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Compute percentage of exits for top-10 exit ops

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Maxime Chevalier-Boisvert 2021-04-08 15:18:18 -04:00 committed by Alan Wu
parent 7108da16e9
commit 032b2ecf4b
1 changed files with 6 additions and 2 deletions
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8
yjit_iface.c
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@ -829,17 +829,21 @@ print_insn_count_buffer(int how_many, int left_pad)
// Sort the exit ops by decreasing frequency
const struct insn_count *sorted_exit_ops = sort_insn_count_array(exit_op_count);
// Compute the longest instruction name
// Compute the longest instruction name and top10_exit_count
size_t longest_insn_len = 0;
size_t top10_exit_count = 0;
for (int i = 0; i < how_many; i++) {
const char *instruction_name = insn_name(sorted_exit_ops[i].insn);
size_t len = strlen(instruction_name);
if (len > longest_insn_len) {
longest_insn_len = len;
}
top10_exit_count += sorted_exit_ops[i].count;
}
fprintf(stderr, "most frequent exit op:\n");
double top10_exit_percent = 100.0 * top10_exit_count / total_exit_count;
fprintf(stderr, "top-%d most frequent exit ops (%.1f%% of exits):\n", how_many, top10_exit_percent);
// Print the top-N most frequent exit counts
for (int i = 0; i < how_many; i++) {