添加两个整数在 cobol 中给出不需要的结果

Question

我正在将文件读入 table，请注意第一行不是 table 的一部分。

1000
MS 1 - Join Grps    Group Project       5             5             
Four Programs       Programming         15            9             
Quiz 1              Quizzes             10            7             
FORTRAN             Programming         25            18            
Quiz 2              Quizzes             10            9             
HW 1 - Looplang     Homework            20            15

在代码中，table表示如下：

01     GRADES.
05         GRADE OCCURS 1 TO 100 TIMES DEPENDING ON RECORD-COUNT INDEXED BY J.
10            ASSIGNMENT-NAME   PIC X(20).
10            CATEGORY          PIC X(20).
10            POINTS-POSSIBLE   PIC 9(14).
10            POINTS-EARNED     PIC 9(14).

我还有一些其他累加器变量指定用于稍后计算 sums/percentages。

01     RECORD-COUNT             PIC 9(8) VALUE 0.
01     TOTAL-EARNED-POINTS      PIC 9(14).
01     TOTAL-POSSIBLE-POINTS    PIC 9(14) VALUE 0.

我的问题是，当我逐行阅读记录时，我想执行以下操作：

ADD POINTS-EARNED(RECORD-COUNT) TO TOTAL-EARNED-POINTS

其中 RECORD-COUNT 是迭代中的当前位置。我希望在第一次迭代后 TOTAL-EARNED-POINTS 的值只是 5，对吗？但是，当我 DISPLAY TOTAL-EARNED-POINTS 的值时，控制台显示：

50000000000000

这是50万亿吗？或者它是数字 5 的一个有趣的表示形式？我怎样才能写这个，以便我可以用它做正确的数学运算来打印正确的成绩报告？

编辑：我知道可能有更好的方法来编写这个程序，但我在尝试编写这个程序之前从未使用过 cobol，而且我可能不会再使用它，或者至少在很长一段时间内不会使用它很久。这是针对 class，所以只要我能正确打印我的输出，我就很好。到目前为止的完整代码：

IDENTIFICATION DIVISION.
PROGRAM-ID. GRADEREPORT.
AUTHOR. JORDAN RENAUD.
DATE-WRITTEN. 09/18/2020.

ENVIRONMENT DIVISION.
INPUT-OUTPUT SECTION.
FILE-CONTROL.
       SELECT GRADES-FILE ASSIGN TO "bill"
           ORGANIZATION IS LINE SEQUENTIAL
           ACCESS IS SEQUENTIAL.

DATA DIVISION.
FILE SECTION.
FD GRADES-FILE.
01     INPUT-TOTAL-POINTS       PIC 9(4).
01     INPUT-GRADES.
05         INPUT-GRADE OCCURS 1 to 100 TIMES DEPENDING ON RECORD-COUNT INDEXED BY I.
10            INPUT-ASSIGNMENT-NAME   PIC X(20).
10            INPUT-CATEGORY          PIC X(20).
10            INPUT-POINTS-POSSIBLE   PIC 9(14).
10            INPUT-POINTS-EARNED     PIC 9(14).

WORKING-STORAGE SECTION.
77     GRADES-FILE-EOF          PIC 9.
01     RECORD-COUNT             PIC 9(8) VALUE 0.
01     TOTAL-EARNED-POINTS      PIC 9(4) VALUE ZERO.
01     TOTAL-POSSIBLE-POINTS    PIC 9(14) VALUE 5.
01     K                        PIC 9(14) VALUE 1.
01     TMP                      PIC 9(14).
01     CURRENT-CATEGORY         PIC X(20).
01     CATEGORY-WEIGHT          PIC X(3).
01     LAST-CATEGORY            PIC X(20).
01     TOTAL-POINTS             PIC 9(4).
01     GRADES.
05         GRADE OCCURS 1 TO 100 TIMES DEPENDING ON RECORD-COUNT INDEXED BY J.
10            ASSIGNMENT-NAME   PIC X(20).
10            CATEGORY          PIC X(20).
10            POINTS-POSSIBLE   PIC 9(14).
10            POINTS-EARNED     PIC 9(14).

