linux - bash script invocation shows inconsistent behavior -


i have written test script( let's call test.sh) in bash. , start of script has following declaration:

#!/bin/bash -e

followed code.

now shows different results when invoke bash test.sh versus ./test.sh. how possible. thought default ./test.sh reads default shell first line of script , invokes that. understanding correct.

the #!/bin/bash -e shebang used when (try to) execute file (that isn’t binary executable) command. line informs kernel executable should used interpret code in script. see shebang wikipedia article more information.

when run ./test.sh, current shell starts new bash sub-process -e option , new process 1 executes commands in script file.

when run bash test.sh, you’re explicitly starting new bash sub-process (this time without -e option) , starts executing commands in file. when encounters shebang in first line, treats same other comment beginning #. (the shebang ignored comment if source file using . test.sh).

to ensure bash shell behaves way, it’s best use set builtin instead of providing option when invoking bash. e.g., ensure shell exits when command returns non-zero status, should include set -e @ or near start of script.

for more in-depth information how commands executed linux kernel, see this answer what difference between running “bash script.sh” , “./script.sh” , execve man page.


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