COBOL气泡仅排序表的最后一个元素

Question

我目前正在学习COBOL，我正在尝试在程序中实现冒泡排序算法。尽管我还是该语言的新手，但是我写的内容在语义和语法上对我来说都是有意义的，但是如果我按该顺序输入5、4、3、2和1，我的帖子排序表就会变成1、5 4，3，2。有人可以向我解释我哪里出错了吗？

       IDENTIFICATION DIVISION.
         PROGRAM-ID. BubbleSort.

       DATA DIVISION.
         WORKING-STORAGE SECTION.
         01 TVAR PIC 9(4).
         01 CNT PIC 9(1) VALUE 1.
         01 CNT2 PIC 9(1) VALUE 1.
         01 ARR.
            05 ARRELEMENT PIC 9(4) OCCURS 5 TIMES.
         01 TABLELENGTH PIC 9(1) VALUE 5.

       PROCEDURE DIVISION.
         DISPLAY "Enter 5 numbers: ".
         PERFORM INPUT-PARA VARYING CNT FROM 1 BY 1 UNTIL CNT>5.

         DISPLAY "Pre Bubble-Sort: ".
         PERFORM PRINT-PARA VARYING CNT FROM 1 BY 1 UNTIL CNT>5.

         PERFORM BBLSORT-PARA.

         DISPLAY "Post Bubble-Sort: ".
         PERFORM PRINT-PARA VARYING CNT FROM 1 BY 1 UNTIL CNT>5.

         STOP RUN.

         INPUT-PARA.
            ACCEPT ARRELEMENT(CNT).

         PRINT-PARA.
            DISPLAY "Table element: "ARRELEMENT(CNT).

         BBLSORT-PARA.
             INITIALIZE CNT CNT2.
             MOVE 1 TO CNT.
             MOVE 2 TO CNT2.
             PERFORM UNTIL CNT>6
               PERFORM UNTIL CNT2>5
               DISPLAY "IF "ARRELEMENT(CNT) " IS > "ARRELEMENT(CNT2)
                 IF (ARRELEMENT((CNT)) > ARRELEMENT((CNT2)))
                 THEN
                   DISPLAY ARRELEMENT(CNT) " IS > "ARRELEMENT(CNT2)
                    MOVE ARRELEMENT(CNT) TO TVAR
                    MOVE ARRELEMENT(CNT2) TO ARRELEMENT(CNT)
                    MOVE TVAR TO ARRELEMENT(CNT2)
                 END-IF
                 DISPLAY "EXIT IF LOOP"
                 ADD 1 TO CNT2 GIVING CNT2
               END-PERFORM
               ADD 1 TO CNT GIVING CNT
             END-PERFORM.

       END PROGRAM BubbleSort.

Answer 1

免责声明：我不是COBOL程序员，但是通过对循环的这些更改，我能够完成这项工作。

PERFORM UNTIL CNT>4
           MOVE 1 TO CNT2
           PERFORM UNTIL CNT2+CNT>5
             ADD 1 TO CNT2 GIVING CNT3
             IF (ARRELEMENT((CNT2)) > ARRELEMENT((CNT3)))
             THEN
               DISPLAY ARRELEMENT(CNT2) " IS > "ARRELEMENT(CNT3)
                MOVE ARRELEMENT(CNT3) TO TVAR
                MOVE ARRELEMENT(CNT2) TO ARRELEMENT(CNT3)
                MOVE TVAR TO ARRELEMENT(CNT2)
             END-IF
             DISPLAY "EXIT IF LOOP"
             ADD 1 TO CNT2 GIVING CNT2
           END-PERFORM
           ADD 1 TO CNT GIVING CNT
         END-PERFORM.

关键是添加一个我用来代表CNT2 + 1的CNT3。在内循环之前，有必要每次将CNT2重新发送为1。然后该算法将每次在循环中将项CNT2与项CNT2 + 1进行比较，而无需引用CNT。

而且，也不需要每次都一直循环到内部循环中的数组末尾。我发现参考Geeks for Geeks on bubble sort

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