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H-1) Which of the following registers cannot be split into high and low bytes? (a) CS (b) AX (c) DS (d) SS (e) BX (f) DX (g) CX (h) SI (i) DI H-2) Name the segment registers.

H-1) Which of the following registers cannot be split into high and low bytes? (a) CS (b) AX (c) DS (d) SS (e) BX (f) DX (g) CX (h) SI (i) DI H-2) Name the segment registers.

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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H-6) Answer the question. If CS =3499H and IP=2500H, find: (a) The logical address (b) The physical address (c) The lower and upper ranges of the code segment Solution: (a) (b) (c) H-7) Answer the question. If DS=7FA2H and the offset is 438EH, (a) Calculate the physical address. (b) Calculate the lower range. (c) Show the logical address. Solution: (a) (b) (c) H-8) The contents of the memory locations after the execution of each instruction: H-8-1) MOV BX,129FH MOV [1450],BX DS: 1450 contains 9FH DS:1451 contains 12H H-8-2) MOV DX, 8C63H MOV [2348],DX DS: 2348 contains 63H DS: 2349 contains 8CH
Transcribed Image Text:H-6) Answer the question. If CS =3499H and IP=2500H, find: (a) The logical address (b) The physical address (c) The lower and upper ranges of the code segment Solution: (a) (b) (c) H-7) Answer the question. If DS=7FA2H and the offset is 438EH, (a) Calculate the physical address. (b) Calculate the lower range. (c) Show the logical address. Solution: (a) (b) (c) H-8) The contents of the memory locations after the execution of each instruction: H-8-1) MOV BX,129FH MOV [1450],BX DS: 1450 contains 9FH DS:1451 contains 12H H-8-2) MOV DX, 8C63H MOV [2348],DX DS: 2348 contains 63H DS: 2349 contains 8CH
H-1) Which of the following registers cannot be split into high and low bytes? (a) CS (b) AX (c) DS (d) SS (e) BX (f) DX (g) CX (h) SI (i) DI H-2) Name the segment registers. H-3) Please study it. Assume that DS is 5000 and the offset is 1950. Calculate the physical address of the byte. Solution: DS offset 5000 1950 The physical address will be 50000+ 1950=51950. 1. Start with DS. 2. Shift DS left. 3. Add the offset. H-4) Please study it. Segment register: Offset register(s): H-5) Please study this example. Example: If CS=24F6H and IP = 634AH, show: (a) The logical address → 24F6:634A (b) The offset address → 634A calculate: CS IP 5000 50 000 950 5 1 DS SI, DI, BX (c) The physical address → 2B2AA (24F60 +634A) (d) The lower range → 24F60 (24F60+0000) (e) The upper range of the code segment → 34F5F (24F60+ FFFF) ES SI, DI, BX SS SP, BP
Transcribed Image Text:H-1) Which of the following registers cannot be split into high and low bytes? (a) CS (b) AX (c) DS (d) SS (e) BX (f) DX (g) CX (h) SI (i) DI H-2) Name the segment registers. H-3) Please study it. Assume that DS is 5000 and the offset is 1950. Calculate the physical address of the byte. Solution: DS offset 5000 1950 The physical address will be 50000+ 1950=51950. 1. Start with DS. 2. Shift DS left. 3. Add the offset. H-4) Please study it. Segment register: Offset register(s): H-5) Please study this example. Example: If CS=24F6H and IP = 634AH, show: (a) The logical address → 24F6:634A (b) The offset address → 634A calculate: CS IP 5000 50 000 950 5 1 DS SI, DI, BX (c) The physical address → 2B2AA (24F60 +634A) (d) The lower range → 24F60 (24F60+0000) (e) The upper range of the code segment → 34F5F (24F60+ FFFF) ES SI, DI, BX SS SP, BP
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