Cmd 为什么SHIFT不改变值？

轮班似乎不起作用，或者我不明白。这是我正在努力制作的剧本

C:>type shiftit.bat
@echo off
echo all is %*
echo 0 is %0
echo 5 is %5
shift /5
echo shifted by 5
echo 0 is %0
echo 1 is %1
echo 2 is %2

然而，我本以为在移位之后，%1将包含“5”或“6”。事实并非如此。似乎

SHIFT/5

命令无效。我错过了什么

12:57:25.78  C:\src\t
C:>cmd /E:ON
Microsoft Windows [Version 6.1.7601]
Copyright (c) 2009 Microsoft Corporation.  All rights reserved.

12:57:29.57  C:\src\t
C:>shiftit.bat 1 2 3 4 5 6 7 8 9
all is 1 2 3 4 5 6 7 8 9
0 is shiftit.bat
5 is 5
shifted by 5
0 is shiftit.bat
1 is 1
2 is 2

---

你不明白，对吧， shift/5将删除参数%5，因此%6变为%5、%7->%6等

要删除%1..%5您必须换班5次。
allargs%*不受移位的影响

请参阅更改的批次

:: shiftit.bat 1 2 3 4 5 6 7 8 9
@echo off
echo all is %*
Echo %%1 IS %1
Echo %%2 IS %2
Echo %%3 IS %3
Echo %%4 IS %4
Echo %%5 IS %5
Echo %%6 IS %6
Echo %%7 IS %7
Echo %%8 IS %8
Echo %%9 IS %9
shift /5
echo shifted by 5

echo all is %*
Echo %%1 IS %1
Echo %%2 IS %2
Echo %%3 IS %3
Echo %%4 IS %4
Echo %%5 IS %5
Echo %%6 IS %6
Echo %%7 IS %7
Echo %%8 IS %8
Echo %%9 IS %9

和样本输出：

> shiftit.bat 1 2 3 4 5 6 7 8 9
all is 1 2 3 4 5 6 7 8 9
%1 IS 1
%2 IS 2
%3 IS 3
%4 IS 4
%5 IS 5
%6 IS 6
%7 IS 7
%8 IS 8
%9 IS 9
shifted by 5
all is 1 2 3 4 5 6 7 8 9
%1 IS 1
%2 IS 2
%3 IS 3
%4 IS 4
%5 IS 6
%6 IS 7
%7 IS 8
%8 IS 9
%9 IS

---

这是绝对正确的。在不使用参数编号或使用参数编号的情况下，使用SHIFT命令一次或多次时，

%*

不会发生更改，这确实是经常发生的意外情况。SHIFT仅影响编号的参数引用，即不带或带有由编写的修饰符的

%1

、

%2

、

%3

，但并非所有参数都引用

%*

。微软的命令文档清楚地记录了这一点。