Use one single Display (-d) command to display the first 256 bytes of the data s
ID: 2268125 • Letter: U
Question
Use one single Display (-d) command to display the first 256 bytes of the data segment in the memory. Cut and paste your command and the output produced by the command on your report file (as indicated on the first page of this lab) 2. B93:0100 8A C8 2E A1 72 91 0A E4-75 3D 8A E8 2E 8B 16 74 0B93:0110 91 83 FB 01 75 02 86 E9-EB 19 2E 8B 34 00 82 0B B93:0120 A1 72 91 0A E4 75 20 8A-C8 2E A1 74 91 0A E4 75 B93:0130 16 8A E8 83 FA 64 73 04-81 C2 6C 07 5B SE B4 FF B93:0140 BO 07 E8 29 FB EB 10 5B-SE B4 FF B0 03 E8 1E FB B93:0150 2E C7 06 B2 90 09 00 5A-59 58 C3 8D 36 4E 91 2E B93:0160 83 3C FF 74 02 EB 0F 1E-52 50 OE 1F B8 00 38 8B 0B93:0170 D6 CD 21 58 5A 1F C3 51-52 33 C9 2E 8A 04 0A CO B93:0180 74 3C 2E 80 3E 4D 91 00-74 0F 80 FB 01 75 04 3C B93:0190 3A 74 30 3C 2E 74 2C EB-0C 3C 2D 74 26 3C 2F 74 B93:01A0 22 3C 2E 74 1E E8 E6 FD-72 1E B4 00 91 BA OA 00 0B93:01B0 F7 E2 OB D2 75 11 03 C1-72 0E 91 46 EB BD 8A D8 OB93:01C0 P8 EB 05 46 F8 EB 01 F9-8B C1 5A 59 C3 50 51 52 9B93:01DO 56 53 56 E8 85 FF 2E P6-44 11 01 SE 75 03 E8 01 OB93:01EO 01 2E C7 06 70 91 00 00-2E C? 06 72 91 00 00 2E B93:01FO C? 06 74 91 00 00 2E C?-06 76 91 00 00 2E C6 06 0B93:0200 4D 91 01 B3 01 E8 6P FR-72 5D 2E A3 70 91 A DB 0B93:0210 74 5F E8 62 FF 72 50 2E-A3 72 91 ØA DB 74 52 B3 OB93:0220 02 E8 53 FF 72 41 2E A3-74 91 0A DB 75 36 2E F6 DB93:0230 06 BD 90 02 75 3B 2E 8B-36 B4 90 80 7C FF 2C 75 B93:0240 30 C6 44 FF 2E 2E C? 06-BD 90 00 00 2E 80 ØE BD B93:0250 90 02 2E 8B ØE 4Explanation / Answer
Answer:-1) C8h (hex value) is the second byte of the first row.
2) The logic address is 0101h for the second byte.
3) Segment address = 0B93h and offset = 0101h. Thus physical address = 0B93*10 + 0101h = 0BA31h.
4) At logical address 0101h the word i.e. 16-bit value stored is 2EC8h. This is a misaligned word. Word address must be an even address but here we have odd address i.e. 0101h.
5) At address CS:120, word stored is 72A1h. This is an aligned word since staring from even memory location.
6) The first byte having msb as 1, is 8Ah = 1000_1010. Assuming unsigned then decimal value is = 27 + 8 + 2 = 138.
The next byte with msb as 1, is C8h = 1100_1000. Assuming 2's complement signed number then decimal value is = -(27) + 26 + 8 = -56.
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