Batch files - number of command line arguments

Question

Just converting some shell scripts into batch files and there is one thing I can\'t seem to find...and that is a simple count of the number of command line arguments.

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Answer 1

The function :getargc below may be what you're looking for.

@echo off
setlocal enableextensions enabledelayedexpansion
call :getargc argc %*
echo Count is %argc%
echo Args are %*
endlocal
goto :eof

:getargc
    set getargc_v0=%1
    set /a "%getargc_v0% = 0"
:getargc_l0
    if not x%2x==xx (
        shift
        set /a "%getargc_v0% = %getargc_v0% + 1"
        goto :getargc_l0
    )
    set getargc_v0=
    goto :eof

It basically iterates once over the list (which is local to the function so the shifts won't affect the list back in the main program), counting them until it runs out.

It also uses a nifty trick, passing the name of the return variable to be set by the function.

The main program just illustrates how to call it and echos the arguments afterwards to ensure that they're untouched:

C:\Here> xx.cmd 1 2 3 4 5
    Count is 5
    Args are 1 2 3 4 5
C:\Here> xx.cmd 1 2 3 4 5 6 7 8 9 10 11
    Count is 11
    Args are 1 2 3 4 5 6 7 8 9 10 11
C:\Here> xx.cmd 1
    Count is 1
    Args are 1
C:\Here> xx.cmd
    Count is 0
    Args are
C:\Here> xx.cmd 1 2 "3 4 5"
    Count is 3
    Args are 1 2 "3 4 5"