When is the CD environment variable updated? - batch-file

Question

The tipp to use the CD environment variable from batch scripts to get the current working directory is commonly posted. But CD won't get updated when calling another batch file. Then the cd command echoes the new path of the other batch file, but %CD% (or !CD!) is not updated. Example:
#echo off
cd %~dp0
echo in %0: CD=%CD%
pause
call X:\testcall.cmd
Save this as C:\testcall.cmd and X:\testcall.cmd, then run C:\testcall.cmd. You should see that the value of CD has not changed. This seems not to dependend on call; none of the following works:
start /D <NEW_DIR> <OTHER_CMD_FILE>
start cmd /c <NEW_DIR>\<OTHER_CMD_FILE>
cmd /c <NEW_DIR>\<OTHER_CMD_FILE>
<NEW_DIR>\<OTHER_CMD_FILE>
cd %~dp0
pushd %~dp0
CD will keep it's old value, while cd (the command) shows the correct directory. Therefore I set CD at the begin of a script:
set CD=%~dp0
...while assuming cmd.exe sets CD only if this variable is yet unset. True?

Answer 1

%CD% is the current directory, while %~dp0 is the directory of the currently-running script (with trailing '\').
Also, don't set an env. var called CD, since it will override the default %CD% pseudo-var, and will be incredibly confusing - see OldNewThing - ERRORLEVEL is not %ERRORLEVEL%.
For example, when running c:\temp\a.cmd, which is:
#echo off
echo Currently running script: %~dpnx0
cd %~dp0
echo scriptDir=%~dp0, CD=%CD%
cd %~dp0a
echo scriptDir=%~dp0, CD=%CD%
set CD=bogus value
echo scriptDir=%~dp0, CD=%CD%
output:
Currently running script: c:\temp\a.cmd
scriptDir=c:\temp\, CD=c:\temp
scriptDir=c:\temp\, CD=c:\temp\a
scriptDir=c:\temp\, CD=bogus value

Answer 2

Diagnosis
You have at some point set the CD variable explicitly. If you do this it will no longer automatically reflect the current working directory. To undo this, set it to empty:
set CD=
It will then begin working again.
Why is this? Well, the automatic CD variable was introduced as a feature. I assume they just didn't want to break pre-existing scripts which already used that varible name. So if you set it explicitly, CMD will assume you are doing so on purpose.
Discussion
Firstly, if the parent process has an explicitly set CD variable, it will be inherited by the child processes.
On the other hand, you shouldn't expect any of these to update the value of %CD% for the parent process:
start /D <NEW_DIR> <OTHER_CMD_FILE>
start cmd /c <NEW_DIR>\<OTHER_CMD_FILE>
cmd /c <NEW_DIR>\<OTHER_CMD_FILE>
These all create new processes, the new process then changes its own working directory. You should not expect this to affect the parent process.
The final one, does not update the working directory at all, unless OTHER_CMD_FILE executes a CD command:
<NEW_DIR>\<OTHER_CMD_FILE>
Just because you executed a script in a different directory does not mean that the script's working directory will change. The script working directory does not have to be set to the location of the script.
Advice
Relying on the working directory being set to anything in particular is generally a bad idea.
You probably want something like this:
SET SCRIPT_DIR=%~dp0
Then use (for example) "%SCRIPT_DIR%\config.txt" to refer to a file in that directory.
Alternatively if you wish to rely on the current directory, use cd /d %~dp0

Answer 3

I build two batch files from your comment
test1.bat - located in C:\temp
#echo off
cd %~dp0
echo File %~f CD=%CD%
call X:\test2.bat
test2.bat - located in *X:*
#echo off
cd %~dp0
echo File %~f CD=%CD%
Starting test1 it from C:\temp, the output is
File C:\temp\test1.bat CD=C:\temp
File X:\test2.bat CD=C:\temp
The result is absolutly correct!
Starting a batch file (or any other program) with an absolute or relative path doesn't change the current directory.
The CD seems to fail, as CD can't change the drive the way you used it.
You need to add a switch CD /d %~dp0

Answer 4

You can set the %cd% variable to whatever you want, the real current directory for the C: drive is stored in the %=c:% variable, and you can't change this with the set command:
#echo off
echo Currently running script: %~dpnx0
cd %~dp0
echo scriptDir=%~dp0, CD=%CD%
set CD=bogus value
echo scriptDir=%~dp0, CD=%CD%, =c:=%=c:%
set =c:=bogus value
echo scriptDir=%~dp0, CD=%CD%, =c:=%=c:%
Output is:
Currently running script: C:\OldDir\a.bat
scriptDir=C:\OldDir\, CD=bogus value
scriptDir=C:\OldDir\, CD=bogus value, =c:=C:\OldDir
syntax error.
scriptDir=C:\OldDir\, CD=bogus value, =c:=C:\OldDir
The =C: variable for a child process is always set from the parent process. If you avoid the setlocal command in your script OR choose endlocal, you can change persistend the current directory for the current cmd session:
C:\OldDir>type script.bat
cd c:\newdir
C:\OldDir>script
C:\OldDir>cd c:\newdir
C:\NewDir>
.
C:\OldDir>type script.bat
setlocal
cd c:\newdir
C:\OldDir>script
C:\OldDir>setlocal
C:\OldDir>cd c:\newdir
C:\OldDir>

Answer 5

Using the script-directory %~dp0 can be a solution, but usually isn't. This works better:
cd >tmpfile
set /P CD= <tmpfile
del tmpfile
This solution is consistent with the CD variable. CD contains the current directory, not ending in a slash character if it is not in the root directory of the current drive. The path printed by cd behaves exactly so.
I'm using this for years and had no problems setting CD explicitly. Later I began using PWD (as in Bash). This variable is not reserved in MSDOS. So my question here was actually: "Can we get rid of these lines, or is this some MSDOS idiom?"
Some have written that setting CD explicitly is bad. Why? MSODS was never properly designed, and is not further developed. Anything you can do with it is legal. There is no bad MSDOS programming - just good and bad hacks. I know no other language where this perspective is legal to such an extent.