x86: Drop interrupt support in 64-bit mode
authorSimon Glass <sjg@chromium.org>
Mon, 16 Jan 2017 14:04:14 +0000 (07:04 -0700)
committerBin Meng <bmeng.cn@gmail.com>
Mon, 6 Feb 2017 03:38:46 +0000 (11:38 +0800)
This is not currently supported, so drop the code.

Signed-off-by: Simon Glass <sjg@chromium.org>
Reviewed-by: Bin Meng <bmeng.cn@gmail.com>
arch/x86/lib/interrupts.c

index dd08402665596ccf8654f682e69ebae565d6a028..d3ae6d9694aaf0d0e074a1fef74568f6a78a507b 100644 (file)
@@ -33,6 +33,8 @@
 #include <common.h>
 #include <asm/interrupt.h>
 
+#if !CONFIG_IS_ENABLED(X86_64)
+
 struct irq_action {
        interrupt_handler_t *handler;
        void *arg;
@@ -118,10 +120,12 @@ void do_irq(int hw_irq)
                }
        }
 }
+#endif
 
 #if defined(CONFIG_CMD_IRQ)
 int do_irqinfo(cmd_tbl_t *cmdtp, int flag, int argc, char * const argv[])
 {
+#if !CONFIG_IS_ENABLED(X86_64)
        int irq;
 
        printf("Spurious IRQ: %u, last unknown IRQ: %d\n",
@@ -139,6 +143,7 @@ int do_irqinfo(cmd_tbl_t *cmdtp, int flag, int argc, char * const argv[])
                                        irq_handlers[irq].count);
                }
        }
+#endif
 
        return 0;
 }