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@ -454,14 +454,14 @@ 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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@ -470,25 +470,38 @@ 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_mutex_lock(&list_lock);
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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_mutex_unlock(&list_lock);
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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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