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UEFI: get the guest entry address explicitly

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Previously with the help from gcc, __builtin_return_address(0) is
used as the guest entry address, but it can't work well with gcc 7.3
and leads to the guest unable to return to UEFI correctly.
Let's get it explicitly.

Signed-off-by: Jack Ren <jack.ren@intel.com>
Acked-by: Eddie Dong <eddie.dong@intel.com>
This commit is contained in:
Jack Ren 2018-04-06 17:27:16 +08:00
parent 4e5029383d
commit ec1cbbb660
1 changed files with 5 additions and 1 deletions
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6
hypervisor/bsp/uefi/efi/boot.c
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@ -40,13 +40,14 @@
EFI_SYSTEM_TABLE *sys_table;
EFI_BOOT_SERVICES *boot;
EFI_RUNTIME_SERVICES *runtime;
extern const uint64_t guest_entry;
static inline void hv_jump(EFI_PHYSICAL_ADDRESS hv_start,
struct multiboot_info *mbi, struct efi_ctx *efi_ctx)
{
hv_func hf;
efi_ctx->rip = (uint64_t)__builtin_return_address(0);
efi_ctx->rip = (uint64_t)&guest_entry;
/* The 64-bit entry of acrn hypervisor is 0x200 from the start
* address of hv image. But due to there is multiboot header,
@ -308,6 +309,9 @@ switch_to_guest_mode(EFI_HANDLE image)
asm volatile ("movq %%r15, %0" : "=r"(efi_ctx->r15));
hv_jump(CONFIG_RAM_START, mbi, efi_ctx);
asm volatile (".global guest_entry\n\t"
"guest_entry:\n\t");
out:
return err;
}