[ALSA] Fix invalid schedule_timeout_interruptible()
Fixed the invalid use of schedule_timeout_interruptible() without checking pending signals. Simply replaced with schedule_timeout(). Suggestions thanks to Jeff Garzik. Signed-off-by: Takashi Iwai <tiwai@suse.de> Signed-off-by: Jaroslav Kysela <perex@suse.cz>
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committed by
Jaroslav Kysela
parent
c12aad6efb
commit
e65365de5b
@@ -109,7 +109,7 @@ void snd_seq_instr_list_free(struct snd_seq_kinstr_list **list_ptr)
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spin_lock_irqsave(&list->lock, flags);
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while (instr->use) {
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spin_unlock_irqrestore(&list->lock, flags);
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schedule_timeout_interruptible(1);
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schedule_timeout(1);
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spin_lock_irqsave(&list->lock, flags);
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}
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spin_unlock_irqrestore(&list->lock, flags);
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@@ -199,7 +199,7 @@ int snd_seq_instr_list_free_cond(struct snd_seq_kinstr_list *list,
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instr = flist;
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flist = instr->next;
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while (instr->use)
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schedule_timeout_interruptible(1);
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schedule_timeout(1);
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if (snd_seq_instr_free(instr, atomic)<0)
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snd_printk(KERN_WARNING "instrument free problem\n");
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instr = next;
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@@ -555,7 +555,7 @@ static int instr_free(struct snd_seq_kinstr_ops *ops,
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SNDRV_SEQ_INSTR_NOTIFY_REMOVE);
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while (instr->use) {
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spin_unlock_irqrestore(&list->lock, flags);
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schedule_timeout_interruptible(1);
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schedule_timeout(1);
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spin_lock_irqsave(&list->lock, flags);
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}
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spin_unlock_irqrestore(&list->lock, flags);
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