Bash – Understanding the ‘declare’ Command

bashcontrol flowdeclarefunctionshell-builtin

After reading ilkkachu's answer to this question I learned on the existence of the declare (with argument -n) shell built in.

help declare brings:

Set variable values and attributes.

Declare variables and give them attributes. If no NAMEs are given,
display the attributes and values of all variables.

-n … make NAME a reference to the variable named by its value

I ask for a general explanation with an example regarding declare because I don't understand the man. I know what is a variable and expanding it but I still miss the man on declare (variable attribute?).

Maybe you'd like to explain this based on the code by ilkkachu in the answer:

#!/bin/bash
function read_and_verify  {
    read -p "Please enter value for '$1': " tmp1
    read -p "Please repeat the value to verify: " tmp2
    if [ "$tmp1" != "$tmp2" ]; then
        echo "Values unmatched. Please try again."; return 2
    else
        declare -n ref="$1"
        ref=$tmp1
    fi
}

Best Answer

In most cases it is enough with an implicit declaration in bash

asdf="some text"

But, sometimes you want a variable's value to only be integer (so in case it would later change, even automatically, it could only be changed to an integer, defaults to zero in some cases), and can use:

declare -i num

declare -i num=15

Sometimes you want arrays, and then you need declare

declare -a asdf   # indexed type

declare -A asdf   # associative type

You can find good tutorials about arrays in bash when you browse the internet with the search string 'bash array tutorial' (without quotes), for example

linuxconfig.org/how-to-use-arrays-in-bash-script

I think these are the most common cases when you declare variables.

Please notice also, that

in a function, declare makes the variable local (in the function)
without any name, it lists all variables (in the active shell)
```
declare
```

Finally, you get a brief summary of the features of the shell built-in command declare in bash with the command

help declare

Related Solutions

Bash – Modify $READLINE_LINE and $READLINE_POINT values inside bash script

Just by the same reason why this won’t work:

$ export a=1
$ bash -c 'echo $a; let a++'
1
$ echo $a
1

Environment variables are heritable, not shareable. Since autocomplete.sh is executed as a new child process, it can read all parent’s variables, but can‘t push new values back.

To modify READLINE_LINE and READLINE_POINT you have to execute your autocomplete in the same process – source and functions will help you.

# autocomplete.sh
# should be sourced from ~/.bashrc or something

autocomplete() {
    echo $READLINE_LINE $READLINE_POINT
    EXPANSION=($(magical_autocomplete $READLINE_LINE))
    #we store the desired value for the line in ${EXPANSION[0]}
    [[ ${#EXPANSION[@]} -gt 1 ]] && echo ${EXPANSION[@]:1}

    READLINE_LINE=${EXPANSION[0]}
    READLINE_POINT=${#READLINE_LINE}
}

Binding:

if [[ -s "$HOME/.bashrc.d/autocomplete.sh" ]]; then
    source "$HOME/.bashrc.d/autocomplete.sh" 
    bind -x '"\t" : autocomplete'
fi

Bash – Does `declare -a A` Create an Empty Array?

That depends whether the corresponding variable has already been declared in the current scope (top-level aka global or current function) before.

If it hasn't been declared in the current scope (and beware that in the top-level scope, the variable may have declared (and assigned) by importing it from the environment), then it declares it (makes it local to the function when in function scope), assigning it a type, but doesn't initialise it, not even to an empty list (declare -p a shows declare -a a, not declare -a a=() as it would if you had declared and/or assigned it with a=()).

If it had already been declared in the current scope (for instance because it was imported as a scalar variable from the environment when in the global scope), then declare -a a would try to convert it to an array.

If it was previously a scalar, then it becomes a ([0]=value-of-the-variable) array. If it was already an array, it is left untouched. If it was an associative array, it fails with a cannot convert associative to indexed array error.

Note that declare a would not convert an array or hash to scalar. bash would not be able to convert a hash/array to scalar anyway. You can use declare +aA a to force a scalar (that would fail with an error if the variable was previously a hash/array in the current scope).

In your case, the variable was probably not already declared in the current scope, so it ended up declared but not assigned which explains why trying to expand it fails under set -u.

That distinction between two declared and assigned/set states of a variable is not specific to bash. In POSIX sh, you can also export a variable or make it readonly without giving it a value.

$ sh -uc 'unset -v var; readonly var; : "$var"'
sh: 1: var: parameter not set

Note that unset both unsets and undeclares the variable. In bash, mksh and yash it may restore the variable from an outer scope.

In zsh, except in sh emulation, using typeset on a variable declares and sets it to an empty value if it was not already set or was set but from a different type (scalar vs array vs associative array).

Best Answer

Related Solutions

Bash – Modify $READLINE_LINE and $READLINE_POINT values inside bash script

Bash – Does `declare -a A` Create an Empty Array?

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