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Answer the following based on the image below; What does this code do with R2 and R3 to compute R4? Answer with one simple line of valid C code of the form: R4 = exp;, where exp is a simple C expression using the the variables R2 and R3. exp must not include C statements like loops or conditionals.

Answer the following based on the image below; What does this code do with R2 and R3 to compute R4? Answer with one simple line of valid C code of the form: R4 = exp;, where exp is a simple C expression using the the variables R2 and R3. exp must not include C statements like loops or conditionals.

Database System Concepts
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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Answer the following based on the image below; What does this code do with R2 and R3 to compute R4? Answer with one simple line of valid C code of the form: R4 = exp;, where exp is a simple C expression using the the variables R2 and R3. exp must not include C statements like loops or conditionals.

The diagram illustrates a microarchitecture datapath, showcasing the flow of instructions and data through various components of a simple processor. Key components and their connections are described as follows: 1. **PC (Program Counter):** Provides the address of the next instruction to fetch from the Instruction Memory. 2. **Instruction Memory:** An instruction is fetched using the Program Counter (PC) as a read address, outputting a 16-bit `Inst` signal. 3. **Control Unit:** Decodes parts of the instruction into control signals such as `Mem`, `Reg Write`, `Branch`, and `ALU Control`. 4. **Register File:** - Inputs: - `Read Addr 1` and `Read Addr 2` for reading data. - `Write Addr` and `Write Data` for writing data. - Outputs: - `Read Data 1` and `Read Data 2` are sent to the ALU. 5. **ALU (Arithmetic Logic Unit):** Processes data according to the `ALU Control` signals. It can output an ALU result or a zero flag. 6. **Data Memory:** - Uses the ALU result as an address. - `Mem Store` signal determines if data is written (Write Enable) or read at this address. 7. **Multiplexers (Mux):** - Decide between different inputs based on control signals. - One mux controls the input to the ALU between `Read Data 2` and a sign-extended immediate value. - Another controls whether the next PC is incremented by 2 or a branch target is selected. 8. **Adder (PC Incrementer):** Adds 2 to the current PC for the next instruction fetch. 9. **Shift Left by 1:** Shifts the immediate value (after Sign extension) left by one bit to form a branch address. 10. **Branch Logic:** - Uses the zero flag from the ALU and `Branch` control signals to determine if a branch is taken. The diagram visually integrates these elements to demonstrate how instruction execution progresses through fetch, decode, and execute stages within the hardware microarchitecture. **HW microarchitecture datapath**
Transcribed Image Text:The diagram illustrates a microarchitecture datapath, showcasing the flow of instructions and data through various components of a simple processor. Key components and their connections are described as follows: 1. **PC (Program Counter):** Provides the address of the next instruction to fetch from the Instruction Memory. 2. **Instruction Memory:** An instruction is fetched using the Program Counter (PC) as a read address, outputting a 16-bit `Inst` signal. 3. **Control Unit:** Decodes parts of the instruction into control signals such as `Mem`, `Reg Write`, `Branch`, and `ALU Control`. 4. **Register File:** - Inputs: - `Read Addr 1` and `Read Addr 2` for reading data. - `Write Addr` and `Write Data` for writing data. - Outputs: - `Read Data 1` and `Read Data 2` are sent to the ALU. 5. **ALU (Arithmetic Logic Unit):** Processes data according to the `ALU Control` signals. It can output an ALU result or a zero flag. 6. **Data Memory:** - Uses the ALU result as an address. - `Mem Store` signal determines if data is written (Write Enable) or read at this address. 