Batch files - number of command line arguments - shell

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.
eg. if you have:
myapp foo bar
In Shell:
$# -> 2
$* -> foo bar
$0 -> myapp
$1 -> foo
$2 -> bar
In batch
?? -> 2 <---- what command?!
%* -> foo bar
%0 -> myapp
%1 -> foo
%2 -> bar
So I've looked around, and either I'm looking in the wrong spot or I'm blind, but I can't seem to find a way to get a count of number of command line arguments passed in.
Is there a command similar to shell's "$#" for batch files?
ps. the closest i've found is to iterate through the %1s and use 'shift', but I need to refernece %1,%2 etc later in the script so that's no good.

Answer 1

Googling a bit gives you the following result from wikibooks:
set argC=0
for %%x in (%*) do Set /A argC+=1
echo %argC%
Seems like cmd.exe has evolved a bit from the old DOS days :)

Answer 2

You tend to handle number of arguments with this sort of logic:
IF "%1"=="" GOTO HAVE_0
IF "%2"=="" GOTO HAVE_1
IF "%3"=="" GOTO HAVE_2
etc.
If you have more than 9 arguments then you are screwed with this approach though. There are various hacks for creating counters which you can find here, but be warned these are not for the faint hearted.

Answer 3

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"

Answer 4

Try this:
SET /A ARGS_COUNT=0
FOR %%A in (%*) DO SET /A ARGS_COUNT+=1
ECHO %ARGS_COUNT%

Answer 5

If the number of arguments should be an exact number (less or equal to 9), then this is a simple way to check it:
if "%2" == "" goto args_count_wrong
if "%3" == "" goto args_count_ok
:args_count_wrong
echo I need exactly two command line arguments
exit /b 1
:args_count_ok

Answer 6

Avoids using either shift or a for cycle at the cost of size and readability.
#echo off
setlocal EnableExtensions EnableDelayedExpansion
set /a arg_idx=1
set "curr_arg_value="
:loop1
if !arg_idx! GTR 9 goto :done
set curr_arg_label=%%!arg_idx!
call :get_value curr_arg_value !curr_arg_label!
if defined curr_arg_value (
echo/!curr_arg_label!: !curr_arg_value!
set /a arg_idx+=1
goto :loop1
)
:done
set /a cnt=!arg_idx!-1
echo/argument count: !cnt!
endlocal
goto :eof
:get_value
(
set %1=%2
)
Output:
count_cmdline_args.bat testing more_testing arg3 another_arg
%1: testing
%2: more_testing
%3: arg3
%4: another_arg
argument count: 4
EDIT: The "trick" used here involves:
Constructing a string that represents a currently evaluated command-line argument variable (i.e. "%1", "%2" etc.) using a string that contains a percent character (%%) and a counter variable arg_idx on each loop iteration.
Storing that string into a variable curr_arg_label.
Passing both that string (!curr_arg_label!) and a return variable's name (curr_arg_value) to a primitive subprogram get_value.
In the subprogram its first argument's (%1) value is used on the left side of assignment (set) and its second argument's (%2) value on the right. However, when the second subprogram's argument is passed it is resolved into value of the main program's command-line argument by the command interpreter. That is, what is passed is not, for example, "%4" but whatever value the fourth command-line argument variable holds ("another_arg" in the sample usage).
Then the variable given to the subprogram as return variable (curr_arg_value) is tested for being undefined, which would happen if currently evaluated command-line argument is absent. Initially this was a comparison of the return variable's value wrapped in square brackets to empty square brackets (which is the only way I know of testing program or subprogram arguments which may contain quotes and was an overlooked leftover from trial-and-error phase) but was since fixed to how it is now.

Answer 7

A reasonably robust solution is to delegate counting to a subroutine invoked with call; the subroutine uses goto statements to emulate a loop in which shift is used to consume the (subroutine-only) arguments iteratively:
#echo off
setlocal
:: Call the argument-counting subroutine with all arguments received,
:: without interfering with the ability to reference the arguments
:: with %1, ... later.
:: NOTE: See comments re /? below.
call :count_args %* >NUL || (echo Usage: ... & exit /b 0)
:: Print the result.
echo %ReturnValue% argument(s) received.
:: Exit the batch file.
exit /b
:: Subroutine that counts the arguments given.
:: Returns the count in %ReturnValue%
:count_args
set /a ReturnValue = 0
:count_args_for
if %1.==. goto :eof
set /a ReturnValue += 1
shift
goto count_args_for
Limitations:
/? among the arguments - except if passed as "/?" - is not supported, because call invariably interprets it as a request to show its command-line help.
The code above detects this case (curiously, call signals failure via its exit code after showing its help) and interprets this case as wanting to show the current batch file's help instead of counting arguments.
Arguments with metacharacters (special characters) are supported, if enclosed in "..." (as is typical), but character-individual ^-escaping in an unquoted argument does not work (because the ^ is already stripped by the time the batch file sees the arguments, and expanding %* then causes the metacharacters to act as such):
:: OK - arguments with special chars. are "..."-enclosed.
:: -> "3 argument(s) received."
test.cmd hi "you & me" "a<b"
:: !! FAILS
:: The last argument uses character-individual ^-escaping.
:: -> error "The system cannot find the file specified."
test.cmd hi "you & me" a^<b
A syntactically invalid argument list is not supported (which is arguably not a problem):
:: !! FAILS
:: Syntactically invalid argument list
:: -> error "The syntax of the command is incorrect."
test.cmd a"

Answer 8

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.