Computer Systems: A Programmer's Perspective (3rd Edition)
3rd Edition
ISBN: 9780134092669
Author: Bryant, Randal E. Bryant, David R. O'Hallaron, David R., Randal E.; O'Hallaron, Bryant/O'hallaron
Publisher: PEARSON
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Question
Chapter 4.4, Problem 4.28PP
A.
Program Plan Intro
Processing stages:
- The processing of an instruction has number of operations.
- The operations are organized into particular sequence of stages.
- It attempts to follow a uniform sequence for all instructions.
- The description of stages are shown below:
- Fetch:
- It uses program counter “PC” as memory address to read instruction bytes from memory.
- The 4-bit portions “icode” and “ifun” of specifier byte is extracted from instruction.
- It fetches “valC” that denotes an 8-byte constant.
- It computes “valP” that denotes value of “PC” plus length of fetched instruction.
- Decode:
- The register file is been read with two operands.
- It gives values “valA” and “valB” for operands.
- It reads registers with instruction fields “rA” and “rB”.
- Execute:
- In this stage the ALU either performs required operation or increments and decrements stack pointer.
- The resulting value is termed as “valE”.
- The condition codes are evaluated and destination register is updated based on condition.
- It determines whether branch should be taken or not in a jump instruction.
- Memory:
- The data is been written to memory or read from memory in this stage.
- The value that is read is determined as “valM”.
- Write back:
- The results are been written to register file.
- It can write up to two results.
- PC update:
- The program counter “PC” denotes memory address to read bytes of instruction from memory.
- It is used to set next instruction’s address.
- Fetch:
B.
Program Plan Intro
Processing stages:
- The processing of an instruction has number of operations.
- The operations are organized into particular sequence of stages.
- It attempts to follow a uniform sequence for all instructions.
- The description of stages are shown below:
- Fetch:
- It uses program counter “PC” as memory address to read instruction bytes from memory.
- The 4-bit portions “icode” and “ifun” of specifier byte is extracted from instruction.
- It fetches “valC” that denotes an 8-byte constant.
- It computes “valP” that denotes value of “PC” plus length of fetched instruction.
- Decode:
- The register file is been read with two operands.
- It gives values “valA” and “valB” for operands.
- It reads registers with instruction fields “rA” and “rB”.
- Execute:
- In this stage the ALU either performs required operation or increments and decrements stack pointer.
- The resulting value is termed as “valE”.
- The condition codes are evaluated and destination register is updated based on condition.
- It determines whether branch should be taken or not in a jump instruction.
- Memory:
- The data is been written to memory or read from memory in this stage.
- The value that is read is determined as “valM”.
- Write back:
- The results are been written to register file.
- It can write up to two results.
- PC update:
- The program counter “PC” denotes memory address to read bytes of instruction from memory.
- It is used to set next instruction’s address.
- Fetch:
C.
Program Plan Intro
Processing stages:
- The processing of an instruction has number of operations.
- The operations are organized into particular sequence of stages.
- It attempts to follow a uniform sequence for all instructions.
- The description of stages are shown below:
- Fetch:
- It uses program counter “PC” as memory address to read instruction bytes from memory.
- The 4-bit portions “icode” and “ifun” of specifier byte is extracted from instruction.
- It fetches “valC” that denotes an 8-byte constant.
- It computes “valP” that denotes value of “PC” plus length of fetched instruction.
- Decode:
- The register file is been read with two operands.
- It gives values “valA” and “valB” for operands.
- It reads registers with instruction fields “rA” and “rB”.
- Execute:
- In this stage the ALU either performs required operation or increments and decrements stack pointer.
- The resulting value is termed as “valE”.
- The condition codes are evaluated and destination register is updated based on condition.
- It determines whether branch should be taken or not in a jump instruction.
- Memory:
- The data is been written to memory or read from memory in this stage.
- The value that is read is determined as “valM”.
- Write back:
- The results are been written to register file.
- It can write up to two results.
- PC update:
- The program counter “PC” denotes memory address to read bytes of instruction from memory.
- It is used to set next instruction’s address.
- Fetch:
D.
Program Plan Intro
Processing stages:
- The processing of an instruction has number of operations.
- The operations are organized into particular sequence of stages.
- It attempts to follow a uniform sequence for all instructions.
- The description of stages are shown below:
- Fetch:
- It uses program counter “PC” as memory address to read instruction bytes from memory.
- The 4-bit portions “icode” and “ifun” of specifier byte is extracted from instruction.
- It fetches “valC” that denotes an 8-byte constant.
- It computes “valP” that denotes value of “PC” plus length of fetched instruction.
- Decode:
- The register file is been read with two operands.
- It gives values “valA” and “valB” for operands.
- It reads registers with instruction fields “rA” and “rB”.
- Execute:
- In this stage the ALU either performs required operation or increments and decrements stack pointer.
- The resulting value is termed as “valE”.
- The condition codes are evaluated and destination register is updated based on condition.
- It determines whether branch should be taken or not in a jump instruction.
- Memory:
- The data is been written to memory or read from memory in this stage.
- The value that is read is determined as “valM”.
- Write back:
- The results are been written to register file.
- It can write up to two results.
- PC update:
- The program counter “PC” denotes memory address to read bytes of instruction from memory.
- It is used to set next instruction’s address.
- Fetch:
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2.1
2. [4pts] Use the following C-Code for the problems below.
int recFunc (int a, int b) {
if (b
= 0)
==
return a;
else
return 1+recFunc (a, b-1);
a. Give the flowchart for the C-Code
b. Convert to MIPS assembly and comment each assembly instruction to indicate
corresponding C-Code.
Problem I ( Assembler )
Provide the assembly implementation of the C - code below . Sub 10 is a function that subtract 10 from a given input x.
