sched: Fix /proc/sched_debug failure on very very large systems
On systems with 4096 cores attemping to read /proc/sched_debug fails because we are trying to push all the data into a single kmalloc buffer. The issue is on these very large machines all the data will not fit in 4mb. A better solution is to not us the single_open mechanism but to provide our own seq_operations and treat each cpu as an individual record. The output should be identical to the previous version. Reported-by: Dave Jones <davej@redhat.com> Signed-off-by: Nathan Zimmer <nzimmer@sgi.com> Cc: Peter Zijlstra <peterz@infradead.org>) [ Whitespace fixlet] [ Fix spello in comment] Signed-off-by: Andrew Morton <akpm@linux-foundation.org> Signed-off-by: Ingo Molnar <mingo@kernel.org>
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@ -269,11 +269,11 @@ static void print_cpu(struct seq_file *m, int cpu)
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{
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unsigned int freq = cpu_khz ? : 1;
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SEQ_printf(m, "\ncpu#%d, %u.%03u MHz\n",
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SEQ_printf(m, "cpu#%d, %u.%03u MHz\n",
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cpu, freq / 1000, (freq % 1000));
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}
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#else
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SEQ_printf(m, "\ncpu#%d\n", cpu);
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SEQ_printf(m, "cpu#%d\n", cpu);
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#endif
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#define P(x) \
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@ -330,6 +330,7 @@ do { \
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print_rq(m, rq, cpu);
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rcu_read_unlock();
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spin_unlock_irqrestore(&sched_debug_lock, flags);
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SEQ_printf(m, "\n");
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}
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static const char *sched_tunable_scaling_names[] = {
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@ -338,11 +339,10 @@ static const char *sched_tunable_scaling_names[] = {
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"linear"
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};
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static int sched_debug_show(struct seq_file *m, void *v)
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static void sched_debug_header(struct seq_file *m)
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{
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u64 ktime, sched_clk, cpu_clk;
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unsigned long flags;
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int cpu;
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local_irq_save(flags);
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ktime = ktime_to_ns(ktime_get());
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@ -384,33 +384,101 @@ static int sched_debug_show(struct seq_file *m, void *v)
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#undef PN
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#undef P
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SEQ_printf(m, " .%-40s: %d (%s)\n", "sysctl_sched_tunable_scaling",
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SEQ_printf(m, " .%-40s: %d (%s)\n",
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"sysctl_sched_tunable_scaling",
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sysctl_sched_tunable_scaling,
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sched_tunable_scaling_names[sysctl_sched_tunable_scaling]);
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for_each_online_cpu(cpu)
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print_cpu(m, cpu);
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SEQ_printf(m, "\n");
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}
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static int sched_debug_show(struct seq_file *m, void *v)
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{
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int cpu = (unsigned long)(v - 2);
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if (cpu != -1)
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print_cpu(m, cpu);
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else
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sched_debug_header(m);
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return 0;
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}
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void sysrq_sched_debug_show(void)
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{
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sched_debug_show(NULL, NULL);
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int cpu;
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sched_debug_header(NULL);
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for_each_online_cpu(cpu)
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print_cpu(NULL, cpu);
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}
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/*
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* This itererator needs some explanation.
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* It returns 1 for the header position.
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* This means 2 is cpu 0.
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* In a hotplugged system some cpus, including cpu 0, may be missing so we have
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* to use cpumask_* to iterate over the cpus.
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*/
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static void *sched_debug_start(struct seq_file *file, loff_t *offset)
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{
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unsigned long n = *offset;
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if (n == 0)
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return (void *) 1;
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n--;
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if (n > 0)
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n = cpumask_next(n - 1, cpu_online_mask);
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else
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n = cpumask_first(cpu_online_mask);
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*offset = n + 1;
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if (n < nr_cpu_ids)
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return (void *)(unsigned long)(n + 2);
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return NULL;
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}
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static void *sched_debug_next(struct seq_file *file, void *data, loff_t *offset)
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{
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(*offset)++;
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return sched_debug_start(file, offset);
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}
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static void sched_debug_stop(struct seq_file *file, void *data)
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{
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}
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static const struct seq_operations sched_debug_sops = {
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.start = sched_debug_start,
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.next = sched_debug_next,
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.stop = sched_debug_stop,
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.show = sched_debug_show,
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};
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static int sched_debug_release(struct inode *inode, struct file *file)
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{
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seq_release(inode, file);
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return 0;
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}
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static int sched_debug_open(struct inode *inode, struct file *filp)
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{
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return single_open(filp, sched_debug_show, NULL);
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int ret = 0;
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ret = seq_open(filp, &sched_debug_sops);
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return ret;
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}
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static const struct file_operations sched_debug_fops = {
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.open = sched_debug_open,
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.read = seq_read,
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.llseek = seq_lseek,
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.release = single_release,
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.release = sched_debug_release,
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};
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static int __init init_sched_debug_procfs(void)
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