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	Merged revisions 11062,11089 via svnmerge from
https://origsvn.digium.com/svn/asterisk/branches/1.2 ........ r11062 | kpfleming | 2006-02-24 22:59:50 -0600 (Fri, 24 Feb 2006) | 3 lines reformat code to fit guidelines remember which translation paths are multi-step paths ........ r11089 | kpfleming | 2006-02-24 23:08:46 -0600 (Fri, 24 Feb 2006) | 2 lines factor the number of translation steps required into translation path decisions, so that equal cost paths that require fewer translations are preferred ........ git-svn-id: https://origsvn.digium.com/svn/asterisk/trunk@11090 65c4cc65-6c06-0410-ace0-fbb531ad65f3
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								translate.c
									
									
									
									
									
								
							
							
						
						
									
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								translate.c
									
									
									
									
									
								
							@@ -1,7 +1,7 @@
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/*
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 * Asterisk -- An open source telephony toolkit.
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 *
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 * Copyright (C) 1999 - 2005, Digium, Inc.
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 * Copyright (C) 1999 - 2006, Digium, Inc.
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 *
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 * Mark Spencer <markster@digium.com>
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 *
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@@ -55,7 +55,8 @@ static AST_LIST_HEAD_STATIC(translators, ast_translator);
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struct ast_translator_dir {
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	struct ast_translator *step;	/*!< Next step translator */
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	int cost;			/*!< Complete cost to destination */
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	unsigned int cost;		/*!< Complete cost to destination */
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	unsigned int multistep;		/*!< Multiple conversions required for this translation */
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};
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struct ast_frame_delivery {
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@@ -275,55 +276,63 @@ static void rebuild_matrix(int samples)
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{
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	struct ast_translator *t;
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	int changed;
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	int x,y,z;
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	int x, y, z;
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	if (option_debug)
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		ast_log(LOG_DEBUG, "Resetting translation matrix\n");
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	bzero(tr_matrix, sizeof(tr_matrix));
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	AST_LIST_TRAVERSE(&translators, t, list) {
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		if(samples)
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		if (samples)
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			calc_cost(t, samples);
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		if (!tr_matrix[t->srcfmt][t->dstfmt].step ||
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		     tr_matrix[t->srcfmt][t->dstfmt].cost > t->cost) {
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		    tr_matrix[t->srcfmt][t->dstfmt].cost > t->cost) {
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			tr_matrix[t->srcfmt][t->dstfmt].step = t;
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			tr_matrix[t->srcfmt][t->dstfmt].cost = t->cost;
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		}
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	}
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	do {
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		changed = 0;
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		/* Don't you just love O(N^3) operations? */
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		for (x=0; x< MAX_FORMAT; x++)				/* For each source format */
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			for (y=0; y < MAX_FORMAT; y++) 			/* And each destination format */
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				if (x != y)				/* Except ourselves, of course */
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					for (z=0; z < MAX_FORMAT; z++) 	/* And each format it might convert to */
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						if ((x!=z) && (y!=z)) 		/* Don't ever convert back to us */
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							if (tr_matrix[x][y].step && /* We can convert from x to y */
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								tr_matrix[y][z].step && /* And from y to z and... */
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								(!tr_matrix[x][z].step || 	/* Either there isn't an x->z conversion */
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								(tr_matrix[x][y].cost + 
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								 tr_matrix[y][z].cost <	/* Or we're cheaper than the existing */
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								 tr_matrix[x][z].cost)  /* solution */
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							     )) {
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								 			/* We can get from x to z via y with a cost that
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											   is the sum of the transition from x to y and
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											   from y to z */
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								 	tr_matrix[x][z].step = tr_matrix[x][y].step;
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									tr_matrix[x][z].cost = tr_matrix[x][y].cost + 
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														   tr_matrix[y][z].cost;
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									if (option_debug)
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										ast_log(LOG_DEBUG, "Discovered %d cost path from %s to %s, via %d\n", tr_matrix[x][z].cost, ast_getformatname(x), ast_getformatname(z), y);
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									changed++;
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								 }
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		for (x = 0; x< MAX_FORMAT; x++) {			/* For each source format */
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			for (y = 0; y < MAX_FORMAT; y++) {		/* And each destination format */
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				if (x == y)				/* Except ourselves, of course */
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					continue;
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				for (z=0; z < MAX_FORMAT; z++) { 	/* And each format it might convert to */
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					if ((x == z) || (y == z))	/* Don't ever convert back to us */
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						continue;
