High-level computer languages are created to be understood by humans. As a result, the keywords and the commands of these languages are easy to understand. Machine languages are harder to understand and operate.
For this assignment, you should assume that the memory cells at addresses F0 to F9 are in the machine described here, and that it contains the hexadecimal bit patterns described in the following table:
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F0
|
20
|
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F1
|
C0
|
|
F2
|
30
|
|
F3
|
F8
|
|
F4
|
20
|
|
F5
|
00
|
|
F6
|
30
|
|
F7
|
F9
|
|
F8
|
FF
|
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F9
|
FF
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1. Explain (in detail) each step of the machine cycle. Show the contents of each of the registers and each of the memory cells after the execution of the code.
2. Compare and contrast machine and high-level languages using resources from the Internet or AIU's library. Be sure to explain why the hexadecimal and binary codes are important for programming in both languages.
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Op-Code
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Operand
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Description
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1
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RXY
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LOAD the register R with the bit pattern found in the memory cell whose address is XY
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2
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RXY
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LOAD the register R with the bit XY
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3
|
RXY
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STORE the bit pattern found in register R in the memory cell whose address is XY
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|
4
|
0RS
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MOVE the bit pattern found in register R to register S
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|
5
|
RST
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ADD the bit patterns in registers S and T as though they were two's complement representations and leave the result in register R
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|
6
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RST
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ADD the bit patterns in registers S and T as though they represented values in floating-point notation and leave the result in register R
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7
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RST
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OR the bit pattern in registers S and T and place the result in register R
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8
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RST
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AND the bit patterns in register S and T and place the result in register R
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|
9
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RST
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Exclusive OR the bit patterns in registers S and T and place the result in register R
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A
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R0X
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ROTATE the bit pattern in register R one bit to the right X times. Each time place the bit that started at the low-order end at the high-order end.
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B
|
RXY
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JUMP to the instruction located in the memory cell at address XY if the bit pattern in register R is equal to the bit pattern in register number 0. Otherwise, continue with the normal sequence of execution.
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|
C
|
000
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HALT execution
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