1. What is the result of executing the following instruction sequence ? ADD AL, BL DAA Assume that AL contains 29H (the BCD code for decimal number 29), BL contain 13H (the BCD code for decimal number 13), and AH has been cleared.

Computer Networking: A Top-Down Approach (7th Edition)
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Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
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Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Procedure:
1. What is the result of executing the following instruction sequence ?
ADD AL, BL
DAA
Assume that AL contains 29H (the BCD code for decimal number 29), BL
contain 13H (the BCD code for decimal number 13). and AH has been
cleared.
2. Repeat the execution of the same instruction above assume that AL
contain 96 H & BLcontain 07 H.
3. What is the result of executing the following instruction sequence?
SUB AL, CL
DAS
Assume that AL contains 32H (the BCD code for decimal number 32), CL
contain 17H (the BCD code for decimal number 17).
4. Repeat the execution of the same instruction above assume that AL
contain 23 H & CL contain 58 H.
5. Load the register (CL) from the memory location [0500H] then subtract
the content of this register from the accumulator (AL). Corect the result as
a (BCD) numbers.
Let [0500H] = 12H & AL = 3FH
Transcribed Image Text:Procedure: 1. What is the result of executing the following instruction sequence ? ADD AL, BL DAA Assume that AL contains 29H (the BCD code for decimal number 29), BL contain 13H (the BCD code for decimal number 13). and AH has been cleared. 2. Repeat the execution of the same instruction above assume that AL contain 96 H & BLcontain 07 H. 3. What is the result of executing the following instruction sequence? SUB AL, CL DAS Assume that AL contains 32H (the BCD code for decimal number 32), CL contain 17H (the BCD code for decimal number 17). 4. Repeat the execution of the same instruction above assume that AL contain 23 H & CL contain 58 H. 5. Load the register (CL) from the memory location [0500H] then subtract the content of this register from the accumulator (AL). Corect the result as a (BCD) numbers. Let [0500H] = 12H & AL = 3FH
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