Assumption :
MyArray base address is store in register $S1.
Feel free to use instruction li or si.
li load an immediate value into a register . For instance, li $S4 5 will copy value 5 into register $S4.
C code
for ( i = 0,1 < 10 , i ++ )
{
MyArray [ i ] = MyArray [ i - 1 ] + MyArray [ i + 1 ] ;
Sub10 ( MyArray [ i ];
}
Sub10 ( x )
{
Return ( x - 10 ) ;
}
Chapter 4 Solutions
Computer Systems: A Programmer's Perspective (3rd Edition)
Ch. 4.1 - Prob. 4.1PPCh. 4.1 - Prob. 4.2PPCh. 4.1 - Prob. 4.3PPCh. 4.1 - Prob. 4.4PPCh. 4.1 - Prob. 4.5PPCh. 4.1 - Prob. 4.6PPCh. 4.1 - Prob. 4.7PPCh. 4.1 - Prob. 4.8PPCh. 4.2 - Practice Problem 4.9 (solution page 484) Write an...Ch. 4.2 - Prob. 4.10PP
Ch. 4.2 - Prob. 4.11PPCh. 4.2 - Prob. 4.12PPCh. 4.3 - Prob. 4.13PPCh. 4.3 - Prob. 4.14PPCh. 4.3 - Prob. 4.15PPCh. 4.3 - Prob. 4.16PPCh. 4.3 - Prob. 4.17PPCh. 4.3 - Prob. 4.18PPCh. 4.3 - Prob. 4.19PPCh. 4.3 - Prob. 4.20PPCh. 4.3 - Prob. 4.21PPCh. 4.3 - Prob. 4.22PPCh. 4.3 - Prob. 4.23PPCh. 4.3 - Prob. 4.24PPCh. 4.3 - Prob. 4.25PPCh. 4.3 - Prob. 4.26PPCh. 4.3 - Prob. 4.27PPCh. 4.4 - Prob. 4.28PPCh. 4.4 - Prob. 4.29PPCh. 4.5 - Prob. 4.30PPCh. 4.5 - Prob. 4.31PPCh. 4.5 - Prob. 4.32PPCh. 4.5 - Prob. 4.33PPCh. 4.5 - Prob. 4.34PPCh. 4.5 - Prob. 4.35PPCh. 4.5 - Prob. 4.36PPCh. 4.5 - Prob. 4.37PPCh. 4.5 - Prob. 4.38PPCh. 4.5 - Prob. 4.39PPCh. 4.5 - Prob. 4.40PPCh. 4.5 - Prob. 4.41PPCh. 4.5 - Prob. 4.42PPCh. 4.5 - Prob. 4.43PPCh. 4.5 - Prob. 4.44PPCh. 4 - Prob. 4.45HWCh. 4 - Prob. 4.46HWCh. 4 - Prob. 4.47HWCh. 4 - Prob. 4.48HWCh. 4 - Modify the code you wrote for Problem 4.47 to...Ch. 4 - In Section 3.6.8, we saw that a common way to...Ch. 4 - Prob. 4.51HWCh. 4 - The file seq-full.hcl contains the HCL description...Ch. 4 - Prob. 4.53HWCh. 4 - The file pie=full. hcl contains a copy of the PIPE...Ch. 4 - Prob. 4.55HWCh. 4 - Prob. 4.56HWCh. 4 - Prob. 4.57HWCh. 4 - Our pipelined design is a bit unrealistic in that...Ch. 4 - Prob. 4.59HW
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- Problem I ( Assembler ) Provide the assembly implementation of the C - code below . Sub 10 is a function that subtract 10 from a given input x. Assumption : MyArray base address is store in register $S1. Feel free to use instruction li or si. li load an immediate value into a register . For instance, li $S4 5 will copy value 5 into register $S4. C code for ( i = 0,1 < 10 , i ++ ) { MyArray [ i ] = MyArray [ i - 1 ] + MyArray [ i + 1 ] ; Sub10 ( MyArray [ i ]; } Sub10 ( x ) { Return ( x - 10 ) ; } Code in Assembly Language: sub10(int): ; Implementation of the sub10() function push rbp mov rbp, rsp mov DWORD PTR [rbp-4], edi mov eax, DWORD PTR [rbp-4] sub eax, 10 pop rbp ret main: ; Main function Implementation push rbp mov rbp, rsp sub rsp, 64 mov…arrow_forwardQuestion 1 (> For the first 14 questions, use the MIPS assembly code: SW $22, 40 ($24). AND $8, $17, $18 ADDI $20, $12, 5 OR $16, $10, $11 Each register contains an initial value of decimal 100 plus its register number. (e.g. register $8 contains 108, register $22 contains 122, etc). The code begins running on a 5-stage MIPS pipelined processor with SW starting in cycle 1. Diagram the instructions within a pipeline diagram to determine which stage each instruction will be in during each cycle. Then answer the next 14 questions, During cycle 3, which instruction will use values from the IF/ID pipeline register? Question 2 During cycle 4, which instruction will use values from the ID/EX pipeline register? Question 3 During cycle 4, what decimal register number will be obtained from ID/EX.RegisterRs? Question 4 During cycle 4, which instruction will use values from the EX/MEM pipeline register? Question 5 ( During cycle 4, what decimal value is obtained from EX/MEM.ALUresult ?arrow_forward15.a) Consider an instruction pipeline with four stages with the stage delays 5 nsec, 6 nsec, 11 nsec, and 8 nsec respectively. The delay of an inter-stage register stage of the pipeline is 1 nsec. What is the approximate speedup of the pipeline in the steady-state underideal conditions as compared to the corresponding non-pipelined implementation? b) Discuss structural hazards and control hazards with examples pls answer both subparrts,,its urgent thanksarrow_forward
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