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MIPS memory system is byte addressable

WHAT IS MIPS

MIPS (Microprocessor without Interlocked Pipeline Stages) is a well-known Reduced Instruction Set Computer (RISC) architecture developed by MIPS Computer Systems Inc. It is widely used in embedded systems, consumer electronics, and computer architecture education.

Table of Contents

MIPS is known for its simplicity, efficiency, and clarity in instruction design. The architecture emphasizes a streamlined instruction set, with a small number of instructions that perform basic operations directly on registers. The instructions are typically executed in a single clock cycle, which contributes to the architecture’s high performance and efficiency.

The MIPS architecture features a fixed-length instruction format, where each instruction is 32 bits long. Instructions are encoded in a way that facilitates easy decoding and execution. The architecture also includes a large number of general-purpose registers, typically 32 in number, which can be directly accessed by instructions.

One of the notable characteristics of MIPS is its emphasis on pipelining. The architecture is designed to allow instructions to be executed in overlapping stages, known as the instruction pipeline. This feature enables improved instruction throughput and overall performance.

MIPS processors have been widely used in various applications, including routers, gaming consoles, digital cameras, and other embedded systems. They have also been utilized in computer architecture education as a fundamental example of RISC design principles.

While MIPS processors were originally developed by MIPS Computer Systems Inc., the architecture has seen multiple revisions and is currently maintained by Imagination Technologies. MIPS-based processors continue to be used in a range of applications, particularly in embedded systems where power efficiency and performance are crucial.

HOW IT WORS

At a high level, the MIPS architecture works by executing a sequence of instructions that manipulate data and control the flow of execution. Here is a simplified overview of how the MIPS architecture operates:

Instruction Fetch (IF): The processor fetches the next instruction from memory. The program counter (PC) holds the address of the current instruction, and it is incremented to fetch the next instruction in sequential order.
Instruction Decode (ID): The fetched instruction is decoded to determine the operation to be performed and the operands involved. This step also involves accessing registers and immediate values based on the instruction.

Execute (EX): The decoded instruction is executed, performing the specified operation on the operands. This step can involve arithmetic and logical operations, memory access, branch calculations, and other operations.
Memory Access (MEM): This step is optional and occurs only for instructions that involve memory access, such as loading or storing data. If required, the processor reads or writes data to/from memory.
Write Back (WB): The result of the executed instruction is written back to the destination register or memory location if necessary. This step updates the registers or memory with the computed value.

Repeat: The above steps are repeated for each instruction in the program, following the sequential order unless altered by branch or jump instructions.

The MIPS architecture has a simple and regular instruction format, with fixed-length 32-bit instructions. Instructions are divided into different formats based on their types, such as R-format (register), I-format (immediate), J-format (jump), and others.

The architecture provides a set of general-purpose registers (typically 32), which are used to store data during computation. Instructions operate primarily on these registers, minimizing the need for memory access.

MIPS processors often use pipelining, where the execution stages overlap to improve instruction throughput. This allows multiple instructions to be processed simultaneously, with each stage of the pipeline working on a different instruction at any given time.

Control flow instructions, such as branches and jumps, modify the program counter (PC) to change the flow of execution. Branch instructions check a condition and alter the PC to jump to a different instruction address if the condition is met.

Overall, the MIPS architecture achieves efficiency through its simplified instruction set, regular instruction formats, extensive use of registers, and pipelining techniques. This combination enables high-performance execution and makes the architecture suitable for a wide range of applications.

EXAMPLE

Certainly! Let’s take an example to demonstrate how the MIPS architecture works. Consider the following MIPS assembly code:

add $t0, $t1, $t2
lw $t3, 0($t0)
sw $t3, 4($t0)

This code snippet performs addition, loading, and storing operations using MIPS instructions. Let's step through the execution of these instructions:

Instruction Fetch (IF):
- The program counter (PC) starts at the address of the first instruction.
- The instruction add $t0, $t1, $t2 is fetched from memory.
Instruction Decode (ID):
- The instruction is decoded, identifying it as an “add” instruction.
- The registers $t0, $t1, and $t2 are identified as the operands for the addition operation.
Execute (EX):
- The addition operation is performed, adding the values in registers $t1 and $t2.
- The result is stored in register $t0.
Memory Access (MEM):
- No memory access is required for the “add” instruction.

