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DOS批处理 & 脚本技术（批处理室） » Why does this batch file end up showing two "one"s?

Floor1　gool123456	Posted 2010-04-12 00:44
初级用户　Posts 76　Credits 89
@echo off set a=one set b=two echo/Before call :swap a b call :Swap a b echo/After call 1 :swap a b call :Swap b a echo/After call 2 :swap b a pause goto :EOF :Swap call set a=%%%1%% call set b=%%%2%% set "%1=%b%" & set "%2=%a%" & goto :EOF ---------------------------------------------------------------------- Displayed result: Before call :swap a b After call 1 :swap a b After call 2 :swap b a I really can't figure it out, please explain..

Floor2　Hanyeguxing	Posted 2010-04-12 03:24
银牌会员　Posts 897　Credits 1,039　From 在地狱中仰望天堂
When it gets to echo/After call 2 :swap b a call set a=%%%1%% call set b=%%%2%% because %1=b, %2=a that is, set a=!b! set b=!a! so as far as the values are concerned, !a!=!b!=one [ Last edited by Hanyeguxing on 2010-4-12 at 14:04 ]

Floor3　gool123456	Posted 2010-04-12 17:18
初级用户　Posts 76　Credits 89
Your reply again, thanks first of all!! Oh, so that's how it is!! I was treating the variables brought in by CALL as values and doing call set on them, no wonder two disappeared... uh