Batch files - Count the command line argument or pass them to the executable

An interesting post that might help us http://stackoverflow.com/questions/1291941/batch-files-number-of-command-line-arguments, if the user is not user how many argument will be passed to the executable. This code might help
#1 Sample
set argC=0
for %%x in (%*) do Set /A argC+=1
echo %argC%

#2 Sample, more robust
@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
Output
C:\> ArgList.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

Why this article is written: Windows Batch files usually support 0-9 arguments.