声明具有进程替换的变量时，能否捕获子外壳程序的退出代码？

Question

Assume set -e is active for all that follows.

This will exit non-zero:

x=$(false)

But, none of these will exit non-zero:

declare x=$(false)
readonly x=$(false)
local x=$(false)

If you want to declare a mutable variable you can just do x=$(false) . But if you want a readonly variable you have to resort to something like this: x=$(false) ; readonly x x=$(false) ; readonly x

However, if you are in a function and you want a local, readonly variable you cannot do

x=$(false) ; local -r x

since you will overwrite the global x.

You could do local x=$(false) and the try to inspect x manually to see if there was an error. Or, you might check for output on stderr. But, neither of these will work in all cases; and, they're cumbersome.

Is there any way to get the exit code from a process substitution when assigning the result?

Answer 1

To check the status of the process substitution, the assignment must be the only thing performed by the statement. In a statement such as local x=$(false) , the value returned is the status of local, which typically succeeds. To create a local readonly variable in a function and check the assignment from a process substitution, you can do:

local x
x=$(false) || echo "assigment of x failed !!" >&2
readonly x