1

I have this bash code:

   local r2g_keep_temp=$(r2g_match_arg "--keep" "${my_args[@]}");
   local r2g_multi_temp=$(r2g_match_arg "--multi" "${my_args[@]}");

   local r2g_multi=[ "$r2g_multi_temp" || "$r2g_keep_temp" ];

I just want r2g_multi to represent a boolean, if either $r2g_multi_temp or $r2g_keep_temp is defined. How can I do that? The above is syntactically invalid.

On the other hand, this is syntactically valid, but not sure if it's correct:

local r2g_multi=$("$r2g_multi_temp" || "$r2g_keep_temp");
1

There are no booleans in the shell scripts.

Best simulation for them are integer values.
For example, 0 for true and 1 for false.

[[ -v variable_name ]] will return 1 if the variable_name was not defined or 0 if it was.

Therefore, you can get your desired behavior with this:

// If one of the variables is defined, return value of this command is 0.
[[ -v r2g_keep_temp ]] || [[ -v r2g_multi_temp ]]

// Save the return value of the last command (0 or 1).
local r2g_multi="$?"

Of courese, you can interpret the numbers however you like, I just wanted to demonstrate one example.


By the way,

local r2g_multi=$("$r2g_multi_temp" || "$r2g_keep_temp");

is not the thing that you wish.

  1. First the variables r2g_multi_temp and r2g_keep_temp will be replaced by their values.
  2. Now the subshell will try to execute the value of the r2g_multi_temp.
  3. If by some miracle that value is a valid bash command there are 2 cases:
    • That command is executed succesfully and its stdout is saved in the r2g_multi.
    • That command failed and subshell invoked the value of the r2g_multi_temp.
      Similar story again, if it is a valid command it will be executed and its stdout will be appended on the possible stdout of the command executed from the value of r2g_keep_temp and everything will be stored in r2g_multi.

All in all, run away from this :D

| improve this answer | |
  • nice, what's the difference between -n $foo and -v $foo? I just heard of using -n to check if a variable is non empty – Alexander Mills May 7 '18 at 23:27
  • "0" and "1" are both defined though, so I need to return either "" or "1" not "0" and "1" – Alexander Mills May 7 '18 at 23:28
  • 1
    foo="", for this variable -n $foo will return 1 while -v foo (note that there is no $) will return 0. I don't understand the second quetion, can you explain it a bit more? – Iskustvo May 7 '18 at 23:57
  • nevermind, I was just confused between -v -z -n – Alexander Mills May 8 '18 at 0:03

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