4 pointsby nonrandomstring10 hours ago1 comment
  • timonoko8 hours ago
    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.
    • timonoko8 hours ago
      Has recursion. OK.

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