Fish shell 脚本：添加一个列表作为函数的参数

Question

I would like to pass a list of variable length as a parameter into a fish function called and defined within a fish script, similar to how $argv is a list on the top level of a fish script I guess.

function main_function
   function f --argument xs
      echo "called f"
      echo $xs     
   end
    set xs "one" "two" "three";
    f $xs
end

calling function --help mentions the --inherit-variable option. While setting the option causes the xs variable within the f's scope to be the full list, it makes a dumb copy at the point where the function is defined. I need the variable to be copied when f is called, however.

function main_function
   set xs "one" "two" "three";
   function f --inherit-variable xs
      echo "called f"
      echo $xs     
   end
   f $xs
end

Answer 1

As @ridiculous_fish mentioned in the comments, $argv is defined by and available in any function, regardless of its "level". So if you need to "copy" a list into the function when it is called (vs. defined ), just pass the item(s) as argument(s) and access using $argv .

For instance:

function outer
    set --show argv
    set xs "one" "two" "three"
    function inner
        set --show argv
    end
    inner $xs
end

Calling outer with no arguments will return:

$argv: set in local scope, unexported, with 0 elements
$argv: set in local scope, unexported, with 3 elements
$argv[1]: |one|
$argv[2]: |two|
$argv[3]: |three|

But calling outer four five size will result in:

$argv: set in local scope, unexported, with 3 elements
$argv[1]: |four|
$argv[2]: |five|
$argv[3]: |six|
$argv: set in local scope, unexported, with 3 elements
$argv[1]: |one|
$argv[2]: |two|
$argv[3]: |three|

Each function's $argv is distinct, even though they are nested.