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linuxbashreturn-value

bash: $? != function return value?

Return values in bash functions are a popular topic, but I have not seen an answer that explains the behavior I'm seeing.

I have a bash function library "pt2026_firmware", including a function "pt2026_start_firmware" that explicitly returns 0 for success, 1 for failure. When I run

bash -x -c "source pt2026_firmware; pt2026_start_firmware; echo $?"

I get:

+ source pt2026_firmware
[...stuff deleted...]
+ '[' 0 -eq 0 ']'
+ return 1
+ echo 0
0

I'm obviously missing something very basic, but I don't understand why I'm getting $?=0 when I "return 1"? Thanks in advance for your help!

Solution

  • If you want $? to be expanded after pt2026_start_firmware runs, you need to use single quotes around your code:

    bash -x -c 'source pt2026_firmware; pt2026_start_firmware; echo $?'

    Otherwise, the echo $? is changed to echo 0 before your new copy of bash even starts.