake=20
  To: EUPHORIA at LISTSERV.MUOHIO.EDU=20
  Sent: Tuesday, October 31, 2000 6:49 PM
  Subject: Re: Find 3: The Reckoning



    ----- Original Message -----=20
    From: Derek Parnell=20
    To: EUPHORIA at LISTSERV.MUOHIO.EDU=20
    Sent: Sunday, October 29, 2000 10:42 AM
    Subject: Re: Find 3: The Reckoning


    Ok Thomas, I'll bite.

    ------
    Derek Parnell
    Melbourne, Australia
    (Vote [1] The Cheshire Cat for Internet Mascot)


      include get.e
      sequence names,curname
      object input
      integer location
      atom wn
      wn=3Drand(3)
      names =3D{"bob","them","worms"}
      curname=3Dnames[wn]
      printf(1,"curname: %s\n",{curname})

      input=3Dgets(0)
      ? curname
      puts(1,"\n")
      ? input
      location =3D find(input[1],names[2])
    This previous line is saying "Find where the first character of =
input inside of the second sequence of names".  Is this what you are =
really trying to find out?
    I suspect you are trying to find out if "input" is one of the three =
possible names. If that is so, try this instead...
      if sequence(input) then
          -- Strip off the trailing newline char.
          input =3D input[1 .. length(input) - 1]
          -- convert to lowercase.
          input =3D lower(input)
          -- See if the input sequence is one of the sequences in names.
          location =3D find(input, names)
          -- "location" should equal 0, 1, 2, or 3 now.
      else
          -- End-Of-File (Ctrl-Z) entered.
          location =3D 0
      end if       =20


      puts(1,"\n")
      ? location


          The "location" variable only gives a correct number if "them" =
is used.=20
    And what is the correct number? What were you expecting to see?
      Nothing else. I've looked at all the documentation I've got and =
either I am quite blind or, I just haven't found my solution. Anywho, if =
you could solve this problem, I'd be most grateful....

    I suspect you are confused by the find() function. It takes two =
parameters, the first can be an atom or a sequence but the second must =
be a sequence. It then scans each element in the second parameter to see =
if it equals the first parameter. If so, it returns the number of the =
element in the second parmeter that equalled the first parameter.

    This means that if the first parameter is a sequence, as is your =
case here, then each of the elements in the second parmeter should be =
sequences too. Also, this is what you have coded for names. However, in =
your coding of find() you used input[1] which is not a sequence but the =
first character of input. It is a single character - an atom. Also, you =
coded the second parameter as names[2] which is a sequence if characters =
- namely "them" . Thus, find() went looking through the characters ''t', =
'h', 'e', and 'm' to see if the first character of input equalled any.

    This would mean that location would be non-zero, 1 to 4, for any =
input that begun with 't', 'h', 'e', or 'm'.

      Das Svendanya
      Thomas


      PS. This should  be the last one.........
    Yeah, sure Thomas. --------I meant the last one on this program, =
but, you're right blink

    What I'm trying to do here is get the placement of an individual =
letter from a word. So if the word is "moor" and the user types in =
'o'.....wait, bad example, there are two o's....would eusphoria see that =
as [2..3]?

    okay, if the user types in 'r', whatever variable I assigned find() =
to will be 4.

    -Hope this clarifies things.....
    -HAPPY HALLOWEEN!!!
    -Thomas

    PS. How'd you like the Halloween-esque sound?


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<DIV><FONT face=3DArial size=3D2>Just kidding....</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
<DIV><FONT face=3DArial size=3D2>Happy Halloween....</FONT></DIV>
<BLOCKQUOTE dir=3Dltr=20
style=3D"PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; =
BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
  <DIV style=3D"FONT: 10pt arial">----- Original Message ----- </DIV>
  <DIV=20
  style=3D"BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: =
black"><B>From:</B>=20
  <A title=3Dpaulk at UNISERVE.COM href=3D"mailto:paulk at UNISERVE.COM">Liona =

