提交 d0ad6392 编写于 作者: T Thomas Gleixner

pci: intr_remap: Remove unused functions

No users.
Signed-off-by: NThomas Gleixner <tglx@linutronix.de>
Reviewed-by: NIngo Molnar <mingo@elte.hu>
Acked-by: NSuresh Siddha <suresh.b.siddha@intel.com>
Cc: David Woodhouse <dwmw2@infradead.org>
Cc: Jesse Barnes <jbarnes@virtuousgeek.org>
上级 a8ef54ae
......@@ -275,28 +275,6 @@ int set_irte_irq(int irq, struct intel_iommu *iommu, u16 index, u16 subhandle)
return 0;
}
int clear_irte_irq(int irq, struct intel_iommu *iommu, u16 index)
{
struct irq_2_iommu *irq_iommu;
unsigned long flags;
spin_lock_irqsave(&irq_2_ir_lock, flags);
irq_iommu = valid_irq_2_iommu(irq);
if (!irq_iommu) {
spin_unlock_irqrestore(&irq_2_ir_lock, flags);
return -1;
}
irq_iommu->iommu = NULL;
irq_iommu->irte_index = 0;
irq_iommu->sub_handle = 0;
irq_2_iommu(irq)->irte_mask = 0;
spin_unlock_irqrestore(&irq_2_ir_lock, flags);
return 0;
}
int modify_irte(int irq, struct irte *irte_modified)
{
int rc;
......@@ -328,31 +306,6 @@ int modify_irte(int irq, struct irte *irte_modified)
return rc;
}
int flush_irte(int irq)
{
int rc;
int index;
struct intel_iommu *iommu;
struct irq_2_iommu *irq_iommu;
unsigned long flags;
spin_lock_irqsave(&irq_2_ir_lock, flags);
irq_iommu = valid_irq_2_iommu(irq);
if (!irq_iommu) {
spin_unlock_irqrestore(&irq_2_ir_lock, flags);
return -1;
}
iommu = irq_iommu->iommu;
index = irq_iommu->irte_index + irq_iommu->sub_handle;
rc = qi_flush_iec(iommu, index, irq_iommu->irte_mask);
spin_unlock_irqrestore(&irq_2_ir_lock, flags);
return rc;
}
struct intel_iommu *map_hpet_to_ir(u8 hpet_id)
{
int i;
......
......@@ -119,8 +119,6 @@ extern int alloc_irte(struct intel_iommu *iommu, int irq, u16 count);
extern int set_irte_irq(int irq, struct intel_iommu *iommu, u16 index,
u16 sub_handle);
extern int map_irq_to_irte_handle(int irq, u16 *sub_handle);
extern int clear_irte_irq(int irq, struct intel_iommu *iommu, u16 index);
extern int flush_irte(int irq);
extern int free_irte(int irq);
extern int irq_remapped(int irq);
......
Markdown is supported
0% .
You are about to add 0 people to the discussion. Proceed with caution.
先完成此消息的编辑!
想要评论请 注册