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Example 2: write assembly program to perform (x-y) and store the result in x, where Ex and y are places in memory contain 32 bit each x= (DS:203µ) (DS:202) (DS:201) (DS:200) (DS:103) (DS:102) (DS:101) (DS:100) Solution : First we subtract the LSB 16 bits of the 32 bit word by using these instructions: MOV AX, [200 H] SUB AX, [100 H] MOV [200 H], AX Next the MSB are subtracted with barrow from LSB: MOV AX, [202 H]

Example 2: write assembly program to perform (x-y) and store the result in x, where Ex and y are places in memory contain 32 bit each x= (DS:203µ) (DS:202) (DS:201) (DS:200) (DS:103) (DS:102) (DS:101) (DS:100) Solution : First we subtract the LSB 16 bits of the 32 bit word by using these instructions: MOV AX, [200 H] SUB AX, [100 H] MOV [200 H], AX Next the MSB are subtracted with barrow from LSB: MOV AX, [202 H]

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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Example 2: write assembly program to perform (x-y) and store the result in x, where x and y are places in memory contain 32 bit each x= (DS:203|) (DS:202) (DS:201) (DS:200) y= (DS:103) (DS:102|) (DS:101) (DS:100) Solution : First we subtract the LSB 16 bits of the 32 bit word by using these instructions: MOV AX, [200 H] SUB AX, [100 H] MOV [200 H], AX Next the MSB are subtracted with barrow from LSB: MOV AX, [202 H] SBB AX, [102н MOV [202 „1, AX H.W: the above example is solved using the direct addressing mode, solve the same example using the indirect addressing mode.
Transcribed Image Text:Example 2: write assembly program to perform (x-y) and store the result in x, where x and y are places in memory contain 32 bit each x= (DS:203|) (DS:202) (DS:201) (DS:200) y= (DS:103) (DS:102|) (DS:101) (DS:100) Solution : First we subtract the LSB 16 bits of the 32 bit word by using these instructions: MOV AX, [200 H] SUB AX, [100 H] MOV [200 H], AX Next the MSB are subtracted with barrow from LSB: MOV AX, [202 H] SBB AX, [102н MOV [202 „1, AX H.W: the above example is solved using the direct addressing mode, solve the same example using the indirect addressing mode.
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