# 1.5 Intro to Assembly Language, MIPS Intro ## Intro to Assembly Language, MIPS Intro ### 汇编语言 * CPU的基本工作:执行大量指令(instructions)。 * 指令是CPU可执行的基本操作。 * 不同的cpu实现不同的指令集。特定CPU实现的指令集是**指令集体系结构(Instruction Set Architecture ISA)**。 * 例子:ARM, Intel x86, MIPS, RISC-V, IBM/Motorola PowerPC (old Mac), Intel IA64, ... ### Instruction Set Architectures 指令集架构 * 早期的趋势是向新的cpu添加越来越多的指令来做复杂的运算， * VAX架构有一个乘法多项式的指令! * RISC 哲学(Cocke IBM, Patterson, Hennessy, 1980年) * 简化指令集计算保持指令集小而简单，使它更容易建立快速的硬件。 * 让软件通过组合更简单的操作来完成复杂的操作。 ### MIPS Architecture MIPS 架构 * MIPS * 建立了第一个商用RISC架构之一的半导体公司。 * MIPS简单、优雅。不陷入细节的纠缠。 * MIPS广泛用于嵌入式应用，x86很少用于嵌入式，嵌入式计算机比PC多 ### Assembly Variables : Registers 寄存器 * 汇编不能使用变量 * 为什么？为了保持硬件简单 * 汇编操作数是寄存器 Assembly Operands are registers * 直接构建到硬件中的特殊位置的数量有限 Limited number of special locations built directly into the hardware * 操作只能在这些上执行! Operations can only be performed on these! * 优点:因为寄存器直接在硬件中，所以速度非常快(比1纳秒还快) ### Number of MIPS Registers * 缺点:由于寄存器在硬件中，所以预先确定了寄存器的数量 * 解决方案: MIPS代码必须非常小心地 put together to efficiently use registers * 32 registers in MIPS * 为什么32 ?越小越快，但太小就不好了。 * 每个MIPS寄存器都是 32 bits wide * Groups of 32 bits called a *word* in MIPS * `MIPS register = 32 bits = 1 word = 4 bytes` ### Names of MIPS Registers * 寄存器的编号是从0到31 * 每个寄存器都可以通过编号或名称进行调用 * Number references: * `$0, $1, $2, … $30, $31` * For now: * `$16 ~ $23 → $s0 ~ $s7` (correspond to C variables对应C变量) * `$ 8 ~ $15 → $t0 ~ $t7` (correspond to temporary variables对应临时变量) * 通常，使用名称可以使您的代码更具可读性 ### C, Java variables vs. registers * 在C语言(以及大多数高级语言)中，变量声明并给定一个类型 * 每个变量只能表示它声明为的类型的值(不能混合和匹配int和char变量)。 * 在汇编语言中，寄存器没有类型; **操作(operation)** 决定如何处理寄存器内容 ### 整数的加法和减法(Addition and Subtraction of Integers) #### 汇编加法 Example：\ `add $s0,$s1,$s2` (in MIPS)\ Equivalent to:\ `a = b + c` (in C)\ where C variables ⇔ MIPS registers are:\ `a ⇔ $s0`\ `b ⇔ $s1`\ `c ⇔ $s2` #### 汇编减法 Example:\ `sub $s3,$s4,$s5` (in MIPS)\ Equivalent to:\ `d = e - f` (in C)\ where C variables ⇔ MIPS registers are:\ `d ⇔ $s3`\ `e ⇔ $s4`\ `f ⇔ $s5` #### Addition and Subtraction of Integers Example 1 How to do the following C statement?\ `a = b + c + d - e;` Break into multiple instructions ``` add $t0,$s1,$s2 # temp = b + c add $t0,$t0,$s3 # temp = temp + d sub $s0,$t0,$s4 # a = temp - e ``` * 单行C可以分解成几行MIPS。 * 注意临时寄存器的使用 * don’t want to modify the variable registers $s * 每行上的 # 符号之后的所有内容都会被忽略(注释) ### Immediates 即时 * Immediates 是数值常数。 * 它们经常出现在代码中，因此对它们有特殊的指令。 * Add Immediate: `addi $s0,$s1,-10` (in MIPS) `f = g - 10` (in C) where MIPS registers $s0,$s1 are associated with C variables f, g * 语法类似于add指令，只是最后一个参数是一个数字，而不是寄存器。