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 last answer was two years ago now, but I needed a version for more than nine command line arguments. May be another one also does...

@echo off
setlocal

set argc_=1
set arg0_=%0
set argv_=

:_LOOP
set arg_=%1
if defined arg_ (
  set arg%argc_%_=%1
  set argv_=%argv_% %1
  set /a argc_+=1
  shift
  goto _LOOP
)
::dont count arg0
set /a argc_-=1
echo %argc_% arg(s)

for /L %%i in (0,1,%argc_%) do (
  call :_SHOW_ARG arg%%i_ %%arg%%i_%%
)

echo converted to local args
call :_LIST_ARGS %argv_%
exit /b


:_LIST_ARGS
setlocal
set argc_=0
echo arg0=%0

:_LOOP_LIST_ARGS
set arg_=%1
if not defined arg_ exit /b
set /a argc_+=1
call :_SHOW_ARG arg%argc_% %1
shift
goto _LOOP_LIST_ARGS


:_SHOW_ARG
echo %1=%2
exit /b

The solution is the first 19 lines and converts all arguments to variables in a c-like style. All other stuff just probes the result and shows conversion to local args. You can reference arguments by index in any function.