How do you call a function defined in .bashrc from the shell?

Question

In my .bashrc, I have a function called hello:

function hello() {    echo "Hello, $1!" } 

I want to be able to invoke hello() from the shell as follows:

$ hello Lloyd 

And get the output:

> Hello, Lloyd! 

What's the trick?

(The real function I have in mind is more complicated, of course.)

EDIT: This is REALLY caused by a syntax error in the function, I think! :(

function coolness() {      if[ [-z "$1"] -o [-z "$2"] ]; then         echo "Usage: $0 [sub_package] [endpoint]";         exit 1;     fi         echo "Hi!" } 

Answer 1

You can export functions. In your ~/.bashrc file after you define the function, add export -f functionname.

function hello() {    echo "Hello, $1!" }  export -f hello 

Then the function will be available at the shell prompt and also in other scripts that you call from there.

Note that it's not necessary to export functions unless they are going to be used in child processes (the "also" in the previous sentence). Usually, even then, it's better to source the function into the file in which it will be used.

Edit:

Brackets in Bash conditional statements are not brackets, they're commands. They have to have spaces around them. If you want to group conditions, use parentheses. Here's your function:

function coolness() {      if [ -z "$1" -o -z "$2" ]; then         echo "Usage: $0 [sub_package] [endpoint]";         exit 1;     fi         echo "Hi!" } 

A better way to write that conditional is:

    if [[ -z "$1" || -z "$2" ]]; then 

because the double brackets provide more capability than the single ones.