Chapter 2 | Instruction¶

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概述

Introduction
Operations of the computer hardware (计算机硬件的操作)
Operands of the computer hardware(计算机硬件的操作数)
Signed and unsigned numbers (有符号和无符号数)
Representing instructions in the computer(计算机中指令的表示)
Logical operations(逻辑操作)
Instructions for making decision(决策指令)
Supporting procedures in computer hardware(计算机对过程的支持)
Instruction addressing (指令的寻址)

信息量 Op > Func 3 > Func 6/7

Mention¶

重点理解

Branch指令进行循环等操作
Procdure尤其是栈的写法
- 递归，斐波那契

区分jal jalr

jal 跳转是基于PC的跳转 UJ type

jalr 相当于jal的寄存器版本跳转范围更加广 I type

PC + 4 都要存回去

立即数的种类
转换code的时候注意rs1，rs2的位置

Introduction¶

存储程序概念： 指令和多种类型的数据不加区分地存储在存储器中并因此易于更改，因为产生了存储程序计算机

Operation¶

一条指令一个操作

Operands¶

RISC - V 32 × 64-bit register file

word: 32 bits

double word : 64 bits

Arithmetic instructions use register operands 必须在寄存器
以64bits的想法理解后续的所有操作

Register¶

C code: f = (g + h) - (i + j); f, …, j in x19, x20, …, x23 Compiled

RISC-V code:
add x5, x20, x21 add x6, x22, x23 sub x19, x5, x6

Memory Operands¶

byte addressed¶

little endian¶

大端小端针对的是一个word, 在内存中有影响
Least-significant byte at least address of a word
c.f. Big Endian: most-significant byte at least address

0x 12345678 word index : 3210

32bits,4bytes

==> 78 is the lsb, 这个byte位于3

word 对齐 Alignment¶

RISV - V不强制要求word aligned

struct {
    int a;
    char b;
    char c[2];
    char d[3]
    float e;
}

example¶

能访问寄存器的只有 load store
这里展示的是字节寻址（8 bits）

Constant or immediate operands¶

避免常数操作时load store 浪费时间

Offer versions of the instruction 
        addi   x22, x22, 4  // x22= x22+ 4 

Constant zero: a register x0

Representing Instructions¶

Mapping registers into numbers map registers x0 to x31 onto registers 0 to 31

RISC-V instructions

Encoded as 32-bit instruction words 每条指令一个word

Small number of formats encoding operation code (opcode), register numbers, …Regularity

sd A,B
ld

R-Format instructions¶

opcode: operation code
rd: destination register number
funct3: 3-bit function code (additional opcode) 例如区分加减法
rs1: the first source register number
rs2: the second source register number
funct7: 7-bit function code (additional opcode)

example

I-Format Instructions¶

load addi

Example：ld x9, 64(x22)
22 (x22) is placed rs1;
64 is placed immediate
9 (x9) is placed rd

S-Format Instructions¶

Example：sd x9, 64(x22)
    22 (x22) is placed rs1;
    64 is placed immediate
    9 (x9) is placed rs2

Logical¶

slli、ori...使用的仍然是I type Instruction，只是imm只用了低6位

Instructions for making decisions¶

Branch 指令¶

通过slli x10，x22，3将 i 左移3位得到地址偏移量

SLT¶

注意这里的大于表示方法, bge>= beq== bne!= A>B B>=A

有符号数比较

blt : less than
bge : greater equal

无符号数 bltu bgeu

Switch/Case¶

jalr ：jump and load register

JumpTable中存储的是目标程序的地址，这个Table的起始地址假设在X6

边界控制
计算Table中的地址
取出Table中的值（一个跳转的地址）
跳转

Basic Blocks¶

Procedure/function¶

jal : jump and link

jal x1,100 跳到 PC + 100 的位置

jarl x1,100(x5) 几乎相同，只是跳转的基址是reg中的值

基本流程
    // Caller !push to stack!
    Place Parameters in a place where the procedure can access them

    Transfer control to the procedure：jump to  

    Acquire the storage resources needed for the procedure

    Perform the desired task

    Place the result value in a place where the calling program can access it 
    // Callee jalr x0,0(x1) 
    // !pull from stack and recover!
    // !注意函数return的值要给某个另外的寄存器，而不是temporary寄存器!
    Return control to the point of origin

Stack

- 当参数多于临时寄存器的个数 - 恢复原来的参数针对Saved value - 注意方向，可以看到下面是低位，所以push要sp - 8*bits

202404082213406

custom-image-17

Communicating with People¶

    strcpy:     addi   sp, sp, -8       # adjust stack for 1 doubleword
                sd      x19, 0(sp)              # save x19
                add    x19, x0, x0          # i  =  0 
     L1:        
                add   x5, x19, x11              # x5 = address of y[ i ]
                lbu   x6, 0(x5)             # x6  =  y [ i ]
                add   x7, x19, x10          # x7 = address of x[ i ] 
                sb    x6, 0(x7)             # x[ i ]  =  y[ i ]
                beq   x6, x0,  L2                # if y[i] == 0 then exit
                addi    x19, x19, 1                # i  =  i  +  1
                jal     x0, L1             # next iteration of loop
     L2:    ld  x19, 0(sp)         # restore saved old s3
            addi    sp, sp, 8      # pop 1 double word from stack
            jalr    x0  0(x1)  # return

注意这里计算address的时候是char类型的

32-Bit Immediate and Addresses¶

lui

寻址¶

PC相对寻址方式

下面两种imm的最低为恒为1

Branches

Jump

这里的jump指的是jal

Synchronization in RISC-V | 同步¶

不想学了

hobbitqia的Notes

A C Sort Example To Put it All Together¶

主要要理解一下loop

题目¶

大小端对齐¶

这里一个byte存储两个16进制数

answer

0x11
0x88