What does `LBUFFER%%(#m)[_a-zA-Z0-9]#` do in zsh?

I need help understanding what the following code in this documentation means.

LBUFFER=${LBUFFER%%(#m)[_a-zA-Z0-9]#} LBUFFER+=${abbreviations[$MATCH]:-$MATCH} 

I learned that LBUFFER contains the left from the cursor. However, there are 3 things that confuse me.

1. What is the %% doing? Is it escaping %?
2. What does (#m)[_a-zA-Z0-9]# do? Is it something like m/[_a-zA-Z0-9]/ in Perl? If so, what is done to the matched string?
3. What is the :- part doing in the second line?

Thanks.

1 Answer

1. ${a%%b} removes the longest occurrence of b (which can be a regular expression), from a.

2. [_a-zA-Z0-9]# is a regular expression matching an alphanumeric character (as indicated by the stuff between []) zero or more times (specified by the # glob operator)

3. (#m) populates the $MATCH variable after evaluating the previous regular expression.

4. The combination of 1, 2, and 3 means that the stretch of alphanumeric characters at the end of the $LBUFFER is deleted and stored in $MATCH.

5. ${a:-b} makes a equal to b only if a is not defined.

If you want to learn more, take a look at the expansion chapter of the zsh manual.