Why do bash variable assignments affect the last exit code $? and where is this documented?

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edited Aug 16, 2016 at 20:31

I was surprised to find out that setting an environment variable modified the last exist code variable $?:

$ false
$ echo $?
1
$ false
$ FOO=BAR
$ echo $?
0
$

The bash manual says only this about $?:

($?) Expands to the exit status of the most recently executed foreground pipeline.

Where is the exit code of a variable being set documented?Where is the exit code of a variable being set documented? Can it fail?

I was surprised to find out that setting an environment variable modified the last exist code variable $?:

$ false
$ echo $?
1
$ false
$ FOO=BAR
$ echo $?
0
$

The bash manual says only this about $?:

($?) Expands to the exit status of the most recently executed foreground pipeline.

Where is the exit code of a variable being set documented? Can it fail?

I was surprised to find out that setting an environment variable modified the last exist code variable $?:

$ false
$ echo $?
1
$ false
$ FOO=BAR
$ echo $?
0
$

The bash manual says only this about $?:

($?) Expands to the exit status of the most recently executed foreground pipeline.

Where is the exit code of a variable being set documented? Can it fail?

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asked Aug 16, 2016 at 20:10

Why do bash variable assignments affect the last exit code $?

I was surprised to find out that setting an environment variable modified the last exist code variable $?:

$ false
$ echo $?
1
$ false
$ FOO=BAR
$ echo $?
0
$

The bash manual says only this about $?:

($?) Expands to the exit status of the most recently executed foreground pipeline.

Where is the exit code of a variable being set documented? Can it fail?