ubuntu function, works when sourced, but not with the bash command - bash

Question

I'm trying to learn how to write some basic functions in Ubuntu, and I've found that some of them work, and some do not, and I can't figure out why.
Specifically, the following function addseq2.sh will work when I source it, but when I just try to run it with bash addseq2.shit doesn't work. When I check with $? I get a 0: command not found. Does anyone have an idea why this might be the case? Thanks for any suggestions!
Here's the code for addseq2.sh:
#!/usr/bin/env bash
# File: addseq2.sh
function addseq2 {
local sum=0
for element in $#
do
let sum=sum+$element
done
echo $sum
}
Thanks everyone for all the useful advice and help!
To expand on my original question, I have two simple functions already written. The first one, hello.sh looks like this:
#!/usr/bin/env bash
# File: hello.sh
function hello {
echo "Hello"
}
hello
hello
hello
When I call this function, without having done anything else, I would type:
$ bash hello.sh
Which seems to work fine. After I source it with $ source hello.sh, I'm then able to just type hello and it also runs as expected.
So what has been driving me crazy is the first function I mentioned here, addseq2.sh. If I try to repeat the same steps, calling it just with $ bash addseq2.sh 1 2 3. I don't see any result. I can see after checking as you suggested with $ echo $?that I get a 0 and it executed correctly, but nothing prints to the screen.
After I source it with $ source addseq2.sh, then I call it just by typing $ addseq2 1 2 3 it returns 6 as expected.
I don't understand why the two functions are behaving differently.

Answer 1

When you do bash foo.sh, it spawns a new instance of bash, which then reads and executes every command in foo.sh.
In the case of hello.sh, the commands are:
function hello {
echo "Hello"
}
This command has no visible effects, but it defines a function named hello.
hello
hello
hello
These commands call the hello function three times, each printing Hello to stdout.
Upon reaching the end of the script, bash exits with a status of 0. The hello function is gone (it was only defined within the bash process that just stopped running).
In the case of addseq2.sh, the commands are:
function addseq2 {
local sum=0
for element in $#
do
let sum=sum+$element
done
echo $sum
}
This command has no visible effects, but it defines a function named addseq2.
Upon reaching the end of the script, bash exits with a status of 0. The addseq2 function is gone (it was only defined within the bash process that just stopped running).
That's why bash addseq2.sh does nothing: It simply defines (and immediately forgets) a function without ever calling it.
The source command is different. It tells the currently running shell to execute commands from a file as if you had typed them on the command line. The commands themselves still execute as before, but now the functions persist because the bash process they were defined in is still alive.
If you want bash addseq2.sh 1 2 3 to automatically call the addseq2 function and pass it the list of command line arguments, you have to say so explicitly: Add
addseq2 "$#"
at the end of addseq2.sh.

Answer 2

When I check with $? I get a 0: command not found
This is because of the way you are checking it, for example:
(the leading $ is the convention for showing the command-line prompt)
$ $?
-bash: 0: command not found
Instead you could do this:
$ echo $?
0
By convention 0 indicated success. A better way to test in a script is something like this:
if addseq.sh
then
echo 'script worked'
else
# Redirect error message to stderr
echo 'script failed' >&2
fi
Now, why might your script not "work" even though it returned 0? You have a function but you are not calling it. With your code I appended a call:
#!/usr/bin/env bash
# File: addseq2.sh
function addseq2 {
local sum=0
for element in $#
do
let sum=sum+$element
done
echo $sum
}
addseq2 1 2 3 4 # <<<<<<<
and I got:
10
By the way, an alternative way of saying:
let sum=sum+$element
is:
sum=$((sum + element))