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+--- Thread: loop issue-Solved (/thread-20091.html)
loop issue-Solved - Amer7 - 06-15-2023 07:24 PM
Hi,
I'm having an issue writing. I need a loop that will compare if the number is >360, then it will subtract -360, then compare if the number is greater than 180 if it is then -180, else +180;
Code:
EXPORT numbers()
BEGIN
LOCAL n1,i,c,b;
INPUT({n1},"numbers", {"number"});
Local Rr2,i,Rr3;
i:=1;
c:=n1;
WHILE i<=c DO
M0(i+1,2):=M0(i,2)+M0(i+1,1);
WHILE M0(i+1,2)>360 DO
M0(i+1,2):=M0(i+1,2)-360;
IF M0(i+1,2)>=180 THEN
M0(i+1,2):=M0(i+1,2)-180;
ELSE IF M0(i+1,2)<180 THEN
M0(i+1,2):=M0(i+1,2)+180;
END;
END;
END;
i:=i+1;
END;
EDITMAT(M0,"m0-mat");
END;-I use a while loop to make it subtract -360 when the number is greater than 360
Then I compare the number IF its >=180 then -180, else +180.
But it's not working
WHILE M0(i+1,2)>360 DO /// While its >360 subtract -360
M0(i+1,2):=M0(i+1,2)-360; ---- Memorize it
IF M0(i+1,2)>=180 THEN ----- Compare if the memorized value is>=180
M0(i+1,2):=M0(i+1,2)-180; ------ THEN -180
ELSE IF M0(i+1,2)<180 THEN -----If the memorized value vas <180
M0(i+1,2):=M0(i+1,2)+180; ADD +180
END;
END;
END;
Example
Result
The program does this:
71+320= Compares it the correct result should be 211 and it is
Next row, 211+10=231 compare it, the result should be 51(231>180 then 231-180=51)
next row, 51+20=71 compare it, result should be 251(71<180 then 71+180=251)...
EDIT SOLUTION
Code:
EXPORT numbers()
BEGIN
LOCAL n1,i,c,b;
INPUT({n1},"numbers", {"number"});
Local Rr2,i,Rr3;
i:=1;
c:=n1;
WHILE i<=c DO
M0(i+1,2):=M0(i,2)+M0(i+1,1);
WHILE M0(i+1,2)>360 DO
M0(i+1,2):=M0(i+1,2)-360;
END;
IF M0(i+1,2)>=180 THEN
M0(i+1,2):=M0(i+1,2)-180
ELSE
M0(i+1,2):=M0(i+1,2)+180;
END;
i:=i+1;
END;
EDITMAT(M0,"m0-mat");
END;RE: loop issue-Solved - HPCarnace - 06-18-2023 02:03 PM
Isn't it easier to use the division algorithm?, that is, take the remainder of the division of the sum of angles with 90 degrees: irem(71+320+10+20+171,90) returns 52
RE: loop issue-Solved - Amer7 - 06-25-2023 12:32 AM
(06-18-2023 02:03 PM)HPCarnace Wrote: Isn't it easier to use the division algorithm?, that is, take the remainder of the division of the sum of angles with 90 degrees: irem(71+320+10+20+171,90) returns 52
probably? This is the first time I hear about division algorithm