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Does the following mean that SYSTEMD-BOOTX64.EFI executes first, then GRUBX64.EFI, which then reads grub.conf?

#> efibootmgr -v
BootCurrent: 0001
Timeout: 1 seconds
BootOrder: 0003,0001,000B,0005,000D,000C,0002
Boot0001* Linux Boot Manager    HD(1,GPT,e0.... \SYSTEMD-BOOTX64.EFI)
Boot0002* UEFI: Built-in EFI Shell  VenMedia(5023...012)..BO
Boot0003* arch HD(1,GPT,e07236...921eed1e,0x800,0x100000)/File(\EFI\ARCH\GRUBX64.EFI)
Boot0005* Hard Drive    BBS(HD,,0x0)......BO
Boot000B* UEFI OS   HD(1,GPT,e07236...000)/File(\EFI\BOOT\BOOTX64.EFI)
Boot000C* Unknown Device    BBS(11,,0x0)..GO..NO........y.G.e.n.e.r.i.c. .S.T.O.R.A.G.E. .D.E.V.I.C.E. 
Boot000D* UEFI: Generic STORAGE DEVICE 0272 PciRoot(0x...
dgo.a
  • 779

1 Answers1

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No, it means that your default boot loader is SYSTEMD-BOOTX64.EFI ( whatever that is ). ARCH\GRUBX64.EFI is only used if you choose it from the boot menu, or the first choice returns an error.

psusi
  • 17,303
  • Doesn't that contradict the BootOrder? I was reading BootOrder as: 0003/"GRUBX64.efi" as the default and 0001/"SYSTEMD-BOOTX64.efi" as the fallback? – dgo.a May 01 '16 at 04:32
  • @dgo.a, oops, you are right... I was looking at BootCurrent, which says you booted the systemd one this time, but indeed, it is not the default. – psusi May 01 '16 at 15:12