Parameter is not visible in function

Question

I am writing a script, where I need to work with parameters.

Here is my foo.sh:

say_hello()
{
  if [ "$1" == "hello" ]
  then
    echo "hello"
  else
    echo "<$1>"
  fi
}

echo "$1"
say_hello

The output looks very strange for me:

hello
<>

Could you explain me why in function I can`t work with params? And how I could pass script params in function?

Parameters to function are different from parameters to shell. — Mihir Luthra, Jul 13 '19 at 17:24
@SeanBright you are completely right. Thanks. Add this comment as an answer and I will choose it. — Laughing_Man, Jul 13 '19 at 17:26

score 2 · Accepted Answer · edited Jul 13 '19 at 17:35

Parameters you pass to functions in your shell are different from parameters passed to the shell itself.

For example, if print_script_args looks like this:

echo $1
echo $2

then ./print_script_args hello world will print this:

hello
world

and if print_function_args looks like this:

foo() {
    echo $1
    echo $2
}

foo bye world

then ./print_function_args hello planet will print this:

bye
world

— the parameters to the script do nothing, because what's printed is the parameters passed to the shell function, namely bye world.

score 0 · Answer 2 · answered Jul 15 '19 at 20:24

You need to call the say_hello function with an input parameter.

Within say_hello, $1 refers to the first argument that was passed to that function. You are calling it without any arguments though, like this:

say_hello

What you want is pass $1 (from the main script context) as a parameter to the say_hello function, like this:

say_hello $1

In the script, $1 refers to the first argument passed to the script when it was run from the command-line. In the function say_hello, $1 refers to the first argument that was passed to the function when it was called from the script.

Parameter is not visible in function

2 Answers2