"EXIT" in batch file cannot overwrite return code of previous commands

1
votes

Running this batchfile from the Windows commandline results in an %errorlevel% of 5 (i.e. running echo %errorlevel% on the commandline after having executing the batch file prints the number 5):

EXIT /B 5

This is fine.

However, running this batchfile results in an %errorlevel% of 0, no matter what:

sleep 1
EXIT /B 5

I want it to return the error code 5. How can I do that?

Note: If I add sys.exit(13) to the sleep.py (see below), then the second batch file will return with an exit code of 13. So my batch file will return with the exit code of the sleep.py script instead of the exit code specified via the EXIT command (which is strange).

sleep.bat:

sleep.py %1

sleep.py:

import sys
import time

if len(sys.argv) == 2:
    time.sleep(int(sys.argv[1]))
1
windowsbatch-fileexit-codereturn-code
No, I'm echoing from the commandline. So on the commandline I first enter "my_batch_script.bat" and let it run. Then I enter "echo %errorlevel%". I just tried delayed expansion but it doesn't seem to work on the commandline at all (echoing !errorlevel! just prints !errorlevel!). - robert

1 Answers

3
votes

sleep is a batch file. When calling a batch file from another batch file this way, control never returns to the caller (akin to exec in POSIX). Use call sleep 1 instead.

Or just sleep with built-in programs:

:sleep1
setlocal
set /a X=%~1 + 1
ping ::1 -n %X% >nul 2>&1
endlocal
goto :eof

:sleep2
timeout /T %1 /nobreak
goto :eof