What does ;; do in sh?

Question

If I try to run a;;b in sh, I get this error:

sh: <number>: Syntax error: ";;" unexpected

If I try a;;;b I get the same error, not ;;; instead of ;;, so I think that ;; means something, although I don't know what does it mean.

Here is an example:

$ echo A;;echo B
sh: 1: Syntax error: ";;" unexpected
$ echo A;;;echo B
sh: 1: Syntax error: ";;" unexpected
$ echo A; ;echo B
sh: 1: Syntax error: ";" unexpected

Here you can see that when I use ; ; instead of ;; the error is different, pertaining to the fact that I used ; without a command before. ;; seems to be a different operator, although I don't know what it applies to.

Answer 1

;; separates statements in a case...esac construct in POSIX shells:

case foo in (a) cmd1; cmd2 ;; (b) cmd3; cmd4; esac

To find out about a command, you can run man that-command. If your pager is less, you can search within the man page by pressing /.

Here, you'd run man sh and search for ;;. Some shells support other operators to separate case statements (like ;& in ksh93/zsh/bash/mksh, &| in zsh/mksh, ;;& in bash).