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shellbackground-process

Order of instructions in shell script

I have this simple shell script:

#!/bin/sh

echo "alpha"

finish()
{
    echo "--------------------------------------------------------------"
}
trap finish EXIT

echo "bravo"

cd /non/existent/path
# cd /non/existent/path && gitk &

echo "charlie"

When run, this gives the following output as expected:

alpha
bravo
./test.sh: 13: cd: can't cd to /non/existent/path
charlie
--------------------------------------------------------------

But then, if I uncomment the line containing the gitk call (i.e. line 14), I get this output:

alpha
bravo
./test.sh: 13: cd: can't cd to /non/existent/path
charlie
--------------------------------------------------------------
./test.sh: 14: cd: can't cd to /non/existent/path

Notice how the line with the dashes is no longer at the end. Any idea why this happens?

I also noticed that if I remove the ampersand at the end of that line, the order is once again as I'd expect. So starting gitk as a background process clearly has an impact, but I don't know how exactly.

Finally, is it possible to modify the script such that I can continue to start gitk as a background process but still have the line with the dashes appear at the end?

Solution

  • Behavior is as expected since You spawn a child process via &. If you wanna Your trap function to be executed after just add wait in end of Your script:

    #!/bin/sh

echo "alpha"

finish()
{
    echo "--------------------------------------------------------------"
}
trap finish EXIT

echo "bravo"

cd /non/existent/path
cd /non/existent/path && gitk &

echo "charlie"
wait

    That will wait on all child processes to be completed and then exits (call trap finish). Output:

    $ ./gautam.sh
alpha
bravo
./gautam.sh: line 13: cd: /non/existent/path: No such file or directory
charlie
./gautam.sh: line 14: cd: /non/existent/path: No such file or directory
--------------------------------------------------------------