[PATCH md ] XOR template selection redo
From: NeilBrown <hidden>
Date: 2004-06-24 04:58:34
From: Josh Litherland <redacted> The XOR routine selection process is often confusing as it often doesn't choose the routine that measures the fastest, as sometime one routine is known to have better cache performance. This patch avoids the comparison when there is a known best choice, and makes the report more meaningful. Signed-off-by: Neil Brown <redacted> ### Diffstat output ./drivers/md/xor.c | 39 +++++++++++++++++++++++++-------------- 1 files changed, 25 insertions(+), 14 deletions(-) diff ./drivers/md/xor.c~current~ ./drivers/md/xor.c
--- ./drivers/md/xor.c~current~ 2004-06-24 14:53:51.000000000 +1000
+++ ./drivers/md/xor.c 2004-06-24 14:53:51.000000000 +1000@@ -108,29 +108,40 @@ calibrate_xor_block(void) } b2 = b1 + 2*PAGE_SIZE + BENCH_SIZE; - printk(KERN_INFO "raid5: measuring checksumming speed\n"); + /* + * If this arch/cpu has a short-circuited selection, don't loop through all + * the possible functions, just test the best one + */ + fastest = NULL; + +#ifdef XOR_SELECT_TEMPLATE + fastest = XOR_SELECT_TEMPLATE(fastest); +#endif + #define xor_speed(templ) do_xor_speed((templ), b1, b2) - XOR_TRY_TEMPLATES; + if (fastest) { + printk(KERN_INFO "raid5: automatically using best checksumming function: %s\n", + fastest->name); + xor_speed(fastest); + } else { + printk(KERN_INFO "raid5: measuring checksumming speed\n"); + XOR_TRY_TEMPLATES; + fastest = template_list; + for (f = fastest; f; f = f->next) + if (f->speed > fastest->speed) + fastest = f; + } + + printk("raid5: using function: %s (%d.%03d MB/sec)\n", + fastest->name, fastest->speed / 1000, fastest->speed % 1000); #undef xor_speed free_pages((unsigned long)b1, 2); - fastest = template_list; - for (f = fastest; f; f = f->next) - if (f->speed > fastest->speed) - fastest = f; - -#ifdef XOR_SELECT_TEMPLATE - fastest = XOR_SELECT_TEMPLATE(fastest); -#endif - active_template = fastest; - printk("raid5: using function: %s (%d.%03d MB/sec)\n", - fastest->name, fastest->speed / 1000, fastest->speed % 1000); - return 0; }