Date:         Fri, 17 May 2002 21:38:51 -0700
Reply-To:     Dale McLerran <stringplayer_2@YAHOO.COM>
Sender:       "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
From:         Dale McLerran <stringplayer_2@YAHOO.COM>
Subject:      Re: Need advice on inefficient code
Comments: To: RICH0850 <rich0850@AOL.COM>
In-Reply-To:  <20020517193803.15945.00000243@mb-ft.aol.com>
Content-Type: text/plain; charset=us-ascii

Richard,

To add to my previous reply, let me suggest that you get rid entirely
of the SUBSTR function and replace it with a bit of code which
addresses each character in the string directly from its memory
location.  The functions which you need to invoke in order to perform
this bit of magic are the ADDR and PEEKC functions.  The ADDR
function returns the memory location of the start of the string.
The PEEKC function allows you to fetch the content at a memory
address as a character variable.  So, for the particular problem
you gave, we would have

  RTITL='The Quick Brown Fox Jumps Over The Lazy Dog';
  RLN=LENGTH(RTITL);
  RLEN=0;
  start = addr(rtitl);
  end = start + rln - 1;
  DO I = start to end;
    rlen + input(peekc(i,1), pixels.);
  end;


Of course, you need to create the PIXEL informat as described in my
previous post.  Between replacing the SUBSTR function with the PEEKC
function and replacing the conditional tests and assignment statements
with a single assignment using an informat, this ought to be very
efficient code.  I hope that Hashman (aka Paul Dorfman) is proud to
know that some of the lessons he has taught over the years have
gotten through.  I am sure that Paul could come up with something
which is more efficient yet than using the input format to look up
the length of each character.  But for off-the-rack code, what I
have presented above should be at or near maximal efficiency.

Dale


--- RICH0850 <rich0850@AOL.COM> wrote:
> Dear Group:
>
> I've got some code that parses a string and returns the number of
> pixels wide
> the string is.  I've determined how wide in pixels each character in
> the
> alphabet is  (according to the font and size I'll be displaying in
> the browser)
> and then I run the string through this to get its total size in
> pixels.
>
> Although the following code does the job it seems very inefficient to
> me
> (especially if I've got a lot of strings).  Any ideas on how to do
> this faster?
>
> RTITL='The Quick Brown Fox Jumps Over The Lazy Dog';
> RLN=LENGTH(RTITL);
> RLEN=0;
> DO I = 1 to RLN;
>    IF SUBSTR(RTITL,I,1) IN
>       ('I','i','l','.',' ')
>       THEN RLEN+2;
>    IF SUBSTR(RTITL,I,1) IN
>       ('/','j')
>       THEN RLEN+3;
>    IF SUBSTR(RTITL,I,1) IN
>       ('r')
>       THEN RLEN+4;
>    IF SUBSTR(RTITL,I,1) IN
>       ('a','c','E','F','f','J','t','x',
>        'z','0','1','2','3','4','5','6','7',
>        '8','9')
>       THEN RLEN+5;
>    IF SUBSTR(RTITL,I,1) IN
>       ('B','b','D','d','e','g','H','h',
>        'K','k','L','N','n','o','P','p','q','S',
>        's','T','U','u','X','Y','Z','&')
>       THEN RLEN+6;
>    IF SUBSTR(RTITL,I,1) IN
>       ('A','C','G','O','Q','R','V','v','y')
>       THEN RLEN+7;
>    IF SUBSTR(RTITL,I,1) IN
>       ('M','w')
>       THEN RLEN+9;
>    IF SUBSTR(RTITL,I,1) IN
>       ('m')
>       THEN RLEN+10;
>    IF SUBSTR(RTITL,I,1) IN
>       ('W')
>       THEN RLEN+11;
> END;
>
> Thanks in advance for any advice.
>
> --Richard


=====
---------------------------------------
Dale McLerran
Fred Hutchinson Cancer Research Center
mailto: dmclerra@fhcrc.org
Ph:  (206) 667-2926
Fax: (206) 667-5977
---------------------------------------

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