Hi Ray :)
Check the COMMON or COM (I believe Rich was correct and it is COMMON
but we all know about *MY* memory :) )
The other thing is to find a copy of the old Mille Borne (sp) game...
as I recall (again with the memory!) it was BASIC and used the common
memory block to pass values... besides... if you find it you can play it as
"rigorous software testing to evaluate passing of variable values via
common memory block methodology"
Hey! it sounds good and will pass weekly status reports that aren't
looked at too closely :) hehehe
Art "I never did this with Mille Borne... Mansion is another story!" Bahrs
P.S. but Mansion is ForTran
=======================================================
Art Bahrs, CISSP Information Security The Regence Group
(503) 225-4992 FAX (503) 220-3806
"Ray Shahan"
<rshahan@REPUB
LICTITLE.COM> To
Sent by: [log in to unmask]
"HP-3000 cc
Systems
Discussion" Subject
<HP3000-L@RAVE Re: [HP3000-L] Business BASIC
N.UTC.EDU> question
08/17/2006
06:47 AM
Please respond
to
"Ray Shahan"
<rshahan@REPUB
LICTITLE.COM>
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Thank you, Bruce, however, the A=B=C construct is exactly what I was trying
to do.
Again, thank you Bruce, and everyone else who contributed.
Ray S.
-----Original Message-----
From: HP-3000 Systems Discussion [mailto:[log in to unmask]] On Behalf
Of Robert Collins
Sent: Thursday, August 17, 2006 7:47 AM
To: [log in to unmask]
Subject: Re: [HP3000-L] Business BASIC question
Hey Ray,
From my technical folks:
You do the ability to define areas by offset and length and to assign
multiple values to the same memory; however, the results are stored in a
different memory location. So a true redefinition in the Cobol sense does
not take place.
For example
In COBOL, if A redefines B which redfines C, then A=B=C. If C is changed,
then A & B both are changed since the same memory is referenced.
But in Basic IF A=B AND B=C then A=B=C is not true, since A,B & C occupy
different memory.
* 10 DIM A$[20], B$[10], C$[10]
* 20 A$="ABCDEFGHIJKLMNOPQRTU"
* 30 B$=A$[1,10] \ C$=A$[11,20]
* 40 print a$,b$,c$
* 50 a$="12345678901234567890"
* 60 print a$,b$,c$
* run
ABCDEFGHIJKLMNOPQRTUABCDEFGHIJKLMNOPQRTU
12345678901234567890ABCDEFGHIJKLMNOPQRTU
So, A, B and C are defined as strings and the value in A$ can be assigned
to B$ and C$, but all still point to separate memory areas and, if A is
given a different value, B and C aren't affected.
Robert Collins
U.S. Sales Consultant
Transoft
Part of Computer Software Group plc
1165 Northchase, Suite 375
Marietta, GA 30067
Phone #: 706-265-4110
Cell #: 770-789-1462
fax #: 706-265-1621
www.transoft.com www.computersoftware.com
-----Original Message-----
From: HP-3000 Systems Discussion [mailto:[log in to unmask]] On Behalf
Of Ray Shahan
Sent: Wednesday, August 16, 2006 12:41 PM
To: [log in to unmask]
Subject: [HP3000-L] Business BASIC question
Hi all,
I've scoured the manual, but can't seem to locate this
concept (and it may be because it's not supported). In COBOL (and other
languages), one can REDEFINE a variables attributes, but I can't seem to
find any such concept for BB. Does anyone know if this can be done in
BB, and if yes, how to do it?
TIA,
Ray S.
<http://www.republictitle.com/>
Ray Shahan
Computer Programmer
REPUBLIC TITLE OF TEXAS, INC. <http://www.republictitle.com/>
2701 W Plano Parkway
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ountry=us&new=1&name=&qty=>
Plano, TX 75075
direct 214.556.0202
main 972.578.8611
fax 972.424.5621
www.republictitle.com <http://www.republictitle.com/>
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