BASH 로컬 변수 할당의 명령 대체의 종료 코드 [복제]
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15-11-2019 - |
해결책
If you look at the man file for local
(which is actually just the BASH builtins man page), it is treated as its own command, which gives an exit code of 0
upon successfully creating the local variable. So local
is overwriting the last-executed error code.
Try this:
function testing { local test; test="$(return 1)"; echo $?; }; testing
EDIT: I went ahead and tried it for you, and it works.
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