  Kerslake</A> </DIV>
  <DIV style=3D"FONT: 10pt arial"><B>To:</B> <A =
title=3DEUPHORIA at LISTSERV.MUOHIO.EDU=20
  =
</A>=20
  </DIV>
  <DIV style=3D"FONT: 10pt arial"><B>Sent:</B> Tuesday, October 31, 2000 =
6:49=20
  PM</DIV>
  <DIV style=3D"FONT: 10pt arial"><B>Subject:</B> Re: Find 3: The =
Reckoning</DIV>
  <DIV><BR></DIV><BGSOUND balance=3D0=20
  src=3D"cid:001c01c043b9$0e09ade0$37b04cd8@euman" volume=3D0>
  <DIV>&nbsp;</DIV>
  <BLOCKQUOTE dir=3Dltr=20
  style=3D"PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; =
BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
    <DIV style=3D"FONT: 10pt arial">----- Original Message ----- </DIV>
    <DIV=20
    style=3D"BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: =
black"><B>From:</B>=20
    <A title=3Ddparnell at BIGPOND.NET.AU =
href=3D"mailto:dparnell at BIGPOND.NET.AU">Derek=20
    Parnell</A> </DIV>
    <DIV style=3D"FONT: 10pt arial"><B>To:</B> <A=20
    title=3DEUPHORIA at LISTSERV.MUOHIO.EDU=20
    =
</A>=20
    </DIV>
    <DIV style=3D"FONT: 10pt arial"><B>Sent:</B> Sunday, October 29, =
2000 10:42=20
    AM</DIV>
    <DIV style=3D"FONT: 10pt arial"><B>Subject:</B> Re: Find 3: The=20
Reckoning</DIV>
    <DIV><BR></DIV>
    <DIV dir=3Dltr style=3D"FONT: 10pt arial"><FONT face=3DArial =
size=3D2>Ok Thomas,=20
    I'll bite.</FONT></DIV>
    <DIV dir=3Dltr style=3D"FONT: 10pt arial"><FONT face=3DArial=20
    size=3D2></FONT>&nbsp;</DIV>
    <DIV dir=3Dltr style=3D"FONT: 10pt arial">------<BR>Derek =
Parnell<BR>Melbourne,=20
    Australia<BR>(Vote [1] The Cheshire Cat for Internet =
Mascot)<BR></DIV>
    <BLOCKQUOTE dir=3Dltr=20
    style=3D"PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; =
BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
      <DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
      <DIV><FONT face=3DArial color=3D#000080 size=3D2>include =
get.e<BR>sequence=20
      names,curname<BR>object input<BR>integer location<BR>atom=20
      wn<BR>wn=3Drand(3)<BR>names=20
      =
      %s\n",{curname})</FONT></DIV>
      <DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
      <DIV><FONT face=3DArial color=3D#000080 =
size=3D2>input=3Dgets(0)<BR>?=20
      curname<BR>puts(1,"\n")<BR>? input</FONT></DIV>
      <DIV><FONT face=3DArial size=3D2>
      <DIV><FONT face=3DArial color=3D#000080 size=3D2>location =3D=20
      find(input[1],names[2])</FONT></DIV></FONT></DIV></BLOCKQUOTE>
    <DIV dir=3Dltr><FONT face=3DArial size=3D2>This&nbsp;previous line =
is saying "Find=20
    where the first character of input inside of the second sequence of=20
    names".&nbsp; Is this what you are really trying to find =
out?</FONT></DIV>
    <DIV dir=3Dltr><FONT face=3DArial size=3D2>I suspect you are trying =
to find out if=20
    "input" is one of the three possible names. If that is so, try this=20
    instead...</FONT></DIV>
    <BLOCKQUOTE dir=3Dltr style=3D"MARGIN-RIGHT: 0px">
      <DIV dir=3Dltr><STRONG><FONT size=3D2>if sequence(input)=20
      then</FONT></STRONG></DIV>
      <DIV dir=3Dltr><STRONG><FONT size=3D2>&nbsp;&nbsp;&nbsp; -- Strip =
off the=20
      trailing newline char.</FONT></STRONG></DIV>
      <DIV dir=3Dltr><STRONG><FONT size=3D2>&nbsp;&nbsp;&nbsp; input =3D =
input[1 ..=20
      length(input) - 1]</FONT></STRONG></DIV>
      <DIV dir=3Dltr><STRONG><FONT size=3D2>&nbsp;&nbsp;&nbsp; -- =
convert to=20
      lowercase.</FONT></STRONG></DIV>
      <DIV dir=3Dltr><STRONG><FONT size=3D2>&nbsp;&nbsp;&nbsp; input =3D =

      lower(input)</FONT></STRONG></DIV>
      <DIV dir=3Dltr><STRONG><FONT size=3D2>&nbsp;&nbsp;&nbsp; -- See if =
the input=20
      sequence is one of the sequences in names.</FONT></STRONG></DIV>
      <DIV dir=3Dltr><STRONG><FONT size=3D2>&nbsp;&nbsp;&nbsp; location =
=3D=20
      find(input, names)</FONT></STRONG></DIV>
      <DIV dir=3Dltr><STRONG><FONT size=3D2>&nbsp;&nbsp;&nbsp; -- =
"location" should=20
      equal 0, 1, 2, or 3 now.</FONT></STRONG></DIV>
      <DIV dir=3Dltr><STRONG><FONT size=3D2>else</FONT></STRONG></DIV>
      <DIV dir=3Dltr><STRONG><FONT size=3D2>&nbsp;&nbsp;&nbsp; -- =
End-Of-File=20
      (Ctrl-Z) entered.</FONT></STRONG></DIV>
      <DIV dir=3Dltr><FONT size=3D2><STRONG>&nbsp;&nbsp;&nbsp; location =
=3D=20
      0</STRONG></FONT></DIV>
      <DIV dir=3Dltr><FONT size=3D2><STRONG>end if</STRONG><FONT=20
      face=3DArial>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =
</FONT></FONT></DIV>
      <DIV dir=3Dltr><STRONG><FONT =
    <BLOCKQUOTE dir=3Dltr=20
    style=3D"PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; =
BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
      <DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
      <DIV><FONT face=3DArial color=3D#000080 size=3D2>puts(1,"\n")<BR>? =

