Why does this command fail to print the correct exit code of a Python command?

Question

In a cmd.exe window, when I run python -c "exit(12)" and then run echo %errorlevel%, it prints 12. After that, when I run python -c "exit(13)" & echo %errorlevel%, it prints 12.

Why does the second command fail to print the correct exit code?

Both commands are being run within the same environment context. If you won't use the & operator and run them in sequence it will produce the expected value. Python is irrelevant here. — BartoszKP
– BartoszKP, Commented Jul 13, 2014 at 10:08
@BartoszKP Actually I need to use Python because I am trying to solve this problem. — userpal
– userpal, Commented Jul 13, 2014 at 10:20
It's irrelevant what you are trying to solve. In this question Python is irrelevant, as it concerns only the way cmd works. I didn't want to edit it out though, as it seems too much, but the question would be more helpful to others if you made it more general. — BartoszKP
– BartoszKP, Commented Jul 13, 2014 at 10:32

Jeremy Allen · Accepted Answer · 2014-07-13 10:23:05Z

3

The substitution of %errorlevel% happens when you run the command. It reflects the value of %errorlevel% when you pressed your 'enter' key.

As an example consider:

> set foo=foo

> echo %foo%
foo

> set foo=bar & echo %foo%
foo

> echo %foo%
bar

You can check the errorlevel like this however:

> python -c 'exit(13)' & if errorlevel 13 echo "its 13"

answered Jul 13, 2014 at 10:23

Jeremy Allen

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