fprintf(fp, "\n");
- if (clause->instruction_count) {
- assert(!clause->bundle_count);
-
- for (unsigned i = 0; i < clause->instruction_count; ++i)
- bi_print_instruction(clause->instructions[i], fp);
- } else {
- assert(clause->bundle_count);
-
- for (unsigned i = 0; i < clause->bundle_count; ++i)
- bi_print_bundle(&clause->bundles[i], fp);
- }
+ for (unsigned i = 0; i < clause->bundle_count; ++i)
+ bi_print_bundle(&clause->bundles[i], fp);
if (clause->constant_count) {
for (unsigned i = 0; i < clause->constant_count; ++i)
struct list_head link;
/* A clause can have 8 instructions in bundled FMA/ADD sense, so there
- * can be 8 bundles. But each bundle can have both an FMA and an ADD,
- * so a clause can have up to 16 bi_instructions. Whether bundles or
- * instructions are used depends on where in scheduling we are. */
+ * can be 8 bundles. */
- unsigned instruction_count;
unsigned bundle_count;
-
- union {
- bi_instruction *instructions[16];
- bi_bundle bundles[8];
- };
+ bi_bundle bundles[8];
/* For scoreboarding -- the clause ID (this is not globally unique!)
* and its dependencies in terms of other clauses, computed during