Substituting (*x).y with x->y

Question

I am coding a C project and as I am unexperienced, everytime I needed to access the element of a pointer to a struct I typed (*x).y (where x is a struct and y an item), which I just realized could be rewritten as -> which would look much nicer. So I first tried to make a regular expression to find all such occurences. I thought /(/*.+/)/. would work, but it gave some strange unwanted occurences. For example (queue_t *queue) would match which I don't understand why.

I'm also unsure what macro could be used if I got the correct regex. I thought making then a macro which deletes the first 2 characters, and then would go to the "end" to replace ). with ->, but then if I have something of the type (*(*a).b)).c I don't know how to find the correct ).

Also are there better alternatives to do what I want other than use regular expressions and macros?

Answer 1

Iwould try it like this:

%s/(\*\(\k*\))\./\1->/gc

The search string is:

• ( opening parenthesis
• \* a literal *
• \(...\) a match group. This can later be addressed with \1
  • \k* any number of keywor characters
• ) closing parenthesis
• \. a literal .

The replace is:

• \1 This is replaced with what is matched by the test within \(...\)
• ->

See: :help /\k, :help /\( and :help /\1.