CODEWITHSUNDEEP
arraysbashshellcountksh
I J
I J

Reputation: 57

Can't get array from cmd line argument bash

For some reason in the code below for loop treats args as array and wc -l can count the lines correctly but I can't get the $(#args[@]) to produce the correct count

function doSomthing() {
local i args a
args=$1;
a=("1" "2" "3" "4");
i=0
  echo wc =`wc -l <<< "$args"`;
  for arg in $args; do
    ((i++))
    echo "$i"
  done;
  echo i = $i
  echo a = ${#a[@]}
  echo args = ${#args[@]}
  echo $args
}

The output of this function is

$> doSomthing $'1\n2\n3\n4'
wc =       4
1
2
3
4
i = 4
a = 4
args = 1
1 2 3 4

Upvotes: 0

Views: 52

Answers (2)

I J
I J

Reputation: 57

Problem solved!

I just needed to put $1 inside parenthesis.
For some reason when I tried it before it did not work but now it does.


Here is the new code:

function doSomthing () {
    local i args a;
    args=( $1 );
    a=("1" "2" "3" "4");
    i=0;
    echo wc =`wc -l <<< "$args"`;
    for arg in $args; do
        ((i++));
        echo "$i";
    done;
    echo i = $i;
    echo a = ${#a[@]};
    echo args = ${#args[@]};
    echo ${args[@]}
}



And here is the new output:

$> doSomthing $'1\n2\n3\n4'
wc = 1
1
i = 1
a = 4
args = 4
1 2 3 4

Cheers ;-)

Upvotes: 0

chepner
chepner

Reputation: 531165

args is not an array; it is simply a string that contains embedded newlines. That means, if you try to treat it as an array, it will appear as if you defined it as

args=( $'1\n2\n3\4' )

not

args=(1 2 3 4)

Upvotes: 4

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