How to avoid extra space character when slurping first expression

Question

When using Paredit, I'm often confronted with the following (pipe char represents cursor):

(foo |bar baz)

I want to enclose bar in parens, so I use paredit-wrap-around (M-() to get this:

(foo (|bar) baz)

The same applies when I want to put bar inside a string instead, I can use M-" to get this:

(foo "|bar" baz)

However, Clojure also uses square brackets [] and curly braces {} for vectors, maps and sets. These don't appear to have a paredit 'wrap around' command. So in this case, e.g. to put bar into a vector, I typically use the [ key to create a new vector:

(foo [|] bar baz)

followed by a paredit-forward-slurp-sexp (C-→), which results in:

(foo [| bar] baz)

I'd like to remove the space character that has been inserted before bar. Is this possible? Why does paredit retain a space char when going from an empty expression to one that contains one item? Shouldn't the first item added to an S-expression always be pushed right against the opening paren?

Answer 1

Paredit has, for quite a while, had paredit-wrap-curly and paredit-wrap-square. If you want to use them, simply bind them to a likely-sounding key, and have at it. If you're using such an old version of paredit that these functions don't exist, you should upgrade (but I don't think you are, since the slurp commands work with them).

Answer 2

The latest development version of paredit binds M-[ to paredit-wrap-square which does exactly what you want. Emacs already binds M-{ is to backward-paragraph, but if you don't ever use that, you could customize paredit to shadow it:

(eval-after-load 'paredit
  '(progn
     (define-key paredit-mode-map (kbd "M-{") 'paredit-wrap-curly)))

If you want to delete all space around the point, you can always type M-\ for delete-horizontal-space.