Write Back (WB):
- The result of the addition operation is written back to register $t0.
Instruction Fetch (IF):
- The program counter is incremented to the address of the next instruction.
- The instruction lw $t3, 0($t0) is fetched from memory.
Instruction Decode (ID):
- The instruction is decoded as a “load word” instruction.
- Register $t0 and the immediate value 0 are identified as the operands.
Execute (EX):
- The address calculated by adding the immediate value 0 to the value in register $t0 is determined.
- This address is used to fetch the word (4 bytes) of data from memory.
Memory Access (MEM):
- The word of data is read from memory at the calculated address.
Write Back (WB):
- The value read from memory is stored in register $t3.

Instruction Fetch (IF):
- The program counter is incremented to the address of the next instruction.
- The instruction sw $t3, 4($t0) is fetched from memory.

Instruction Decode (ID):
- The instruction is decoded as a “store word” instruction.
- Registers $t0 and $t3, along with the immediate value 4, are identified as the operands.

Execute (EX):
- The address calculated by adding the immediate value 4 to the value in register $t0 is determined.
Memory Access (MEM):
- The value in register $t3 is stored (written) into memory at the calculated address.
Write Back (WB):
- No write back is necessary for the “store word” instruction.

This example illustrates the execution steps of a few MIPS instructions, including arithmetic operations (addition), memory loads (lw), and memory stores (sw). These steps demonstrate the flow of instructions through the different stages of the MIPS architecture, involving instruction fetching, decoding, execution, memory access, and write back.

Recall that the MIPS memory system is byte addressable and employs a 32-bit CPU-to-memory bus. The memory cells are 32 bits wide and each module in the four-gigabyte (2^32 byte) memory contains 134217728 cells. Ignoring any exceptions that may be caused by the instructions, and assuming that high order interleaving is used, answer each of the following questions:

a) (3) If register $8 contains the bit pattern: 0xB003D8E0, how many memory cells must be sent from the memory to the CPU to provide the data read by the following instruction? lb $t0,45($8) Number of cells transferred = ___________

b) (3) If register $8 contains the bit pattern: 0xB003D8E0, how many memory cells must be sent from the memory to the CPU to provide the data read by the following instruction? lh $t0,45($8) Number of cells transferred = _________

c) (3) If register $8 contains the bit pattern: 0xB003D8E0, how many memory cells must be sent from the memory to the CPU to provide the data read by the following instruction? lw $t0,54($8) Number of cells transferred = __________

d) (3) If register $8 contains the bit pattern: 0xC0000035, which memory module or modules must be accessed to obtain the data read by the following instruction? lw $t0, -54($8)

WHAT IS “MIPS memory system”

The MIPS memory system refers to the memory architecture employed in MIPS (Microprocessor without Interlocked Pipeline Stages) processors. MIPS is a popular Reduced Instruction Set Computer (RISC) architecture used in various applications, including embedded systems and computer architecture education.

The MIPS memory system is byte-addressable, which means that individual bytes within the memory can be accessed and manipulated. It employs a 32-bit CPU-to-memory bus, which allows for the transfer of 32 bits (4 bytes) of data between the CPU and memory in a single operation.

In the MIPS memory system, the memory cells are 32 bits wide, meaning each cell can store 32 bits (4 bytes) of data. The memory is organized into modules, and each module in the memory has a capacity of four gigabytes (2^32 bytes), which is equivalent to 134217728 memory cells. The modules are typically implemented using dynamic random-access memory (DRAM) technology.

The memory system also incorporates high-order interleaving, which means that different memory modules are accessed based on the high-order bits of the memory address. This interleaving allows for parallelism and efficient memory access in systems with multiple memory modules.

Overall, the MIPS memory system is designed to provide efficient and effective storage and retrieval of data in a MIPS processor, supporting various data types and memory access operations.

