Understanding semicolons in Bash functions

I know I can write a function at a Bash command line like so:

$ test() { echo 'test' ; }

I also found that I can write the } on a separate line, and omit ;:

$ test () { echo 'test'
> }

But I cannot omit ; when inputting on the same line:

$ test () { echo 'test' }
> 
> 

Here, the interface for inputting a multi-line function doesn't exit until I either cancel it manually, input a ; (resulting in an immediate syntax error), or input a } (resulting in an extra } in the function (now the } becomes an argument to echo instead).

On the other hand, if I want the process to be backgrounded, I need not and indeed cannot use ;:

$ test () { echo 'test' & }

creates the function immediately which works as expected, but

$ test () { echo 'test' & ;

is an immediate syntax error (giving no chance to input the balancing }), and is still a syntax error with a } on the same line. I tried swapping the & and ; but with no meaningful effect.

What does ; actually do in a Bash function?

Consequently: Why isn't } on the same line recognized as a closing brace for the function without either ; or & at the end of the command? Why is it recognized as such on a separate line? Why do I need ; for an ordinary command invocation, but not for a backgrounded one?

posted 3 days ago

CC BY-SA 4.0

Karl Knechtel‭

406 reputation 21 8 64 10