how to check exit status with substitution of command result in bash

Home>2019-12-08 15:18

I checked it out, but I can't do it easily.
Suppose you make the following substitution in a Bash script.

function () {
  result = $(command)
}

At this time, the result ofcommandis assigned toresult, and the exit status of thiscommandis checked at the same time.

In the article here , you can write it as follows It actually worked well. However, ifresultis a local variable, it will not work. Why is this so?

function () {
  status = 0
  result = $(command) || status = $?
  echo $status # This is displayed as 128
}

function () {
  local status = 0
  local result = $(command) || status = $?
  echo $status # will always be 0
}

Answer # 1
local itself seems to be a command, and when it succeeds in creating a local variable, it returns 0, so it seems to overwrite the exit status.
So, it was necessary to create a local variable with local first and then assign it.
```
function () {
  local status = 0
  local result = $(command) || status = $?
  echo $result # will always be 0
}
```

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