      location</FONT></DIV>
      <DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
      <DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
      <DIV><FONT face=3DArial size=3D2>&nbsp;&nbsp;&nbsp; The "location" =
variable=20
      only gives a correct number if "them" is used. =
</FONT></DIV></BLOCKQUOTE>
    <DIV dir=3Dltr><FONT face=3DArial size=3D2>And what is the=20
    <STRONG>correct</STRONG> number? What were you expecting to=20
see?</FONT></DIV>
    <BLOCKQUOTE dir=3Dltr=20
    style=3D"PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; =
BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
      <DIV><FONT face=3DArial size=3D2>Nothing else. I've looked at all =
the=20
      documentation I've got and either I am quite blind or, I just =
haven't=20
      found my solution. Anywho, if you could solve this problem, I'd be =
most=20
      grateful....</FONT></DIV>
      <DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV></BLOCKQUOTE>
    <DIV dir=3Dltr><FONT face=3DArial size=3D2>I suspect you are =
confused by=20
    the&nbsp;<STRONG>find()</STRONG> function. It takes two parameters, =
the=20
    first can be an atom or a sequence but the second must be a =
sequence. It=20
    then scans each element in the second parameter to see if it equals =
the=20
    first parameter. If so, it returns the number of the element in the =
second=20
    parmeter that equalled the first parameter.</FONT></DIV>
    <DIV dir=3Dltr><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
    <DIV dir=3Dltr><FONT face=3DArial size=3D2>This means that if the =
first parameter=20
    is a sequence, as is your case here, then each of the elements in =
the second=20
    parmeter should be sequences too. Also, this is what you have coded =
for=20
    <EM>names</EM>. However, in your coding of <STRONG>find()</STRONG> =
you used=20
    <EM>input[1]</EM> which is <STRONG>not</STRONG> a sequence but the =
first=20
    character of input. It is a single character -&nbsp;an atom. Also, =
you coded=20
    the second parameter as <EM>names[2]</EM> which is a sequence if =
characters=20
    - namely <EM>"them"</EM> . Thus, <STRONG>find()</STRONG> went =
looking=20
    through the characters ''t', 'h', 'e', and 'm' to see if the first =
character=20
    of <EM>input</EM> equalled any.</FONT></DIV>
    <DIV dir=3Dltr><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
    <DIV dir=3Dltr><FONT face=3DArial size=3D2>This would mean that =
<EM>location</EM>=20
    would be non-zero, 1 to 4,&nbsp;for any input that begun with 't', =
'h', 'e',=20
    or 'm'.</FONT></DIV>
    <BLOCKQUOTE dir=3Dltr=20
    style=3D"PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; =
BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
      <DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
      <DIV><FONT face=3DArial size=3D2>Das Svendanya</FONT></DIV>
      <DIV><FONT face=3DArial size=3D2>Thomas</FONT></DIV>
      <DIV>&nbsp;</DIV>
      <DIV>&nbsp;</DIV>
      <DIV><FONT face=3DArial size=3D2>PS. This <EM>should</EM>&nbsp; be =
the last=20
      one.........</FONT></DIV></BLOCKQUOTE>
    <DIV dir=3Dltr><FONT face=3DArial size=3D2>Yeah, sure&nbsp;Thomas. =
--------I meant=20
    the last one on this program, but, you're right blink</FONT></DIV>
    <DIV dir=3Dltr>&nbsp;</DIV>
    <DIV dir=3Dltr><FONT face=3DArial size=3D2>What I'm trying to do =
here is get the=20
    placement of an individual letter from a word. So if the word is =
"moor" and=20
    the user types in 'o'.....wait, bad example, there are two =
o's....would=20
    eusphoria see that as [2..3]?</FONT></DIV>
    <DIV dir=3Dltr>&nbsp;</DIV>
    <DIV dir=3Dltr><FONT face=3DArial size=3D2>okay, if the user types =
in 'r',=20
    whatever variable I assigned find() to will be 4.</FONT></DIV>
    <DIV dir=3Dltr>&nbsp;</DIV>
    <DIV dir=3Dltr><FONT face=3DArial size=3D2>-Hope this clarifies=20
    things.....</FONT></DIV>
    <DIV dir=3Dltr><FONT face=3DArial size=3D6>-H<FONT =
color=3D#ff8000>A</FONT>P<FONT=20
    color=3D#ff8000>P</FONT>Y <FONT color=3D#ff8000>H</FONT>A<FONT=20
    color=3D#ff8000>L</FONT>L<FONT color=3D#ff8000>O</FONT>W<FONT=20
    color=3D#ff8000>E</FONT>E<FONT color=3D#ff8000>N</FONT>!<FONT=20
    color=3D#ff8000>!</FONT>!</FONT></DIV>
    <DIV dir=3Dltr><FONT face=3DArial size=3D2>-Thomas</FONT></DIV>
    <DIV dir=3Dltr>&nbsp;</DIV>
    <DIV dir=3Dltr><FONT face=3DArial size=3D2>PS. How'd you like the =
Halloween-esque=20
    sound?</FONT><FONT face=3DArial=20

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