Hey Grok, Can you write the simplest recursive fibonacci-generator in COBOL?

       IDENTIFICATION DIVISION.
       PROGRAM-ID. FIBONACCI.
       DATA DIVISION.
       WORKING-STORAGE SECTION.
       01  NUM         PIC 9(10).
       01  RESULT      PIC 9(10).
       PROCEDURE DIVISION.
       MAIN-PARA.
           MOVE 10 TO NUM
           CALL 'FIB' USING NUM RESULT
           DISPLAY 'FIBONACCI OF ' NUM ' IS ' RESULT
           STOP RUN.
       IDENTIFICATION DIVISION.
       PROGRAM-ID. FIB.
       DATA DIVISION.
       LINKAGE SECTION.
       01  N           PIC 9(10).
       01  FIB-RESULT  PIC 9(10).
       PROCEDURE DIVISION USING N FIB-RESULT.
           IF N <= 1
               MOVE N TO FIB-RESULT
           ELSE
               COMPUTE FIB-RESULT = FUNCTION FIB(N - 1) + FUNCTION FIB(N - 2)
           END-IF
           GOBACK.
       END PROGRAM FIB.
       END PROGRAM FIBONACCI.

timonoko 1y

Has recursion. OK.

There was a time, when I learned Fortran does not allow recursion. "You stupid or what?" said the professor.

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