도움이 되었습니까?

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.

라이센스 : CC-BY-SA ~와 함께 속성
제휴하지 않습니다 StackOverflow
scroll top