I have one root directory from where I am running master batch file, which call another batch file(2nd batch file) present in sub directory of root directory.

After the execution of 2nd batch file, would like to remain in root directory:

@echo off 
set root_dir=C:\Users\milan\Desktop 

cd "%root_dir%\2nd_dir"
call 2nd_batch.bat

echo %cd%

Last echo command should show me the patch of root directory.I tried doing this

cd "\%root_dir%\2nd_dir\"

But it didn't work.

  • 1
    Why not just do call "2nd_dir\2nd_batch.bat"? Alternatively, look up the help for the pushd and popd commands.
    – Berend
    Nov 26 '19 at 8:19
  • @Berend, Thanks for reply.To be honest , would like to get what have been answered here, superuser.com/questions/1062271/…. "don't forget the first \ at the beginning of the cd ", its not just working with variable name
    – Milan
    Nov 26 '19 at 8:23

The problem you're facing is with how the call instruction works.

When you use the call instruction in a batch file, it will include that batch file into the current script and run its code. That means that all changes the new script makes to your environment are carried over to the previous batch file, as you have noticed.

You can use the start command instead of the call command to start the new batch file in a new process, thus not carrying over its changes to the previous batchfile.

Alternatively, you can store the current folder before executing the new script and return back to it after the script is finished.

Below are 2 examples. The :: row is a remark. You can copy it inside your script or ommit it if you want.

:: example that uses start
cd "%root_dir%\2nd_dir"
start 2nd_batch.bat

echo %cd%

:: example that stores and sets the path

:: store current folder for retrieval
set masterfolder=%cd%
cd "%root_dir%\2nd_dir"
call 2nd_batch.bat

:: restore folder
cd /d %masterfolder%

echo %cd%
  • 1
    You can use the start command instead of the call command to start the new batch file in a new process, thus not carrying over its changes to the previous batchfile. You must use start /w.
    – Akina
    Nov 26 '19 at 18:19

Use setlocal for to save current settings (current drive, folder, environ, ...), and endlocal for to restore them.


@echo off
echo test1-1 : %cd%
call test2.bat
echo test1-2 : %cd%
echo test1-3 : %cd%


echo test2-1 : %cd%
cd \
echo test2-2: %cd%

enter image description here

PS. setlocal/endlocal may be nested.


Open cmd.exe and do the following queries.


Problem solved.

A lot of answers can be found by doing queries on commands Listed when you use the help /? query.

PUSHD allows you to change directory in a similar way to CD, so:

PUSHD "DirectoryPath"

However, PUSHD stores the previous directory for recall by POPD.

So in the situation you described, you PUSHD to the directory used by or needed for your other batch file, then when you need to return to the previous directory, just use


And voilá, you're there.

To resolve this using PUSHD / POPD:

In your main batch file:

PUSHD "PathFor2ndBatch"
call 2nd_batch.bat
echo %cd%

If the aim is to have the second batch file execute in the directory of your main batch file:

PUSHD "PathFor2ndBatch"
CALL 2nd_Batch.bat

In the start of your 2nd batch:


You can also start your batch file with /D switch:

start /?
START [/D path]
path        Starting directory

And so, running the below code will make the batch file run inside C:\users directory.

start /D C:\users C:\Windows\System32\RCWM\rmdir.bat

Although, like Akina mentioned in the comments, start /w is shorter for this purpose.

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