x86: Drop interrupt support in 64-bit mode

This is not currently supported, so drop the code.

Signed-off-by: Simon Glass <[email protected]>
Reviewed-by: Bin Meng <[email protected]>

diff --git a/arch/x86/lib/interrupts.c b/arch/x86/lib/interrupts.c
index dd08402665596ccf8654f682e69ebae565d6a028..d3ae6d9694aaf0d0e074a1fef74568f6a78a507b 100644 (file)
--- a/arch/x86/lib/interrupts.c
+++ b/arch/x86/lib/interrupts.c
@@ -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;
 }