Bash - check if argument is string

Question

I am attempting to check if an argument is an array with the following code:

if [[ $(declare -p $1) ]] != *-a*;

Here $1 is a string with the value "123". I get the following error message from bash:

`arrays.bash: line 23: declare: 123: not found

This code works if I pass an array as an argument but not a string. I want to verify that the argument is either an array or an associative array. I have no concern with the contents at this point, I only want the type. Any ideas on how to do this?

$1 is always a string. Always. It cannot be anything else. That also means it cannot be an array. — Charles Duffy
– Charles Duffy, Commented Jan 29, 2023 at 0:45
One could pass a string with the name of an array in $1, but it's still a string. — Charles Duffy
– Charles Duffy, Commented Jan 29, 2023 at 0:46
Anyhow -- the code in your question checks what kind of variable the string in $1 refers to, with the assumption that the string is in fact a variable name; it's falling when you aren't passing a variable name at all. This is to be expected. — Charles Duffy
– Charles Duffy, Commented Jan 29, 2023 at 0:47
...regardless, you aren't providing a minimal reproducible example showing us how you're defining the array and how you're trying to pass it; from the question I don't even know that you're using a real array at all (we get a lot of n00bs using strings that contain whitespace and calling them arrays), much less how you're attempting to pass it. — Charles Duffy
– Charles Duffy, Commented Jan 29, 2023 at 1:04
Again, we need an actual minimal reproducible example. I don't know when you say "passing c" you mean yourfunc c or yourfunc "$c" or yourfunc "${c[@]}" or something else. Runnable code is far more precise than English-language descriptions of code. — Charles Duffy
– Charles Duffy, Commented Jan 29, 2023 at 1:12

Diego Torres Milano · Accepted Answer · 2023-01-29 01:51:30Z

1

After all, why worry about the types, if you are relying on it perhaps your approach is wrong or you may need a strong-typed language

% v=1
% declare -p v
declare -- v="1"
% echo $v
1
% echo ${v[@]}
1
% v[1]=2
% declare -p v
declare -a v=([0]="1" [1]="2")
% echo ${v[@]}
1 2

answered Jan 29, 2023 at 1:51

Diego Torres Milano

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Charles Duffy · Accepted Answer · 2023-01-31 22:34:18Z

The Error In The Question

You called yourfunction "$a" instead of yourfunction a, when a=123. Don't do that: You need to pass the name of the variable, not its value.

General Solution: Bash 5.x+

Bash 5 has a new feature called parameter transformation, whereby ${parameter@operator} can perform a variety of actions; one of these is checking the type of the parameter.

myfunc() {
  [[ -v "$1" ]] || { echo "No variable named $1 exists" >&2; return 1; }
  case ${!1@a} in
    *a*) echo "Array";;
    *A*) echo "Associative array";;
    *i*) echo "Integer";;
    "")  echo "Default string";;
    *)   echo "Other/unknown flag set: ${!1@a}";;
  esac
}

Older Solution

myfunc() {
  local typedesc
  typedesc=$(declare -p "$1" 2>/dev/null) || {
    echo "No variable named $1 is set" >&2
    return 1
  }
  case $typedesc in
    "declare -a"*) echo "Array";;
    "declare -A"*) echo "Associative array";;
    "declare -i"*) echo "Integer";;
    "declare --"*) echo "Regular (default) string variable";;
    *)             echo "Other/unrecognized type";;
  esac
}

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