Show the following problem in a diagram(all 6 steps).1.The PC contains 250, the address of the first instruction. This instruction (the value 1940 inhexadecimal) is loaded into the instruction register IR, and the PC is incremented. Note that thisCPU Registers300 PC AC 1940 IR process involves the use of a memory address register and a memory buffer register. For simplicity, these intermediate registers are ignored.2. The first 4 bits (first hexadecimal digit) in the IR indicate the opcode.00012 = 110 =Load AC from memory00102 = 210= Store AC to memory01012 = 510 = Add to AC from memoryThe remaining 12 bits (three hexadecimal digits) specify the address from which data(saymemory location 941contains the data 0003) are to be loaded.3. The next instruction (5941) is fetched from location 251, and the PC is incremented.4. The old contents of the AC and the contents of location 941 (say the data is 0004)are added,and the result is stored in the AC.5. The next instruction (2941) is fetched from location 252, and the PC is incremented.6. The contents of the AC are stored in location 941.
Show the following problem in a diagram(all 6 steps).1.The PC contains 250, the address of the first instruction. This instruction (the value 1940 inhexadecimal) is loaded into the instruction register IR, and the PC is incremented. Note that thisCPU Registers300 PC AC 1940 IR process involves the use of a memory address register and a memory buffer register. For simplicity, these intermediate registers are ignored.2. The first 4 bits (first hexadecimal digit) in the IR indicate the opcode.00012 = 110 =Load AC from memory00102 = 210= Store AC to memory01012 = 510 = Add to AC from memoryThe remaining 12 bits (three hexadecimal digits) specify the address from which data(saymemory location 941contains the data 0003) are to be loaded.3. The next instruction (5941) is fetched from location 251, and the PC is incremented.4. The old contents of the AC and the contents of location 941 (say the data is 0004)are added,and the result is stored in the AC.5. The next instruction (2941) is fetched from location 252, and the PC is incremented.6. The contents of the AC are stored in location 941.
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Show the following problem in a diagram(all 6 steps).
1.The PC contains 250, the address of the first instruction. This instruction (the value 1940 in
hexadecimal) is loaded into the instruction register IR, and the PC is incremented. Note that this
CPU Registers
300 PC
AC
1940 IR process involves the use of a memory address register and a memory buffer register. For
1.The PC contains 250, the address of the first instruction. This instruction (the value 1940 in
hexadecimal) is loaded into the instruction register IR, and the PC is incremented. Note that this
CPU Registers
300 PC
AC
1940 IR process involves the use of a memory address register and a memory buffer register. For
simplicity, these intermediate registers are ignored.
2. The first 4 bits (first hexadecimal digit) in the IR indicate the opcode.
00012 = 110 =Load AC from memory
00102 = 210= Store AC to memory
01012 = 510 = Add to AC from memory
The remaining 12 bits (three hexadecimal digits) specify the address from which data(say
memory location 941contains the data 0003) are to be loaded.
3. The next instruction (5941) is fetched from location 251, and the PC is incremented.
4. The old contents of the AC and the contents of location 941 (say the data is 0004)are added,
and the result is stored in the AC.
5. The next instruction (2941) is fetched from location 252, and the PC is incremented.
6. The contents of the AC are stored in location 941.
2. The first 4 bits (first hexadecimal digit) in the IR indicate the opcode.
00012 = 110 =Load AC from memory
00102 = 210= Store AC to memory
01012 = 510 = Add to AC from memory
The remaining 12 bits (three hexadecimal digits) specify the address from which data(say
memory location 941contains the data 0003) are to be loaded.
3. The next instruction (5941) is fetched from location 251, and the PC is incremented.
4. The old contents of the AC and the contents of location 941 (say the data is 0004)are added,
and the result is stored in the AC.
5. The next instruction (2941) is fetched from location 252, and the PC is incremented.
6. The contents of the AC are stored in location 941.
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