如何在 COBOL 中将十进制值转换为字符串
给出以下代码：
VAR X PIC S9(7)V9(2).
VAR Y PIC X(15)
以下代码有编译错误。
MOVE X TO Y. * compile error.
错误消息类似于“无法将非整数数字移动到字母数字变量”
有没有办法在不使用其他变量（例如显示变量）的情况下解决此问题？
Given following code:
VAR X PIC S9(7)V9(2).
VAR Y PIC X(15)
Following code having compile error.
MOVE X TO Y. * compile error.
the error message is something like this "cannot move non-integer numbers to alphanumeric variable"
is there any way to fix this issue without making use of another variables (e.g. display vars) ?
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修复有时涉及 REDEFINES，如下所示：
请注意，
X占用的存储空间少于Y，因此X可以REDEFINE Y，但反之则不然。 Y 现在占用相同的物理存储，因此可以删除MOVE，如以下程序和输出所示：输出：
由于
X和可以很快看到，结果可能不是您所希望的，因为Y不包含人类可读的格式化数字（即带有前导符号和小数点的数字）。所以我给你的建议是，不要尝试将 COBOL 弯曲成它不是的东西。按照语言的预期用途使用该语言。您可能需要做的是以下内容：
哪些输出：
是的，上面的程序在中间使用了一个附加变量将数字格式转换为显示格式，但这正是该语言的设计方式。啰嗦但非常直接。在练习结束时，变量
Y包含一些对正常人眼可读且有意义的内容。The fix sometimes involves REDEFINES as in:
Notice that
Xoccupies less storage thanYsoXcanREDEFINE Ybut not the other way round. Since bothXandYnow occupy the same physical storage, theMOVEcan be dropped as the following program and output illustrate:Output:
As you can quickly see, the result is probably not what you were hoping for since
Ydoes not contain a human readable formatted number (i.e. one with a leading sign and a decimal point).So my advice to you is don't try bend COBOL into something that it is not. Use the language as it was intended to be used. What you probably need to do is something along the lines of:
Which outputs:
Yes, the above program uses an additional variable in the middle to convert numeric format to display format, but that is exactly how the language was designed. Long winded but very straight forward. At the end of the exercise variable
Ycontains something that is readable and makes sense to the normal human eye.取决于您希望 Y 包含什么，否。
尝试一下，看看什么可以给您带来您想要的答案。
警告，这只是徒手的，我还没有尝试编译它。
Depending on what you want Y to contain, no.
Try it and see what gives you the answer you want.
Caveat, this was just freehand, I haven't tried compiling it.
T 给出右对齐，U 给出左对齐。由于两个 REDEFINES 占用了完整的 15 个字节，因此在 MOVE 到 T 或 U 之前无需考虑 Y 中的任何值。
输出为：
T gives right-alignment, U gives left-alignment. As both REDEFINES occupy the full 15 bytes, there is no need to take account of any value in Y prior to the MOVEs to T or U.
Output is: