Some time ago there was a loader here which used a hacked version of bootsect.exe to install the grubloader (the grubbootloader was named w7ldr). i compared the hacked bootsect.exe (called bootinst.exe) to the original one as i would like to understand what was done. Some things i do not understand and i hope someone here can explain this to me. Thanks in advance Code: comparing bootinst_w7ldr.exe with BOOTSECT.EXE i get this: replaces bootmgr with w7ldr with two spaces at the end 0000D9A1: 57 w 42 b 0000D9A2: 37 7 4F o 0000D9A3: 4C l4F o 0000D9A4: 44 d 54 t 0000D9A5: 52 r4D m 0000D9A6: 20 47 g 0000D9A7: 20 52 r replaces bootmgr with w7ldr, one space at the beginning and one at the end (why?) 0000D9AE: 20 42 b 0000D9AF: 57 w 4F o 0000D9B0: 37 7 4F o 0000D9B1: 4C l 54 t 0000D9B2: 44 d 4D m 0000D9B3: 52 r 47 g 0000D9B4: 20 52 r replaces bootmgr with w7ldr with two spaces at the end 0000DB69: 57 w 42 b 0000DB6A: 37 7 4F o 0000DB6B: 4C l 4F o 0000DB6C: 44 d 54 t 0000DB6D: 52 r 4D m 0000DB6E: 20 47 g 0000DB6F: 20 52 r replaces bootmgr with w7ldr, one space at the beginning and one at the end (why?) 0000DBAE: 20 42 b 0000DBAF: 57 w 4F o 0000DBB0: 37 7 4F o 0000DBB1: 4C l 54 t 0000DBB2: 44 d 4D m 0000DBB3: 52 r 47 g 0000DBB4: 20 52 r what does this do? Can anyone explain? 0000E25C: 57 4E 0000E25E: 37 54 0000E3B2: 5A 00 0000E3B6: 5C 02 0000E3C8: 00 5A 0000E3CC: 02 5C this is only some text, and can be ignored (replaces MS copyright with own text) 000133C4: 68 4D 000133C6: 61 69 000133CA: 6B 72 000133CC: 65 6F 000133CE: 64 73 000133D0: 20 6F 000133D2: 62 66 000133D4: 79 74 000133D6: 20 AE 000133D8: 55 20 000133DA: 6E 57 000133DC: 74 69 000133DE: 65 6E 000133E0: 72 64 000133E2: 6D 6F 000133E4: 65 77 000133E6: 6E 73 000133E8: 73 AE 000133EA: 63 20 000133EC: 68 4F 000133EE: 20 70 000133F0: 20 65 000133F2: 20 72 000133F4: 20 61 000133F6: 20 74 000133F8: 20 69 000133FA: 20 6E 000133FC: 20 67 00013400: 20 53 00013402: 20 79 00013404: 20 73 00013406: 20 74 00013408: 20 65 0001340A: 20 6D