[COBOL] Sum multiples of 3 and 5

Last night I posted the following code on RosettaCode as another solution to the Sum multiples of 3 and 5 challenge. (Mine is the third one down after all the more-mathematical solutions.)

My solution is as brute-force as it gets, using only adds and comparisons.

       IDENTIFICATION DIVISION.
       PROGRAM-ID. SUM35.

       DATA DIVISION.
       WORKING-STORAGE SECTION.
       01  THREE-COUNTER   USAGE BINARY-CHAR value 1.
           88 IS-THREE VALUE 3.
       01  FIVE-COUNTER    USAGE BINARY-CHAR value 1.
           88 IS-FIVE VALUE 5.
       01  SUMMER          USAGE BINARY-DOUBLE value zero. 
       01  I               USAGE BINARY-LONG.
       01  N               USAGE BINARY-LONG.

       PROCEDURE DIVISION.
       10-MAIN-PROCEDURE.
           MOVE 1000000000 TO N.
           MOVE 1 TO I.
           PERFORM 20-INNER-LOOP WITH TEST AFTER UNTIL I >= N.
           DISPLAY SUMMER.
           STOP RUN.
       20-INNER-LOOP.
           IF IS-THREE OR IS-FIVE 
               ADD I TO SUMMER END-ADD
               IF IS-THREE
                   MOVE 1 TO THREE-COUNTER
               ELSE
                   ADD 1 TO THREE-COUNTER
               END-IF
               IF IS-FIVE
                   MOVE 1 TO FIVE-COUNTER
               ELSE    
                   ADD 1 TO FIVE-COUNTER
               END-IF
           ELSE
               ADD 1 TO FIVE-COUNTER END-ADD
               ADD 1 TO THREE-COUNTER END-ADD
           END-IF.
           ADD 1 TO I.
           EXIT.
       END PROGRAM SUM35.

The above code compiles on GnuCOBOL 2.0 and MicroFocus Visual COBOL 2.3. In the latter environment, I was able to get a run time of 7.3 seconds for the 1,000,000,000 iterations (AMD A6-5200 APU running at 2.00 GHz.)

I ended up in COBOL via the usual circuitous route through other programming languages after figuring out a solution using mLite. The next postings will demonstrate the mLite and perhaps others.

© Copyright Bruce M. Axtens., 2015