英文:
What does `exec {logOutFd}>&1 {logErrFd}>&2` do?
问题
我不理解这一行：
exec {logOutFd}>&1 {logErrFd}>&2
这里没有变量引用，看起来只是一个空操作，就像没有参数运行exec一样。发生了什么？
英文:
I don't understand this line:
exec {logOutFd}>&1 {logErrFd}>&2
There are no variable dereferences, and it seems to be just a no-op, like running exec with no parameters. What's going on?
答案1
得分: 4
exec没有参数时用于启动持续到脚本结束的重定向。
根据Bash手册中的重定向部分：
> 每个可能由文件描述符号前导的重定向都可以由形式为*{varname}的单词前导。在这种情况下，对于除>&-和<&-之外的每个重定向操作符，Shell将分配一个大于10的文件描述符号并将其分配给{varname}*。
因此，这将stdout和stderr复制到新的文件描述符，并分别将变量$logOutFd和$logErrFd设置为这些描述符。
这允许稍后的代码在stdout或stderr被重定向的情况下写入原始的stdout或stderr，使用：
echo stdout message >&$logOutFd
英文:
exec with no parameters is used to start redirections that last for the rest of the script.
From the Bash manual section on redirection:
> Each redirection that may be preceded by a file descriptor number may instead be preceded by a word of the form {varname}. In this case, for each redirection operator except >&- and <&-, the shell will allocate a file descriptor greater than 10 and assign it to {varname}.
So this is duplicating stdout and stderr to new file descriptors, and setting the variables $logOutFd and $logErrFd to these descriptors, respectively.
This allows later code that executes while stdout or stderr are redirected to write to the original stdout or stderr, with
echo stdout message >&$logOutFd
答案2
得分: -1
Reviewing other preceding script: https://github.com/search?q=repo%3ANixOS%2Fnixpkgs%20%20logOutFd&type=code
And reading:
They are using complex i/o redirections to return the stream pointer from some files previously open (8 and 9 position) to the classic stdout and stderr
exec 8>&1 9>&2
英文:
Reviewing other preceding script: https://github.com/search?q=repo%3ANixOS%2Fnixpkgs%20%20logOutFd&type=code
And reading:
They are using complex i/o redirections to return the stream pointer from some files previously open (8 and 9 position) to the classic stdout and stderr
exec 8>&1 9>&2
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