BookmarkSubscribeRSS Feed
esirjhm
Calcite | Level 5

The following code seems to have two different results with a 10.5 format when the two values in hexidecimal are identical.  What am I missing?

DATA TEST;                                                 

  HMWID = 62.7;                                             

  LIN_FT = 1;                                               

  RUN;                                                      

DATA AVAL (KEEP = K FT);                                   

  ARRAY LAR{240:799} _TEMPORARY_;                           

  SET TEST END=END;                                         

  PUTLOG HMWID=10.5  +3 HMWID=HEX16.;                       

  HMWID = HMWID * 10;                                       

  PUTLOG 'STORING VALUE '  HMWID=10.5 +3   HMWID=HEX16.;    

  LAR{HMWID} + LIN_FT;                                      

  IF END;                                                   

  DO J = 240 TO (799-40);                                   

     K  = J;                                                

     FT = LAR{J};                                           

     IF FT THEN DO;                                         

        OUTPUT AVAL;                                        

        PUTLOG 'RESTORING VALUE ' K=10.5 +3 HMWID=HEX16.;   

     END;                                                   

  END;                                                      

  RUN; 

Output results...                                                    

HMWID=62.70000    HMWID=423EB33333333333               

STORING VALUE HMWID=627.00000    HMWID=43272FFFFFFFFFFF

RESTORING VALUE K=626.00000    HMWID=43272FFFFFFFFFFF  

1 REPLY 1
DLing
Obsidian | Level 7

Doesn't look like anything is wrong. Don't unerstand why you're showing for RESTORING K not being 627.

What's posted looks like an excerpt from a much larger program, perhaps the culprit is elsewhere and got edited out.

By the way, display format, like 10.5, have nothing to do with internal precision.

1 DATA TEST;

2 HMWID = 62.7;

3 LIN_FT = 1;

4 RUN;

NOTE: The data set WORK.TEST has 1 observations and 2 variables.

NOTE: DATA statement used (Total process time):

real time 0.01 seconds

cpu time 0.01 seconds

 

5

6 DATA AVAL (KEEP = K FT);

7 ARRAY LAR{240:799} _TEMPORARY_;

8 SET TEST END=END;

9 PUTLOG HMWID=10.5 +3 HMWID=HEX16.;

10

11 HMWID = HMWID * 10;

12 PUTLOG 'STORING VALUE ' HMWID=10.5 +3 HMWID=HEX16.;

13 LAR{HMWID} + LIN_FT;

14

15 IF END;

16

17 DO J = 240 TO (799-40);

18 K = J;

19 FT = LAR{J};

20 IF FT THEN DO;

21 OUTPUT AVAL;

22 PUTLOG 'RESTORING VALUE ' K=10.5 +3 HMWID=HEX16.;

23 END;

24 END;

25 RUN;

HMWID=62.70000 HMWID=404F59999999999A

STORING VALUE HMWID=627.00000 HMWID=4083980000000000

RESTORING VALUE K=627.00000 HMWID=4083980000000000

NOTE: There were 1 observations read from the data set WORK.TEST.

NOTE: The data set WORK.AVAL has 1 observations and 2 variables.

NOTE: DATA statement used (Total process time):

real time 0.03 seconds

cpu time 0.01 seconds

sas-innovate-2026-white.png



April 27 – 30 | Gaylord Texan | Grapevine, Texas

Registration is open

Walk in ready to learn. Walk out ready to deliver. This is the data and AI conference you can't afford to miss.
Register now and lock in 2025 pricing—just $495!

Register now

How to Concatenate Values

Learn how use the CAT functions in SAS to join values from multiple variables into a single value.

Find more tutorials on the SAS Users YouTube channel.

SAS Training: Just a Click Away

 Ready to level-up your skills? Choose your own adventure.

Browse our catalog!

Discussion stats
  • 1 reply
  • 977 views
  • 0 likes
  • 2 in conversation