Bash variable declaration vs assignment

Question

The declare keyword is optional in bash, so my_var="foobar" is the same as declare my_var="foobar". Does bash/dash actually assign new value to variable or it just declares new variable and hides existing variable with the same name?

#!/bin/sh

GetValue ()
{
    echo "foobar";
    return 1;
}

var1="$(GetValue)" && echo "Successed!" || echo "Failed!";  # Failed!
declare var2="$(GetValue)" && echo "Successed!" || echo "Failed!";  # Successed!

In the snippet above, although both variables were successfully declared and assigned, the use of declare makes a difference between the status code. Why?

shellcheck/SC2155 explains this exact case. A tip: Use shellcheck.net (and better - its editor integrations) to check script sanity, which is usually faster than typing a question here. (Note: I am not dissuading you from asking questions. Just introducing to a tool that I find quite useful.) — anishsane
– anishsane, Commented Aug 28, 2019 at 4:33
Despite its name, declare really is just used to set attributes on names or (with -p) display the value assigned to a name. If var2 is undefined, it remains undefined after declare var2. — chepner
– chepner, Commented Aug 28, 2019 at 13:01

Inian · Accepted Answer · 2019-08-28 12:51:11Z

3

The function GetValue along with printing a string to stdout also returns a code 1. So assignment to the variable var1 from the function, though sets the value to foobar, sets an exit code 1 which is a non-zero code causing the part after the || to be invoked.

But remember the same happens in the case with the declare, but the reason you see Successed is because the exit code 1 from the function is suppressed when the declare built-in acts on the variable. So

var2="$(GetValue)"  # sets a non-zero exit code 1
declare var2        # sets a success exit code 0

Since the last foreground command's exit code is the one used in the evaluation of success/failure, the part after &&, which is echo "Successed!" gets invoked.

So, your claim of using declare in the variable assignment makes a difference in the exit code is incorrect, because these two still remain the same as they both set the exit code 1 which triggers the false case.

var1="$(GetValue)" && echo "Successed!" || echo "Failed!";
var2="$(GetValue)" && echo "Successed!" || echo "Failed!";

Suggest to always separate the initialization/declaration when using the built-ins local, declare and export

var=5
declare -i var

is preferred over declare -i var=5.

See Why does local/declare sweep the return code of a command? for more reference on the subject.

edited Aug 28, 2019 at 12:51

answered Aug 28, 2019 at 4:06

Inian

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1 Comment

Stéphane Jul 31 at 22:58

Is there a good reason affectation of the value is done before using declare on that variable? I’d do the declare -i var first, especially with -i, as could change what value is affected in some case, like affecting 4+5, which would be evaluated to 9 in case the variable is declared an integer, but the string 4+5 itself if a “standard” variable. I noticed the the declare -p (or declare -fp) could effectively gives the affectation first, before the declare key word is used. What the subtle bash-candy there?

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