I successfully created a multibootable USB using WinSetupFromUSB, allowing me to boot into an ISO for MS-DOS 6.22 and installer ISOs for Windows 7 32- and 64-Bit from a single bootable USB. However, while the Windows 7 ISOs are both accessible from the Windows Boot Manager (bootmgr), MS-DOS requires grub4dos to be loaded.

This means that my bootable USB is split into two separate boot menus; the grub4dos menu, which is initially booted into and contains the menu entry for booting into DOS 6.22, as well as another two (first and second half of Windows 7 installers) for booting into bootmgr, from where both the Windows 7 entries can be found.

I don't like this way of doing things, and I'd really like to unify all four entries into one single menu under bootmgr. Is this at all possible?

1 Answer 1


Windows boot manager cannot boot into ISO's.

Windows boot manager(bootmgr) can load directly "winload.exe"(Vista and later), ntldr(Windows XP) and boot sector code (512 bytes).

Using boot sector code loader for bootmgr you can chainload any foreign OS loader like GRUB. (see Dual boot Windows 7 and Linux/UNIX).

You can use GRUB/grub4dos to boot ISO's directly.

grub4dos has to be chain loaded by Windows boot manager - you could create 3 entries, 2 direct loaders for Windows 7 extracted ISO's, and one for MS-DOS (chainloading grub4dos which loads MS-DOS).


It is possible to use GRUB to load ISO's (of Windows 7) directly also MS-DOS, use GRUB to create a single boot menu for every item you have.

  • It's hard to understand what you're saying here, especially in the last half of the answer. Could you make it clearer? Jul 17, 2017 at 0:51
  • @Hashim, have you read "Dual boot Windows 7 and Linux/Unix" from link provided? Do you understand Windows boot process? Do you understand Linux boot process? (And who the hell is voting down - must be some stupid person with no or little understanding of Windows boot process!)
    – snayob
    Jul 21, 2017 at 9:47

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