Moving

[vinod kumar]

Hi, The following is my program:

D A S 10S 0 INZ(-6)
D B S 10S 0 INZ(3)
D C S 10S 0
D D S 10A
C EVAL C = 100 * (A/B)
C C DSPLY
C MOVE C D
C D DSPLY
C SETON LR 

In this above program the calculated value for C = -200,But when am moving this value to character field D, the value 000000020} is getting assigned. I need the value 0000000200 to be get assigned.
Can you please help me.
Thank you.

[KenM]

Re: Moving

It looks like the negative sign is getting encoded into te character field. If you do not need to represent the number as negative. just test it and make it positive.

[Billw]

Re: Moving

Hi, The following is my program:

D A S 10S 0 INZ(-6)
D B S 10S 0 INZ(3)
D C S 10S 0
D D S 10A
C EVAL C = 100 * (A/B)
C C DSPLY
C MOVE C D
C D DSPLY
C SETON LR

In this above program the calculated value for C = -200,But when am moving this value to character field D, the value 000000020} is getting assigned. I need the value 0000000200 to be get assigned.
Can you please help me.
Thank you.

A negative number contains characters in the low order to denite a negative number. You are displaying it without edtting it on the screen.

change the move line:
eval d = %Editc(C : 'L')
now when it displays you will see 200-
obviusly you don't care about nagtive numbers (but your bank does)
eval d = %editc(C : 'X')
you will then see 0000000200
or make it a positve number then display it
eval c = c *-1

then display d
Million ways to get their
Bill

[arrow483]

Re: Moving

or, use eval d = %editc(C : 'Z') and any negative sign gets ignored. If you don't like the blanks that the edit code leaves, go one step further
eval d = %trim(%editc(C : 'Z'))

[Jerry G]

A dated thread, but for anyone else encountering this (as I am, 13 years later): You need to test for negativity. In free-format:

C = 100 * (A/B); 
IF C >= *ZERO;
  D = %CHAR(C);         // or D =       %EDITC(  C:'X');
ELSE;
  D = '-' + %CHAR(- C); // or D = '-' + %EDITC(- C:'X');
ENDIF;

%EDITC(C:'X') returns a string of the same length as C, if you don't want the potentially-left-truncated result of %CHAR.

[jamief]

This would work also -- added indicator to tell you if negative if required down the code:

d Answer          s             10s 0                                    
d Character10     s             10                                       
d Negative        s               n   inz('0')                           
d Number1         s              6s 0 inz(-6)                            
d Number2         s              6s 0 inz(3)                             

       *inlr = *on;                                                      

       Answer = (100 * (Number1/Number2));                               
       if Answer <> %abs(Answer);                                        
        Negative = *on;                                                  
       endif;                                                            
       Character10 = %editc(%abs(Answer):'X');