PROCEDURE DIVISION.
       OPEN INPUT GRADES-FILE.
       READ GRADES-FILE INTO TOTAL-POINTS.
       DISPLAY TOTAL-EARNED-POINTS
       PERFORM UNTIL GRADES-FILE-EOF = 1
           READ GRADES-FILE
              AT END SET
              GRADES-FILE-EOF TO 1
              NOT AT END
                 ADD 1 TO RECORD-COUNT
                 MOVE INPUT-GRADES TO GRADE(RECORD-COUNT)
                 SET TOTAL-EARNED-POINTS UP BY POINTS-EARNED(RECORD-COUNT)
                 DISPLAY TOTAL-EARNED-POINTS
           END-READ
       END-PERFORM.
       CLOSE GRADES-FILE.
       DISPLAY TOTAL-EARNED-POINTS.
       SORT GRADE ASCENDING CATEGORY.
       MOVE CATEGORY(1) TO LAST-CATEGORY.
       PERFORM RECORD-COUNT TIMES
           MOVE CATEGORY(K) TO CURRENT-CATEGORY
           IF CURRENT-CATEGORY = LAST-CATEGORY THEN
              DISPLAY "SAME CATEGORY"
           ELSE
              DISPLAY "NEW CATEGORY"
              MOVE LAST-CATEGORY TO CURRENT-CATEGORY
           END-IF
           SET K UP BY 1
       END-PERFORM
       DISPLAY GRADES.
       STOP RUN.

编辑 2：在实施给定答案以将输入从文件转换为数字形式后，table 的第一行读起来很好，但从那时起它都是空白值。这是 READ 块的新代码，我不确定是否有更有效的方法来读取和转换组字段中的特定字段，但这是我认为应该完成的方式。

PERFORM UNTIL GRADES-FILE-EOF = 1
           READ GRADES-FILE
              AT END SET
              GRADES-FILE-EOF TO 1
              NOT AT END
                 ADD 1 TO RECORD-COUNT

                 MOVE INPUT-ASSIGNMENT-NAME(RECORD-COUNT) TO ASSIGNMENT-NAME(RECORD-COUNT)
                 DISPLAY INPUT-ASSIGNMENT-NAME(RECORD-COUNT)
                 DISPLAY ASSIGNMENT-NAME(RECORD-COUNT)

                 MOVE INPUT-CATEGORY(RECORD-COUNT) TO CATEGORY(RECORD-COUNT)
                 DISPLAY INPUT-CATEGORY(RECORD-COUNT)
                 DISPLAY CATEGORY(RECORD-COUNT)

                 MOVE FUNCTION NUMVAL (INPUT-POINTS-POSSIBLE(RECORD-COUNT)) TO POINTS-POSSIBLE(RECORD-COUNT)
                 DISPLAY INPUT-POINTS-POSSIBLE(RECORD-COUNT)
                 DISPLAY POINTS-POSSIBLE(RECORD-COUNT)

                 MOVE FUNCTION NUMVAL (INPUT-POINTS-EARNED(RECORD-COUNT)) TO POINTS-EARNED(RECORD-COUNT)
                 DISPLAY INPUT-POINTS-EARNED(RECORD-COUNT)
                 DISPLAY POINTS-EARNED(RECORD-COUNT)

                 COMPUTE TOTAL-EARNED-POINTS = TOTAL-EARNED-POINTS + POINTS-EARNED(RECORD-COUNT)
                 DISPLAY TOTAL-EARNED-POINTS
           END-READ
       END-PERFORM.

Answer 1

is it a funny looking representation of the number 5?

不，这是一个未经检查的致命异常：EC-DATA-INCOMPATIBLE.

原因：
您的数据定义与 record-definition 不匹配：

10 POINTS-EARNED PIC 9(14).

会是

 "00000000000005"

没有

 "5             "

看起来更好的定义是

10 SOME-POSSIBILY-NUMERIC-DATA PIC X(14).

如果您按照标签的建议使用 GnuCOBOL，那么将 -debug 添加到编译命令中，您将看到致命异常停止程序（COBOL 标准定义默认情况下所有异常检查都处于关闭状态，在我的意见：因为遗留和性能，但至少对于开发和测试来说，激活它们是非常合理的[在大多数情况下，让程序中止而不是在测试结束时做错误的数学更合理]）。

与任何计算机语言一样，您应该非常确定拥有有效数据（永远不要信任外部数据，无论它是区块链的一部分还是您阅读的文本文件中).

How can I write this so that I can do proper mathematics with it to print a proper grade report?

如果你想使用“错误数据被忽略”（可能在这里是合适的）只需转换它：

MOVE FUNCTION NUMVAL (SOME-POSSIBILY-NUMERIC-DATA)
  TO POINTS-EARNED(RECORD-COUNT)

否则进行显式检查（完全是数字 [自己检查]，或者 left/right FUNCTION TEST-NUMVAL 可能有空格的数字）并停止 program/skip 错误行DISPLAY ... UPON SYSERR 或任何适合您的内容。

添加两个整数在 cobol 中给出不需要的结果

Adding two integers giving unwanted result in cobol

cobol

gnucobol