pan/bi: Remove more artefacts of 2-pass scheduling
A clause is, by definition, already scheduled. Signed-off-by: Alyssa Rosenzweig <alyssa.rosenzweig@collabora.com> Part-of: <https://gitlab.freedesktop.org/mesa/mesa/-/merge_requests/5260>
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@ -469,17 +469,8 @@ bi_print_clause(bi_clause *clause, FILE *fp)
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fprintf(fp, "\n");
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if (clause->instruction_count) {
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assert(!clause->bundle_count);
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for (unsigned i = 0; i < clause->instruction_count; ++i)
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bi_print_instruction(clause->instructions[i], fp);
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} else {
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assert(clause->bundle_count);
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for (unsigned i = 0; i < clause->bundle_count; ++i)
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bi_print_bundle(&clause->bundles[i], fp);
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}
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for (unsigned i = 0; i < clause->bundle_count; ++i)
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bi_print_bundle(&clause->bundles[i], fp);
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if (clause->constant_count) {
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for (unsigned i = 0; i < clause->constant_count; ++i)
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@ -339,17 +339,10 @@ typedef struct {
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struct list_head link;
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/* A clause can have 8 instructions in bundled FMA/ADD sense, so there
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* can be 8 bundles. But each bundle can have both an FMA and an ADD,
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* so a clause can have up to 16 bi_instructions. Whether bundles or
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* instructions are used depends on where in scheduling we are. */
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* can be 8 bundles. */
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unsigned instruction_count;
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unsigned bundle_count;
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union {
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bi_instruction *instructions[16];
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bi_bundle bundles[8];
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};
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bi_bundle bundles[8];
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/* For scoreboarding -- the clause ID (this is not globally unique!)
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* and its dependencies in terms of other clauses, computed during
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