Ankit Kumar February 2016

bash script invocation shows inconsistent behavior

I have written a test script( let's call it test.sh) in bash. and the start of the script has following declaration:

#!/bin/bash -e

followed by some code.

Now it shows different results when i invoke it as bash test.sh versus ./test.sh. How is it possible. I thought by default ./test.sh reads the default shell from the first line of script and invokes it with that. Is my understanding correct.

Answers


Anthony Geoghegan February 2016

The #!/bin/bash -e shebang is only used when you (try to) execute a file (that isn’t a binary executable) as a command. This line informs the kernel which executable should be used to interpret the code in the script. See the shebang Wikipedia article for more information.

When you run ./test.sh, your current shell starts a new Bash sub-process with the -e option and this new process is the one that executes the commands in the script file.

When you run bash test.sh, you’re explicitly starting a new Bash sub-process (this time without the -e option) and it starts executing the commands in the file. When it encounters the shebang in the first line, it just treats it the same as any other comment beginning with #. (The shebang is also ignored as just another comment if you source the file using . test.sh).

To ensure that the Bash shell behaves a certain way, it’s best to use the set builtin instead of providing an option when invoking Bash. E.g., to ensure the shell exits when a command returns a non-zero status, you should include set -e at or near the start of the script.


For more in-depth information about how commands are executed with a Linux kernel, see this answer to What is the difference between running “bash script.sh” and “./script.sh” and the execve man page.

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