Looking for a replace-in-string function in elisp

Question

I'm looking for an equivalent of replace-regexp-in-string that just uses literal strings, no regular expressions.

(replace-regexp-in-string "." "bar" "foo.buzz") => "barbarbarbarbarbarbarbar"

But I want

(replace-in-string "." "bar" "foo.buzz") => "foobarbuzz"

I tried various replace-* functions but can't figure it out.

Edit

In return for the elaborate answers I decided to benchmark them (yea, I know all benchmarks are wrong, but it's still interesting).

The output of benchmark-run is (time, # garbage collections, GC time):

(benchmark-run 10000
  (replace-regexp-in-string "." "bar" "foo.buzz"))

  => (0.5530160000000001 7 0.4121459999999999)

(benchmark-run 10000
  (haxe-replace-string "." "bar" "foo.buzz"))

  => (5.301392 68 3.851943000000009)

(benchmark-run 10000
  (replace-string-in-string "." "bar" "foo.buzz"))

  => (1.429293 5 0.29774799999999857)

replace-regexp-in-string with a quoted regexp wins. Temporary buffers do remarkably well.

Edit 2

Now with compilation! Had to do 10x more iteration:

(benchmark-run 100000
  (haxe-replace-string "." "bar" "foo.buzz"))

  => (0.8736970000000001 14 0.47306700000000035)

(benchmark-run 100000
  (replace-in-string "." "bar" "foo.buzz"))

  => (1.25983 29 0.9721819999999983)

(benchmark-run 100000
  (replace-string-in-string "." "bar" "foo.buzz"))

  => (11.877136 86 3.1208540000000013)

haxe-replace-string is looking good

Answer 1

Try this:

(defun replace-in-string (what with in)
  (replace-regexp-in-string (regexp-quote what) with in nil 'literal))