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					if (tr_matrix[x][y].step &&	/* We can convert from x to y */
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					    tr_matrix[y][z].step &&	/* And from y to z and... */
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					    (!tr_matrix[x][z].step || 	/* Either there isn't an x->z conversion */
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					     (tr_matrix[x][y].cost + 
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					      tr_matrix[y][z].cost <	/* Or we're cheaper than the existing */
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					      tr_matrix[x][z].cost)	/* solution */
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						    )) {
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						/* We can get from x to z via y with a cost that
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						   is the sum of the transition from x to y and
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						   from y to z */
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						tr_matrix[x][z].step = tr_matrix[x][y].step;
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						tr_matrix[x][z].cost = tr_matrix[x][y].cost + 
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							tr_matrix[y][z].cost;
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						tr_matrix[x][z].multistep = 1;
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						if (option_debug)
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							ast_log(LOG_DEBUG, "Discovered %d cost path from %s to %s, via %d\n", tr_matrix[x][z].cost, ast_getformatname(x), ast_getformatname(z), y);
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						changed++;
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					}
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				}
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			}
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		}
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	} while (changed);
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}
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/*! \brief CLI "show translation" command handler */
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static int show_translation(int fd, int argc, char *argv[])
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{
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@@ -444,14 +453,14 @@ int ast_translator_best_choice(int *dst, int *srcs)
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	int bestdst = 0;
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	int cur = 1;
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	int besttime = INT_MAX;
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	int beststeps = INT_MAX;
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	int common;
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	if ((common = (*dst) & (*srcs))) {
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		/* We have a format in common */
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		for (y=0; y < MAX_FORMAT; y++) {
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		for (y = 0; y < MAX_FORMAT; y++) {
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			if (cur & common) {
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				/* This is a common format to both.  Pick it if we don't have one already */
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				besttime = 0;
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				bestdst = cur;
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				best = cur;
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			}
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@@ -460,25 +469,38 @@ int ast_translator_best_choice(int *dst, int *srcs)
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	} else {
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		/* We will need to translate */
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		AST_LIST_LOCK(&translators);
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		for (y=0; y < MAX_FORMAT; y++) {
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			if (cur & *dst)
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				for (x=0; x < MAX_FORMAT; x++) {
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					if ((*srcs & (1 << x)) &&			/* x is a valid source format */
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					    tr_matrix[x][y].step &&			/* There's a step */
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					    (tr_matrix[x][y].cost < besttime)) {	/* It's better than what we have so far */
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						best = 1 << x;
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						bestdst = cur;
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						besttime = tr_matrix[x][y].cost;
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					}
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		for (y = 0; y < MAX_FORMAT; y++) {
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			if (!(cur & *dst))
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				continue;
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			for (x = 0; x < MAX_FORMAT; x++) {
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				if ((*srcs & (1 << x)) &&			/* x is a valid source format */
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				    tr_matrix[x][y].step) {			/* There's a step */
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					if (tr_matrix[x][y].cost > besttime)
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						continue;			/* It's more expensive, skip it */
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					if (tr_matrix[x][y].cost == besttime &&
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					    tr_matrix[x][y].multistep >= beststeps)
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						continue; 			/* It requires the same (or more) steps,
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										   skip it */
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					/* It's better than what we have so far */
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					best = 1 << x;
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					bestdst = cur;
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					besttime = tr_matrix[x][y].cost;
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					beststeps = tr_matrix[x][y].multistep;
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				}
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			}
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			cur = cur << 1;
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		}
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		AST_LIST_UNLOCK(&translators);
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	}
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	if (best > -1) {
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		*srcs = best;
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		*dst = bestdst;
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		best = 0;
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	}
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	return best;
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}
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