why pipe output into a bash function does not work

Question

There are many discussions on this: Pipe output to bash function

I just want to know why:

  #bin/sh
  function myfunc () {
      echo $1
  }

  ls -la | myfunc

will give empty line. May I ask that why isn't our output of ls not treated as $1 as the function? What is the mechanism behind this?

If we try:

  #bin/sh
  function myfunc () {
      i=${*:-$(</dev/stdin)}
      echo $i
  }

  ls -la | myfunc

Then we have:

total 32 drwxr-xr-x 6 phil staff 204 Sep 11 21:18 . drwx------+ 17 phil staff 578 Sep 10 21:34 .. lrwxr-xr-x 1 phil staff 2 Sep 10 21:35 s1 -> t1 lrwxr-xr-x 1 phil staff 2 Sep 10 21:35 s2 -> t2 lrwxr-xr-x 1 phil staff 2 Sep 10 21:35 s3 -> t3 -rwxr-xr-x 1 phil staff 96 Sep 11 21:39 test.sh

which does not keep the actual format of ls -la (with \n).

What is the correct/proposed way to pass a command output to your function as a parameter as it is?

Thanks

Update +John Kugelman

  #bin/sh
  function myfunc () {
    cat | grep "\->"  | while read line
    do
        echo $line
    done


    cat | grep "\->"  | while read line
    do
        echo "dummy"
    done
      }

  ls -la | myfunc

This will only print once. What if we would like to use the result twice (store it as a variable possible?)

Thanks,

Answer 1

There are two different ways functions can receive input:

• Command-line arguments: $1, $2, etc.
• Standard input.

When you pipe output from one command to another, it's received on stdin, not as arguments. To read it you could do one of these:

myfunc() {
    cat
}

myfunc() {
    local line
    while IFS= read -r line; do
        printf '%s\n' "$line"
    done
}

ls -la | myfunc

If you want to leave your function as is and it expects $1 to be set, you'll need to change from a pipe to command substitution.

myfunc() {
    echo "$1"
}

myfunc "$(ls -la)"

Notice the abundant use of double quotes. Make sure you write echo "$1" with quotes or else the newlines will be mangled.