comp -3字段中二进制文件的转换

Question

我正在尝试转换一个图片S9 (9)V99 comp-3字段，它已经存储在二进制文件中。它是这样的显示: 6/PS X'000000002000‘。

这个数字应该是这样的。我正试图找到正确的方法，将其转换为PP-9 (9)V99字段。

谢谢

Answer 1

它基本上与Decode a Binary Coded Decimal相同

您创建了一个comp-3，再加上一个小数位数，并做'pic x‘移动。

  01  WS-AMT-IN                 PIC S9(009)V99  COMP-3.
  01  WS-AMT-IN-X  REDEFINES                 
      WS-AMT-IN                 PIC  X(006). 

  01  WS-AMT-OUT1               PIC S9(009)V999  COMP-3.
  01  REDEFINES  WS-AMT-OUT1               
      03 WS-AMT-OUT1-X          PIC  X(006).   
      03                        PIC  s9 comp-3 value zero.

 01  WS-AMT-OUT-2               PIC S9(009)V99  COMP-3.


  Move  X'000000002000'         to WS-AMT-IN-X  
  Move  WS-AMT-IN-X             to WS-AMT-OUT1-x 
  Move  WS-AMT-OUT1             to WS-AMT-OUT-2 

Answer 2

下面是我解决这个问题的方法：

01 WS-ZONENUM11.                                
   05 WS-ZONE9NUM                  PIC 9(009).  
   05 WS-ZONE2NUM                  PIC 9(002).  
01 WS-ZONENUM11-RED    REDEFINES WS-ZONENUM11   
                                   PIC 9(09)V99.

01 WS-AMT-OUT                         PIC -9(009).99. 

01 WS0900-AMT-IN               COMP-3 PIC S9(009)V99.
01 WS0900-AMT-IN-RED  REDEFINES                 
         WS0900-AMT-IN                PIC  X(006).   


 MOVE WS0900-AMT-IN-RED         TO  WS-WS0900-AMT-IN.
 MOVE WS-ZONEX5NM               TO  WS-ZONE9NUM.          
 MOVE WS-ZONEX1NM               TO  WS-ZONE2NUM.          

 MOVE WS-ZONENUM11-RED          TO  WS-AMT-OUT.

WS现在显示为_00000020.00，其中_是符号(这里的符号将始终是空的，因为它不在二进制数量字段中。

Answer 3

IDENTIFICATION DIVISION.
PROGRAM-ID. XYZ.
DATA DIVISION.
WORKING-STORAGE SECTION.
01 WS-BIN PIC 9(10).
01 TEMP PIC Z(10).
01 EXP PIC 9(2) VALUE 0.
01 R1 PIC 9(10).
01 C PIC 9(10).
PROCEDURE DIVISION.

DISPLAY " ENTER THE BINARY NUMBER ".
ACCEPT WS-BIN.

PARA1.
DIVIDE WS-BIN BY 10 GIVING WS-BIN REMAINDER R1.
COMPUTE C = C + R1 * 2 ** EXP
ADD 1 TO EXP.
IF WS-BIN NOT = 0
GO TO PARA1
ELSE GO TO PARA2.
PARA2.

MOVE C TO TEMP
DISPLAY " THE DECIMAL NUMBER IS " TEMP.
STOP RUN.