Mercurial > hg > octave-lyh
view scripts/miscellaneous/etime.m @ 5068:ecfba79456a2
[project @ 2004-11-04 20:52:23 by jwe]
author | jwe |
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date | Thu, 04 Nov 2004 20:52:23 +0000 |
parents | c08cb1098afc |
children | 4c8a2e4e0717 |
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## Copyright (C) 1996, 1997 John W. Eaton ## ## This file is part of Octave. ## ## Octave is free software; you can redistribute it and/or modify it ## under the terms of the GNU General Public License as published by ## the Free Software Foundation; either version 2, or (at your option) ## any later version. ## ## Octave is distributed in the hope that it will be useful, but ## WITHOUT ANY WARRANTY; without even the implied warranty of ## MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU ## General Public License for more details. ## ## You should have received a copy of the GNU General Public License ## along with Octave; see the file COPYING. If not, write to the Free ## Software Foundation, 59 Temple Place - Suite 330, Boston, MA ## 02111-1307, USA. ## -*- texinfo -*- ## @deftypefn {Function File} {} etime (@var{t1}, @var{t2}) ## Return the difference (in seconds) between two time values returned from ## @code{clock}. For example: ## ## @example ## t0 = clock (); ## # many computations later... ## elapsed_time = etime (clock (), t0); ## @end example ## ## @noindent ## will set the variable @code{elapsed_time} to the number of seconds since ## the variable @code{t0} was set. ## @end deftypefn ## ## @seealso{tic, toc, clock, and cputime} ## Author: jwe function secs = etime (t1, t0) if (nargin != 2) usage ("etime (t1, t0)"); endif if (isvector (t1) && length (t1) == 6 && isvector (t0) && length (t0) == 6) if (t1 (1) != t0 (1)) error ("etime: can't handle timings over year boundaries yet"); endif ## XXX FIXME XXX -- could check here to ensure that t1 and t0 really do ## make sense as vectors returned from clock(). days_in_months = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]; if (is_leap_year (t1 (1))) days_in_months (2) = days_in_months (2) + 1; endif d1 = sum (days_in_months (1:(t1 (2) - 1))) + t1 (3); d0 = sum (days_in_months (1:(t0 (2) - 1))) + t0 (3); s1 = 86400 * d1 + 3600 * t1 (4) + 60 * t1 (5) + t1 (6); s0 = 86400 * d0 + 3600 * t0 (4) + 60 * t0 (5) + t0 (6); secs = s1 - s0; else error ("etime: args are not 6-element vectors"); endif endfunction