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?
1 answer
The shell uses ; to reliably end a statement, with foreground execution. In some cases, a line break also ends a statement with foreground (except that when the statement is not obviously complete, a line break does not end it). (I don't know if "statement" is the technical term.)
When you want to end a statement, and execute it in the background, use & instead of ;.
I personally always use ; as if I were writing C code, unless running some one-liner interactively. IMO, the semicolons add readability.
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