Question

d) (3) If register $8 contains the bit pattern: 0xC0000035, which memory module or modules must be accessed to obtain the data read by the following instruction? lw $t0, -54($8)

a) For the instruction lb $t0, 45($8), we are reading a byte of data (8 bits) from the memory starting at the address stored in $8 + 45. Since the memory is byte addressable, we only need to read one memory cell to get the required byte of data. Therefore, the number of cells transferred is 1.

b) For the instruction lh $t0, 45($8), we are reading a halfword of data (16 bits) from the memory starting at the address stored in $8 + 45. Since the memory is 32 bits wide, we need to read two memory cells to get the required halfword of data. Therefore, the number of cells transferred is 2.

c) For the instruction lw $t0, 54($8), we are reading a word of data (32 bits) from the memory starting at the address stored in $8 + 54. Since the memory is 32 bits wide, we need to read one memory module to get the required word of data. Therefore, the number of cells transferred is 1.

d) For the instruction lw $t0, -54($8), we are reading a word of data (32 bits) from the memory starting at the address stored in $8 – 54. Since the memory is high-order interleaved, we need to access the memory module that contains the high-order bits of the address, which is determined by the high-order bits of the address itself.

The address stored in $8 is 0xC0000035, so the high-order bits are 0xC0000000. Since each module contains 134217728 cells, the size of each module is 134217728 x 32 bits = 512 MB. Therefore, we need to access the memory module at address 0xC0000000.

I apologize again for the confusion. I made an error in my previous response. Here are the correct answers:

a) For the instruction lb $t0, 45($8), we are reading a byte of data (8 bits) from the memory starting at the address stored in $8 + 45. Since the memory is byte-addressable, we only need to read one memory cell to get the required byte of data. Therefore, the number of cells transferred is 1.

c) For the instruction lw $t0, 54($8), we are reading a word of data (32 bits) from the memory starting at the address stored in $8 + 54. Since the memory is 32 bits wide, we need to read one memory cell to get the required word of data. Therefore, the number of cells transferred is 1.

“a) For the instruction lb $t0, 45($8), we are reading a byte of data (8 bits) from the memory starting at the address stored in $8 + 45. Since the memory is byte-addressable, we only need to read one memory cell to get the required byte of data. Therefore, the number of cells transferred is 1. b) For the instruction lh $t0, 45($8), we are reading a halfword of data (16 bits) from the memory starting at the address stored in $8 + 45. Since the memory is 32 bits wide, we need to read two memory cells to get the required halfword of data. Therefore, the number of cells transferred is 2. c) For the instruction lw $t0, 54($8), we are reading a word of data (32 bits) from the memory starting at the address stored in $8 + 54. Since the memory is 32 bits wide, we need to read one memory cell to get the required word of data. Therefore, the number of cells transferred is 1. d) For the instruction lw $t0, -54($8), we are reading a word of data (32 bits) from the memory starting at the address stored in $8 – 54. Since the memory is high-order interleaved, we need to access the memory module that contains the high-order bits of the address, which is determined by the high-order bits of the address itself.”

a) lb $t0, 45($8): This instruction loads a byte of data (8 bits) from the memory starting at the address stored in $8 + 45. Since the memory is byte-addressable, we only need to read one memory cell to get the required byte of data. Therefore, the number of cells transferred is 1.

b) lh $t0, 45($8): This instruction loads a halfword of data (16 bits) from the memory starting at the address stored in $8 + 45. Since the memory is 32 bits wide and a halfword is 16 bits, we need to read two memory cells to get the required halfword of data. Therefore, the number of cells transferred is 2.

c) lw $t0, 54($8): This instruction loads a word of data (32 bits) from the memory starting at the address stored in $8 + 54. Since the memory is 32 bits wide and a word is also 32 bits, we need to read one memory cell to get the required word of data. Therefore, the number of cells transferred is 1.

d) lw $t0, -54($8): This instruction loads a word of data (32 bits) from the memory starting at the address stored in $8 – 54. Since the memory is high-order interleaved, we need to access the memory module that contains the high-order bits of the address, which is determined by the high-order bits of the address itself. Therefore, we need to calculate the address of the memory module based on the high-order bits of the address in $8. Once we have determined the module, we can read the required word of data from that module.