7. **Multiplexers (Mux):** - Decide between different inputs based on control signals. - One mux controls the input to the ALU between `Read Data 2` and a sign-extended immediate value. - Another controls whether the next PC is incremented by 2 or a branch target is selected. 8. **Adder (PC Incrementer):** Adds 2 to the current PC for the next instruction fetch. 9. **Shift Left by 1:** Shifts the immediate value (after Sign extension) left by one bit to form a branch address. 10. **Branch Logic:** - Uses the zero flag from the ALU and `Branch` control signals to determine if a branch is taken. The diagram visually integrates these elements to demonstrate how instruction execution progresses through fetch, decode, and execute stages within the hardware microarchitecture. **HW microarchitecture datapath**
**Exercise: Register Value Evaluation** **Objective:** Execute the given assembly code with initial values and determine the final values of the registers when the execution reaches the `HALT` instruction. **Initial Conditions:** - Register R2 holds value 5 - Register R3 holds value 3 **Assembly Code:** ``` 0: AND R2, R2, R4 2: AND R3, R3, R5 4: BEQ R5, R0, 3 6: SUB R5, R1, R5 8: ADD R4, R4, R4 A: JMP 2 C: HALT # Stops execution. ``` **Register Definitions:** - **R2:** Initially 5 - **R3:** Initially 3 - **R4:** Undefined at start - **R5:** Undefined at start - **R0:** Constant 0 (commonly used in assembly language) **Execution Steps:** 1. **Instruction 0:** `AND R2, R2, R4` Compute the bitwise AND between R2 and R4, store in R4. (R4 starts undefined) 2. **Instruction 2:** `AND R3, R3, R5` Compute the bitwise AND between R3 and R5, store in R5. (R5 starts undefined) 3. **Instruction 4:** `BEQ R5, R0, 3` If R5 equals R0, branch to instruction at address 3 + current instruction address = 7. Since R5 is undefined, branch will not be taken. 4. **Instruction 6:** `SUB R5, R1, R5` Subtract R1 from R5 and store the result in R5. 5. **Instruction 8:** `ADD R4, R4, R4` Add R4 to itself and store in R4. 6. **Instruction A:** `JMP 2` Jump to instruction at address 2. 7. Repeat from instruction 2 until halted. **Final Register Values:** - **R2:** Unknown final value; remains 5 during execution - **R3:** Unknown final value; remains 3 during execution - **R4:** Depends on computation results - **
Transcribed Image Text:**Exercise: Register Value Evaluation** **Objective:** Execute the given assembly code with initial values and determine the final values of the registers when the execution reaches the `HALT` instruction. **Initial Conditions:** - Register R2 holds value 5 - Register R3 holds value 3 **Assembly Code:** ``` 0: AND R2, R2, R4 2: AND R3, R3, R5 4: BEQ R5, R0, 3 6: SUB R5, R1, R5 8: ADD R4, R4, R4 A: JMP 2 C: HALT # Stops execution. ``` **Register Definitions:** - **R2:** Initially 5 - **R3:** Initially 3 - **R4:** Undefined at start - **R5:** Undefined at start - **R0:** Constant 0 (commonly used in assembly language) **Execution Steps:** 1. **Instruction 0:** `AND R2, R2, R4` Compute the bitwise AND between R2 and R4, store in R4. (R4 starts undefined) 2. **Instruction 2:** `AND R3, R3, R5` Compute the bitwise AND between R3 and R5, store in R5. (R5 starts undefined) 3. **Instruction 4:** `BEQ R5, R0, 3` If R5 equals R0, branch to instruction at address 3 + current instruction address = 7. Since R5 is undefined, branch will not be taken. 4. **Instruction 6:** `SUB R5, R1, R5` Subtract R1 from R5 and store the result in R5. 5. **Instruction 8:** `ADD R4, R4, R4` Add R4 to itself and store in R4. 6. **Instruction A:** `JMP 2` Jump to instruction at address 2. 7. Repeat from instruction 2 until halted. **Final Register Values:** - **R2:** Unknown final value; remains 5 during execution - **R3:** Unknown final value; remains 3 during execution - **R4:** Depends on computation results - **
Expert Solution
Step 1: Here is the one simple line of valid C code:

To compute R4 using R2 and R3:

R4 = R2 * R3;

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