bash's local keyword misconception

[adrelanos]

There is a common misconception about bash's local keyword. I think in #1089 (comment) it's at work.

    #local code=$(declare -p "$array")  # Original line; now replaced with line below
    local code=$(declare -p "$array" 2> /dev/null || true)  # This was line causing error

local will always return 0. Yes. That's it. Code talks...

script:

#!/bin/bash
set -x
set -e
test_function() {
   local somevar=$(this-fails)
   true "does not matter, script goes on"
}
test_function

output:

+ set -e
+ test_function
++ this-fails
./a: line 5: this-fails: command not found
+ local somevar=
+ true 'does not matter, script goes on'

If you preferred it fail, you could not write local and the command in question int the same line.

local somevar
somevar=$(this-fails)

In the example above it probably doesn't matter either way. My point is, if you are using local elsewhere, where non-zero exit codes matter, consider fixing this. That's all. Once agreed on this, this is closeable, unless you think this needs further work.

@nrgaway

[unman]

I've reviewed all bash files under marmarek - I don't see one where this is an issue.
Suggest close.

[unman]

I've reviewed all bash files under marmarek - I don't see one where this is an issue.
Suggest close.

[nrgaway]

Interesting. I was wondering why that statement did not fail; that's the reason I mentioned it. In that case I would not want it to fail anyway.

Thanks for the heads up!

[nrgaway]

Interesting. I was wondering why that statement did not fail; that's the reason I mentioned it. In that case I would not want it to fail anyway.

Thanks for the heads up!