Show, using the table below, the execution trace of that program by filling in the contents of every register and memory location after the fetch cycle and after the execute cycle of every instruction. All values are in hexadecimal. Instruction Cycle PC AC Initially 03BE 73BF CTR PTR Location: Location: Location: Location: 3BE 3BF 3C0 5D3 SD B300 F43A 0C8 0007 000A 0001 FFFF 0001 FFFF Fetch SE B300 F43A 0C8 0007 Execute SE 0007 F43A 0C8 0007 0001 FFFF Fetch Execute CFFF Fetch Execute *** *** *** *** ... ... *** *** *** *** ... www ... ... 000A 000A ... *** *** *** *** ...

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
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Show, using the table below, the execution trace of that program by filling in the contents of
every register and memory location after the fetch cycle and after the execute cycle of every
instruction. All values are in hexadecimal.
Instruction Cycle PC AC
CTR PTR
Initially
SD
B300 F43A 0C8
Fetch SE B300 F43A OC8
Execute SE 0007 F43A 0C8
03BE
Fetch
Execute
73BF
CFFF
Fetch
Execute
***
***
***
***
***
***
***
***
Location: Location: Location: Location:
3BE
3BF
3C0
5D3
0007
000A
0001
FFFF
0007
000A
0001
FFFF
0007
000A
0001
FFFF
.....
***
***
***
***
4**
Transcribed Image Text:Show, using the table below, the execution trace of that program by filling in the contents of every register and memory location after the fetch cycle and after the execute cycle of every instruction. All values are in hexadecimal. Instruction Cycle PC AC CTR PTR Initially SD B300 F43A 0C8 Fetch SE B300 F43A OC8 Execute SE 0007 F43A 0C8 03BE Fetch Execute 73BF CFFF Fetch Execute *** *** *** *** *** *** *** *** Location: Location: Location: Location: 3BE 3BF 3C0 5D3 0007 000A 0001 FFFF 0007 000A 0001 FFFF 0007 000A 0001 FFFF ..... *** *** *** *** 4**
Please answer all the following questions.
1. In a hypothetical computer, the processor has four registers: an 8-bit Program Counter (PC),
a 16-bit Accumulator (AC), a 16-bit Counter (CTR), and a 12-bit Pointer (PTR). The
memory is divided into words each of which is 16-bit long. Each word can hold either an
instruction or a piece of data. For each instruction X, the four most significant bits (denoted
by X15-12) represent an opcode. The rest of the instruction (denoted by X11-0) can be either
an address or a value of an operand. The table below explains some of the instructions
supported by the processor.
Opcode
(binary)
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
Operation
Load AC from a memory location whose address is X11-0.
Store AC into a memory location whose address is X11-0.
Add to AC the contents of a memory location whose address is the value of
PTR. The result is saved to AC.
1100
Subtract from AC the contents of a memory location whose address is the
value of PTR. The result is saved to AC.
Multiply AC times X₁1-0. The result is saved to AC.
Divide AC by X11-0. The result is saved to AC.
Reset AC.
Load PTR with X₁1-0.
Increment PTR (by one).
Load CTR with X₁1-0.
Decrement CTR (by one).
If CTR is not 0, branch to an instruction whose address is X7-0 (i.e., load PC
with X7-0), else continue normally (i.e., do not change PC).
Halt execution.
Given the following program:
Address
(Hexadecimal)
05D
05E
05F
060
061
062
063
064
065
066
Contents
(Hexadecimal)
03BE
73BF
9002
2FFF
8000
A333
BF60
5003
15D3
CFFF
Transcribed Image Text:Please answer all the following questions. 1. In a hypothetical computer, the processor has four registers: an 8-bit Program Counter (PC), a 16-bit Accumulator (AC), a 16-bit Counter (CTR), and a 12-bit Pointer (PTR). The memory is divided into words each of which is 16-bit long. Each word can hold either an instruction or a piece of data. For each instruction X, the four most significant bits (denoted by X15-12) represent an opcode. The rest of the instruction (denoted by X11-0) can be either an address or a value of an operand. The table below explains some of the instructions supported by the processor. Opcode (binary) 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 Operation Load AC from a memory location whose address is X11-0. Store AC into a memory location whose address is X11-0. Add to AC the contents of a memory location whose address is the value of PTR. The result is saved to AC. 1100 Subtract from AC the contents of a memory location whose address is the value of PTR. The result is saved to AC. Multiply AC times X₁1-0. The result is saved to AC. Divide AC by X11-0. The result is saved to AC. Reset AC. Load PTR with X₁1-0. Increment PTR (by one). Load CTR with X₁1-0. Decrement CTR (by one). If CTR is not 0, branch to an instruction whose address is X7-0 (i.e., load PC with X7-0), else continue normally (i.e., do not change PC). Halt execution. Given the following program: Address (Hexadecimal) 05D 05E 05F 060 061 062 063 064 065 066 Contents (Hexadecimal) 03BE 73BF 9002 2FFF 8000 A333 BF60 5003 15D3 CFFF
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