【发布时间】:2013-04-25 17:13:31
【问题描述】:
为什么我不能在 COBOL 中执行这个嵌套的执行循环?
如果我把 END-PERFORM.在任何一行都比我在退出程序之前的最后一行更早 - 它有效。但我需要程序每次都显示 INPUT C 值。在外部执行循环中。快把我逼疯了。
PROCEDURE DIVISION USING INPUTC CIPHER.
COMPUTE CIPHERMAX = CIPHER.
MULTIPLY -1 BY CIPHER
---> PERFORM VARYING CIPHER FROM 0 BY 1
UNTIL CIPHERMAX = CIPHER
DISPLAY 'This is loop number: ' CIPHER
INSPECT INPUTC CONVERTING
"avcdefghijklmnopqrstuvwxyz" to "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
COMPUTE CONVERTNUM = FUNCTION MOD (CIPHER, 26)
INSPECT FUNCTION REVERSE(INPUTC) TALLYING LENGTHNUM FOR LEADING SPACES
COMPUTE LENGTHNUM = LENGTH OF CIPHER - LENGTHNUM
---> PERFORM UNTIL SENTRY = LENGTHNUM
IF ((FUNCTION ORD(INPUTC(SENTRY:1)) + CONVERTNUM) > (FUNCTION ORD('Z')))
MOVE FUNCTION CHAR((FUNCTION ORD(INPUTC(SENTRY:1)) + CONVERTNUM) - 26) TO RECHAR
ELSE
MOVE FUNCTION CHAR(FUNCTION ORD(INPUTC(SENTRY:1)) + CONVERTNUM) TO RECHAR
END-IF
IF (((FUNCTION ORD(INPUTC(SENTRY:1))) >= (FUNCTION ORD('A'))) AND
((FUNCTION ORD(INPUTC(SENTRY:1))) <= (FUNCTION ORD('Z'))))
IF ((FUNCTION ORD(INPUTC(SENTRY:1)) + CONVERTNUM) > (FUNCTION ORD('Z')))
INSPECT INPUTC(SENTRY:1) REPLACING ALL INPUTC(SENTRY:1) BY RECHAR
ELSE
INSPECT INPUTC(SENTRY:1) REPLACING ALL INPUTC(SENTRY:1) BY RECHAR
END-IF
ELSE
INSPECT INPUTC(SENTRY:1) REPLACING ALL INPUTC(SENTRY:1) BY INPUTC(SENTRY:1)
END-IF
COMPUTE SENTRY = SENTRY + 1
---> END-PERFORM
DISPLAY INPUTC.
COMPUTE LOOPI = LOOPI + 1
--->END-PERFORM.
EXIT PROGRAM.
END PROGRAM SOLVE.
【问题讨论】:
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您有一个 IF,其 ELSE 具有相同的代码。您还有一个带有 INSPECT 的 ELSE,它自己替换了一个字符。为什么要使用 INSPECT ... REPLACING 处理单字节字段?为什么不只是移动? LOOPI 有什么用?为什么不 ADD 1 TO SENTRY 而不是 COMPUTE?你乘-1,减...从零会更快。但是随后您立即在 PERFORM 中将 CIPHER 设置为 0,那么为什么要否定呢?我怀疑还有更多,但你用 FUNCTION on FUNCTION 让整个事情变得难以理解。
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另外,在
avcdefghijklmnopqrstuvwxyz中,您的意思可能是abc...