alexander Posted May 16, 2011 Report Posted May 16, 2011 I was kind of surprised when i asked some of my friends pointing them at a set of hex, that they couldn't place it, even though it really is not difficult. So continuing that experiment, can you place this hex (there really are a couple of hints there): 0000000 48eb 0090 8e02 bcd7 7a00 a0bb 8b07 8ece0000010 8edb f3c3 eaa4 005d 07a0 0010 0001 7c000000020 0000 0000 0000 0000 0000 660c 558b b4080000030 c642 1f06 7c00 c032 8966 2216 be00 02030000040 00ff 2000 0001 0000 0200 90fa f690 80c20000050 0275 80b2 59ea 007c 3100 8ec0 8ed8 bcd00000060 2000 a0fb 7c40 ff3c 0274 c288 be52 7d7f0000070 34e8 f601 80c2 5474 41b4 aabb cd55 5a130000080 7252 8149 55fb 75aa a043 7c41 c084 05750000090 e183 7401 6637 4c8b be10 7c05 44c6 01ff00000a0 8b66 441e c77c 1004 c700 0244 0001 896600000b0 085c 44c7 0006 6670 c031 4489 6604 448900000c0 b40c cd42 7213 bb05 7000 7deb 08b4 13cd00000d0 0a73 c2f6 0f80 ea84 e900 008d 05be c67c00000e0 ff44 6600 c031 f088 6640 4489 3104 88d200000f0 c1ca 02e2 e888 f488 8940 0844 c031 d0880000100 e8c0 6602 0489 a166 7c44 3166 66d2 34f70000110 5488 660a d231 f766 0474 5488 890b 0c440000120 443b 7d08 8a3c 0d54 e2c0 8a06 0a4c c1fe0000130 d108 6c8a 5a0c 748a bb0b 7000 c38e db310000140 01b8 cd02 7213 8c2a 8ec3 4806 607c b91e0000150 0100 db8e f631 ff31 f3fc 1fa5 ff61 42260000160 be7c 7d85 40e8 eb00 be0e 7d8a 38e8 eb000000170 be06 7d94 30e8 be00 7d99 2ae8 eb00 47fe0000180 5552 2042 4700 6f65 006d 6148 6472 44200000190 7369 006b 6552 6461 2000 7245 6f72 007200001a0 01bb b400 cd0e ac10 003c f475 00c3 000000001b0 0000 0000 0000 0000 f232 1549 0000 018000001c0 0001 fe83 083f 003f 0000 348a 0002 000000001d0 0901 fe82 fb3f 34c9 0002 9133 003b 000000001e0 fc01 fe83 ffff c5fc 003d 8d05 5716 000000001f0 0000 0000 0000 0000 0000 0000 0000 aa55 Quote
phillip1882 Posted May 17, 2011 Report Posted May 17, 2011 never read hex before, so, no.looks like jsut a bunch of gibberish to me. Quote
alexander Posted May 17, 2011 Author Report Posted May 17, 2011 you dont have to read it to be able to place where it came from, length and last word give it away just fine :) ... Quote
CarlNGraham Posted May 17, 2011 Report Posted May 17, 2011 I've been working working with data and even entering Z80 machine code in hex since 1975, but I don't get this. Quote
alexander Posted May 18, 2011 Author Report Posted May 18, 2011 so wait hold on, the fact that its 512 bytes and ends with the word aa55 doesn't say anything...? maybe dumping strings will offer another clue? (no address numbers or spaces for your convenience)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 Quote
alexander Posted May 18, 2011 Author Report Posted May 18, 2011 close, well rather too precise :) Its a dump of the first 512 bytes of my hard drive (partition table is 64 bytes starting at 01BE), and it ends with aa55, that, ladies and gentlemen is a master boot record (440 bytes of code with option of up to 446 bytes running into the next 2 sections, 4 bytes disk sig (optional), 2 bytes null (usually), 64 bytes partition table, 2 bytes MBR signature (0xAA55) always) I dunno, i guess having written a tiny OS in school (i was bored, ok) it is kind of a no-brainer, and 0xAA55 is kind of forever imprinted in my brain, but there were so many hints, exactly 512 bytes is not an often occurring value and if you dump the strings, you will see a grub signature... hmm... peculiar... Props Mr. Turtle :) ^_^ Quote
CarlNGraham Posted May 18, 2011 Report Posted May 18, 2011 Eeer, if you say so :)I but these things do very between implementation a bit.I think Arc MQX used 0xFEAA557F (I think, not sure if that was for flash B) only) Quote
alexander Posted May 18, 2011 Author Report Posted May 18, 2011 You can do away with MBR all together, and plenty of systems dont use it, notably sparc, power pc machines, intel macs, ibm power systems and embedded devices just to name a few, but if you were to write an OS over a week or so, you would start with a simple, documented, exampled, limited, good ol' MBR... Quote
Turtle Posted May 18, 2011 Report Posted May 18, 2011 close, well rather too precise :) ...Props Mr. Turtle :) ^_^ danke schön. :) afterall, you did say it was not really difficult and stuff like that is right up my alley don't ya know. :P Quote
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