Another implementation for strrep

[Thomas Treichl]

Hi,

I did some tests with the strrep function and I think that I was able to 
implement a very short and very fast code that does depend very little on string 
sizes and the number of strings that have to be replaced:

  octave:1> tic, for CNT = 1:10000, strrep ("This is a % test string", ...
  > "%", "AAAA"); end, toc
    Elapsed time is 10.135750 seconds.
  octave:2> tic, for CNT = 1:10000, _strrep_ ("This is a % test string", ...
  > "%", "AAAA"); end, toc
    Elapsed time is 4.794319 seconds.
  octave:3> tic, for CNT = 1:10000, strrep ("This is a %placeholder%", ...
  > "%placeholder%", "AAAA"); end, toc
    Elapsed time is 14.827904 seconds.
  octave:4> tic, for CNT = 1:10000, _strrep_ ("This is a %placeholder%", ...
  > "%placeholder%", "AAAA"); end, toc
    Elapsed time is 4.650185 seconds.
  octave:5> tic, for CNT=1:10000, strrep ("%placeholder% AA %placeholder%", ...
  > "%placeholder%", "placeholder"); end, toc
    Elapsed time is 16.657799 seconds.
  octave:6> tic, for CNT=1:10000,_strrep_("%placeholder% AA %placeholder%", ...
  > "%placeholder%", "placeholder"); end, toc
    Elapsed time is 5.782204 seconds.

and I also think that this implementation is exactly doing the same then today's 
 implementation of octave:

  octave:9> _strrep_ ("%A %A %", "%", "A")
  ans = AA AA A
  octave:10> _strrep_ ("%A %A %", "", "A")
  ans = %A %A %
  octave:11> _strrep_ ("%A %A %", "A", "")
  ans = % % %
  octave:12> _strrep_ ("%A %A %", "", "")
  ans = %A %A %
  octave:13> _strrep_ ("", "%A", "%B")
  ans =

Code file attached.
Bye, Thomas

function t = _strrep_ (s, x, y)

  if (nargin != 3), print_usage (); endif

  if (! (ischar (s) && ischar (x) && ischar (y)))
    error ("strrep: all arguments must be strings");
  endif

  [a, b] = regexp (s, x);
  b = [1, b+1]; a = [a-1, (length (s))];
  t = s(b(1):a(1));
  for cnt = 2:length (a)
    t = [t, y, s(b(cnt):a(cnt))];
  end

endfunction