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I’m been having some weird problems with bash lately. While trying to simplify my script, I came up with this small piece of code:
$ o(){ echo | while read -r; do return 0; done; echo $?;}; o
0
$ o(){ echo | while read -r; do return 1; done; echo $?;}; o
1
return should have exited the function without printing $?, shouldn’t it? Well, then I checked if I can return from a pipe alone:
$ echo | while read -r; do return 1; done bash: return: can only `return' from a function or sourced script
The same happens without a while loop:
$ foo(){ : | return 1; echo "This should not be printed.";}
$ foo
This should not be printed.
Is there something I’m missing here? A Google search brought nothing about this! My bash version is 4.2.37(1)-release on Debian Wheezy.
Answer
Related: https://stackoverflow.com/a/7804208/4937930
It’s not a bug that you cannot exit a script or return from a function by exit or return in subshells. They are executed in another process and not affecting the main process.
Besides that, I suppose you are seeing undocumented behaviors of bash on (probably) undefined spec. In a function, no errors are asserted for return at top level of subshell commands and it just behaves like exit.
IMHO it’s a bash bug for the inconsistent behavior of return depending on whether the main statement is in a function or not.
#!/bin/bash
o() {
# Runtime error, but no errors are asserted,
# each $? is set to the return code.
echo | return 10
echo $?
(return 11)
echo $?
# Valid, each $? is set to the exit code.
echo | exit 12
echo $?
(exit 13)
echo $?
}
o
# Runtime errors are asserted, each $? is set to 1.
echo | return 20
echo $?
(return 21)
echo $?
# Valid, each $? is set to the exit code.
echo | exit 22
echo $?
(exit 23)
echo $?
Output:
$ bash script.sh 10 11 12 13 script.sh: line 20: return: can only `return' from a function or sourced script 1 script.sh: line 22: return: can only `return' from a function or sourced script 1 22 23