Question

5) Assuming you have a 4 bit version of our ALU with the same instructions and control operations, write a sequence of instructions to perform the following C code. X, Y and Z are variables in memory. Z = X & (2 + ~Y); 6) Imagine our ALU has been increased in size to work on 3 bit inputs and outputs, but has the same set of instructions as before. Follow the given instruction sequence to determine the 3 bit contents of each register A--C at the end of the code. If you know the bit is a 1 or 0, write a 1 or 0 respectively. If you know the bit is the middle bit of the original value of A, write A1 for that bit (as opposed to A2 or A0). Eg. if the only instruction was ROLA, C would contain A1A0A2. SWAP ROLA A ? C SWAP A ? 1102 AND Final values: A B C

          5) Assuming you have a 4 bit version of our ALU with the same instructions and control operations, write a sequence of instructions to perform the following C code. X, Y and Z are variables in memory.
Z = X & (2 + ~Y);
6) Imagine our ALU has been increased in size to work on 3 bit inputs and outputs, but has the same set of instructions as before. Follow the given instruction sequence to determine the 3 bit contents of each register A--C at the end of the code. If you know the bit is a 1 or 0, write a 1 or 0 respectively. If you know the bit is the middle bit of the original value of A, write A1 for that bit (as opposed to A2 or A0). Eg. if the only instruction was ROLA, C would contain A1A0A2.
SWAP
ROLA
A ? C
SWAP
A ? 1102
AND
Final values:
A
B
C

Show more…

5) Assuming you have a 4 bit version of our ALU with the same instructions and control operations, write a sequence of instructions to perform the following C code. X, Y and Z are variables in memory.
Z = X (2 + Y);
6) Imagine our ALU has been increased in size to work on 3 bit inputs and outputs, but has the same set of instructions as before. Follow the given instruction sequence to determine the 3 bit contents of each register A–C at the end of the code. If you know the bit is a 1 or 0, write a 1 or 0 respectively. If you know the bit is the middle bit of the original value of A, write A1 for that bit (as opposed to A2 or A0). Eg. if the only instruction was ROLA, C would contain A1A0A2.
SWAP
ROLA
A ? C
SWAP
A ? 1102
AND
Final values:
A
B
C

Added by Jose S.

Computer Science and Information Technology

Trishna Knowledge Systems 2018 Edition

Instant Answer

Solved by Expert Akash M

07/19/2023

Step 1

Here is the final state of each register after executing the given instruction sequence: ``` A110 // Load the value 110 into register A AND // Perform bitwise AND operation with the value in the accumulator (which is 000), result is 000 SWAP // Swap the values of Show more…

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5) Assuming you have a 4-bit version of our ALU with the same instructions and control operations, write a sequence of instructions to perform the following C code. X, Y, and Z are variables in memory. Z = X & (2 + ~Y); 6) Imagine our ALU has been increased in size to work on 3-bit inputs and outputs, but has the same set of instructions as before. Follow the given instruction sequence to determine the 3-bit contents of each register A--C at the end of the code. If you know the bit is a 1 or 0, write a 1 or 0 respectively. If you know the bit is the middle bit of the original value of A, write A1 for that bit (as opposed to A2 or A0). Eg. if the only instruction was ROLA, C would contain A1A0A2. SWAP ROLA A ← C SWAP A ← 1102 AND Final values: A B C

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Transcript

00:07 Perform the given c code.

00:09 Perform the given c code.

00:23 Perform the given c code.

00:24 Z equals to x and 2 plus.

00:29 By using a 4 bit alu we need to break down the operations of individual instructions.

00:36 For the 3 bit alu we will assume similar instruction as the 4 bit alu.

00:40 So 4 bit alu instruction says first is add.

01:01 Add two numbers to store the result in the designation register.

01:04 Second is not.

01:12 Then is not.

01:17 That is perform bit twice.

01:19 Not on the input and store the result in the designation register.

01:25 It is not.

01:29 Third is and.

01:34 Perform bit twice and between the two inputs and store the result in the designation register.

01:40 Fourth is swap.

01:45 Swap the contents of two registers.

01:47 Fifth is rola.

01:56 Rotate the bits of the input register a one position to the left.

02:00 The given instruction sequence for the 4 bit alu.

02:06 Instruction sequence.

02:14 First load x into register a.

02:27 Load y into register b.

02:30 Perform bit twice...

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