\ `add $s0,$s1,$zero` (in MIPS)\ `f = g` (in C) **Overflow handling in MIPS MIPS中的溢出处理** 提示:由于计算机的精度有限，Overflow occurs when there is a “mistake” in arithmetic * 一些语言检测溢出 (Ada)，一些不检测 (大多数C) * MIPS的解决方案是两种算术指令 (2 kinds of arithmetic instructions: ): * 这些会导致检测到溢出 * add(add) * add immediate (addi) * subtract (sub) * 这些不会导致溢出检测 * add unsigned (addu) * add immediate unsigned (addiu) * subtract unsigned (subu) * 编译器选择合适的 arithmetic * MIPS C compilers produce addu, addiu, subu **Data Transfer: Load from and Store to memory** ![img](https://github.com/chenyuxiang0425/CS-NoteBook/tree/8db905d215070b88362514c97d364c1aa08a4949/CS61C//img/dataTransfer.png) ### 内存地址以字节(Bytes)为单位 * 很多数据都小于32位，但很少有小于8位的——如果所有数据都是8位的倍数效果很好 * 8 bit 称为 a byte (1 word= 4 bytes) * 内存地址以字节(Bytes)为单位,而不是word * Word addresses are 4 bytes apart 字地址之间相隔4个字节 * Word address is same as address of leftmost byte (i.e. Big-endian) Word address 与最左边 byte 的地址相同 (大端法) ![img](https://github.com/chenyuxiang0425/CS-NoteBook/tree/8db905d215070b88362514c97d364c1aa08a4949/CS61C//img/memoryAddressAreInBytes.png) ### Transfer from Memory to Register && Transfer from Register to Memory ```c int A[100]; g = h + A[3]; ``` Using Load Word (lw) in MIPS: ``` lw $t0, 12($s3) # Temp reg $t0 gets A[3] add $s1,$s2,$t0 # g = h + A[3] # Note: # $s3 – base register (pointer) 基本寄存器(指针) # 12 – offset in bytes 偏移量，以字节为单位 # 偏移量必须是常数 at assembly time ``` ```c int A[100]; A[10] = h + A[3]; ``` Using Store Word (sw) in MIPS: ``` lw $t0, 12($s3) # Temp reg $t0 gets A[3] add $t0,$s2,$t0 # Temp reg $t0 gets h + A[3] sw $t0,40($s3) # A[10] = h + A[3] # Note: # $s3 – base register (pointer) 基本寄存器(指针) # 12,40 – offset in bytes 偏移量，以字节为单位 # $s3+12和$s3+40必须是4的倍数,当你使用的是lw ``` ## Loading and Storing bytes * 除了 word 数据传输(lw, sw)， MIPS有 byte 数据传输: * load byte: lb * store byte: sb * Same format as lw, sw * E.g., `lb $s0, 3($s1)` * contents of memory location(内存地址的内容) with address = sum of “3” + contents of register (寄存器的内容) $s1 is copied to the low byte position of register (寄存器的低字节位置) $s0. ![img](https://github.com/chenyuxiang0425/CS-NoteBook/tree/8db905d215070b88362514c97d364c1aa08a4949/CS61C//img/loadingAndStoringBytes.png) ### Speed of Registers vs. Memory Given that * Registers: 32 words (128 Bytes) * Memory: Billions of bytes (2 GB to 8 GB on laptop) and the RISC principle is… * Smaller is faster 寄存器比内存快多少? 大约快100-500倍! * in terms of latency(延迟) of one access ### exercise We want to translate `*x = *y +1` into MIPS (x, y pointers stored in: $s0 $s1) A: `addi $s0,$s1,1` B: `lw $s0,1($s1)`\ `sw $s1,0($s0)` C: `lw $t0,0($s1)`\ `addi $t0,$t0,1`\ `sw $t0,0($s0)` D: `sw $t0,0($s1)`\ `addi $t0,$t0, 1`\ `lw $t0,0($s0)` E: `lw $s0,1($t0)`\ `sw $s1,0($t0)` Answer: \[correct=c]\ We must first load the contents of y’s pointee into a temp variable \[$t0], which is 5. Then we have to store $t0 back into x’s pointee, which is 6. ### MIPS Logical Instructions 对 word 内的 bit 字段进行操作非常有用 | Logical operations | C operators | Java operators | MIPS instructions | | :----------------: | :----------: | :------------: | :---------------------: | | Bit-by-bit AND | & | & | and | | Bit-by-bit OR | \| | \| | or | | Bit-by-bit NOT | \~ | \~ | not | | Shift left | << | << | sll(shift left logical) | | Shift right | >> | >>> | srl | ### Logic Shifting * Shift Left: `sll $s1,$s2,2 #s1=s2<<2` * Store in $s1 the value from $s2 shifted 2 bits to the left (they fall off end), **inserting 0’s** on right; << in C. * Before: 0000 0002hex 0000 0000 0000 0000 0000 0000 0000 0010two * After: 0000 0008hex 0000 0000 0000 0000 0000 0000 0000 10**00**two What arithmetic effect does shift left have? Shift Right: **srl** is opposite shift; >> * 向右移位运算向右移动n位(将高阶符号位插入空位) * For example, if register $s0 contained 1111 1111 1111 1111 1111 1111 1110 0111two= -25ten * If executed sra $s0, $s0, 4, result is: 1111 1111 1111 1111 1111 1111 1111 1110two= -2ten * 不幸的是，这和除以2^n是不一样的 * Fails for odd(奇数) negative(负数) numbers * C的算术语义是 division 应该四舍五入到0,但现在的结果是-2 ### Computer Decision Making * MIPS: if-statement instruction is `beq register1,register2,L1` * means: go to statement labeled L1 `if (value in register1) == (value in register2)` ….otherwise, go to next statement * beq 表示 equal 分支 * bne 表示 not equal ### Types of Branches * branch——控制流的更改 * Conditional Branch——根据比较结果改变控制流 * 分支如果相等(beq) * 分支如果不相等(bne) * Unconditional Branch——always branch * jump(j) ### Example if Statement ``` f → $s0 g → $s1 h → $s2 i → $s3 j → $s4 ``` ```c if (i == j) f = g + h; ``` ``` bne $s3,$s4,Exit add $s0,$s1,$s2 Exit: ``` * May need to negate branch condition ### Example if-else Statement ```c if (i == j) f = g + h; else f =g - h ``` ``` bne $s3,$s4,Else add $s0,$s1,$s2 j Exit Else: sub $s0,$s1,$s2 Exit: ``` ### Inequalities in MIPS * 到目前为止，我们只测试了等式(==和!=在C语言中)。一般的程序也需要测试<和>。 * 引入MIPS不等式指令:“Set on Less Than” Syntax: `slt reg1,reg2,reg3` Meaning: ```c if (reg2 < reg3) reg1 = 1; else reg1 = 0; ``` “set” means “change to 1”,\ “reset” means “change to 0”. ### Inequalities in MIPS Cont. 我们如何使用它?手工编译: ```c if (g < h) goto Less; #g:$s0, h:$s1 ``` ``` slt $t0,$s0,$s1 # $t0 = 1 if g=1)) ``` ### Loops in C/Assembly Simple loop in C;\ A\[] is an array of ints ```c do { g = g + A[i]; i = i + j; } while (i != h); ``` Use this mapping: g, h, i, j, \&A\[0]\ $s1, $s2, $s3, $s4, $s5 ``` Loop: sll $t1,$s3,2 # $t1= 4*i addu $t1,$t1,$s5 # $t1=addr A+4i lw $t1,0($t1) # $t1=A[i] add $s1,$s1,$t1 # g=g+A[i] addu $s3,$s3,$s4 # i=i+j bne $s3,$s2,Loop # goto Loop if i!=h ``` answer: do ... while loop ### Conclusion: * 32 register in MIPS * `MIPS register = 32 bits = 1 word = 4 bytes` * 计算机 words 和 vocabulary 分别称为指令和指令集 * 严格的格式:1 operation, 2 source operands, 1 destination * add,sub,mul,div,and,or,sll,srl,sra * lw,sw,lb,sb to move data to/from registers from/to memory * beq, bne, j, slt, slti for decision/flow control * 从 C 到 MIPS "算术表达式，数组访问，if-then-else" 的简单映射 --- # Agent Instructions: Querying This Documentation If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question. 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