Shell Script 中带有 Shebang (#!/bin/bash) 的“-xe”代表什么？

Question

I have copied a shell script and it is pretty much simple but on the first line with shebang, the -xe flag is specified.

#!/bin/bash -xe

Can anyone explain what does it do? I have searched but could not find any helping material.

Answer 1

This is adding shell options to the shebang:

• -x , -o xtrace : Trace display executed statements.
• -e , -o errexit : Exit on error, but inconsistently for legacy POSIX compliance.

#!/bin/bash -xe

This shebang setting combines two discouraged bad practices with their own set of issues:

1. First bad practice is setting shell options in the shebang; because it means they will be ignored when invoking the script directly with: bash script_name , rather than making the script file executable, and invoking ./script_name .
   Options shall be set explicitly in code with set -x , or preferably use their more straightforward long name (example: set -o xtrace instead of set -x ).
2. Second bad practice is using the -e option or set -o errexit which is only there for legacy POSIX compliance, but acts so inconsistently in different situations; that it causes more bugs than it avoids.

Here are some references about those set -e , set -o errexit quirks and issues:

• BashPitfalls § 60.1. errexit - Greg Wooledge
• Why doesn't sh with errexit work with a command list? - Stack Overflow