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1. Most of you got wrong with the MIPS-1 Homework Question 1. Redo it for extra credit. The question 1-In chapter 4, MARIE instructions are 16-bit long (4-bit opcode and 12-bit address). So, MARIE can address 2^12 (=4KB) memory locations. But MIPS instructions are 32-bit long (with 6-bit opcode) and it says MIPS can address up to 4GB memory locations, which requires 32-bit of address bits. How is it possible? MIPS can address 4GB of memory despite having a 32-bit instruction length because it utilizes various addressing modes that don't require a full 32-bit address to be embedded in each instruction. By using 32-bit addresses as inputs to the memory system and combining different fields within the instruction, such as register values and immediate values, MIPS can form effective memory addresses without needing a 32-bit address field in every instruction. 2. The following table is the answer for the MIPS-2 Question 1. Explain this answer in detail. Hexadecimal Address Value 0x1000 65 0x1004 00 0x1008 00 0x1000 00 0x1010 72 0x1014 00 Hexadecimal Value sub $t2, $t0,$t1 sub $t2, $t2, $t2 6E 30 00 00 61 19 Hexadecimal Value 6F 00 00 00 43 00 Hexadecimal Value 4A 73 OB 14 27 27 3. Explain why the following answer is not correct for the MIPS-2 Homework Question 3b?

1. Most of you got wrong with the MIPS-1 Homework Question 1. Redo it for extra credit. The question 1-In chapter 4, MARIE instructions are 16-bit long (4-bit opcode and 12-bit address). So, MARIE can address 2^12 (=4KB) memory locations. But MIPS instructions are 32-bit long (with 6-bit opcode) and it says MIPS can address up to 4GB memory locations, which requires 32-bit of address bits. How is it possible? MIPS can address 4GB of memory despite having a 32-bit instruction length because it utilizes various addressing modes that don't require a full 32-bit address to be embedded in each instruction. By using 32-bit addresses as inputs to the memory system and combining different fields within the instruction, such as register values and immediate values, MIPS can form effective memory addresses without needing a 32-bit address field in every instruction. 2. The following table is the answer for the MIPS-2 Question 1. Explain this answer in detail. Hexadecimal Address Value 0x1000 65 0x1004 00 0x1008 00 0x1000 00 0x1010 72 0x1014 00 Hexadecimal Value sub $t2, $t0,$t1 sub $t2, $t2, $t2 6E 30 00 00 61 19 Hexadecimal Value 6F 00 00 00 43 00 Hexadecimal Value 4A 73 OB 14 27 27 3. Explain why the following answer is not correct for the MIPS-2 Homework Question 3b?

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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1. Most of you got wrong with the MIPS-1 Homework Question 1. Redo it for extra credit. The question 1-In chapter 4, MARIE instructions are 16-bit long (4-bit opcode and 12-bit address). So, MARIE can address 2^12 (=4KB) memory locations. But MIPS instructions are 32-bit long (with 6-bit opcode) and it says MIPS can address up to 4GB memory locations, which requires 32-bit of address bits. How is it possible? MIPS can address 4GB of memory despite having a 32-bit instruction length because it utilizes various addressing modes that don't require a full 32-bit address to be embedded in each instruction. By using 32-bit addresses as inputs to the memory system and combining different fields within the instruction, such as register values and immediate values, MIPS can form effective memory addresses without needing a 32-bit address field in every instruction. 2. The following table is the answer for the MIPS-2 Question 1. Explain this answer in detail. Address Hexadecimal Value 0x1000 65 0x1004 00 0x1008 00 0x1000 00 0x1010 72 0x1014 00 Hexadecimal Value sub $t2, $t0,$t1 sub $t2, $t2, $t2 6E 30 00 00 61 19 Hexadecimal Value 6F 00 00 00 43 00 Hexadecimal Value 4A 73 OB 14 27 27 3. Explain why the following answer is not correct for the MIPS-2 Homework Question 3b?
Transcribed Image Text:1. Most of you got wrong with the MIPS-1 Homework Question 1. Redo it for extra credit. The question 1-In chapter 4, MARIE instructions are 16-bit long (4-bit opcode and 12-bit address). So, MARIE can address 2^12 (=4KB) memory locations. But MIPS instructions are 32-bit long (with 6-bit opcode) and it says MIPS can address up to 4GB memory locations, which requires 32-bit of address bits. How is it possible? MIPS can address 4GB of memory despite having a 32-bit instruction length because it utilizes various addressing modes that don't require a full 32-bit address to be embedded in each instruction. By using 32-bit addresses as inputs to the memory system and combining different fields within the instruction, such as register values and immediate values, MIPS can form effective memory addresses without needing a 32-bit address field in every instruction. 2. The following table is the answer for the MIPS-2 Question 1. Explain this answer in detail. Address Hexadecimal Value 0x1000 65 0x1004 00 0x1008 00 0x1000 00 0x1010 72 0x1014 00 Hexadecimal Value sub $t2, $t0,$t1 sub $t2, $t2, $t2 6E 30 00 00 61 19 Hexadecimal Value 6F 00 00 00 43 00 Hexadecimal Value 4A 73 OB 14 27 27 3. Explain why the following answer is not correct for the MIPS-2 Homework Question 3b?
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