CSC 320 Chapter #2, Spring 2015

Chapter 2 Instructions: Language of the Computer Chapter 2 — Instructions: Language of the Computer — 2 Instruction Set � The repertoire of instructions of a computer � Different computers have different instruction sets � But with many aspects in common � Early computers had very simple instruction sets � Simplified implementation � Many modern computers also have simple instruction sets §2.1 Introduction Chapter 2 — Instructions: Language of the Computer — 3 The MIPS Instruction Set � Used as the example throughout the book � Stanford MIPS commercialized by MIPS Technologies (www.mips.com) � Large share of embedded core market � Applications in consumer electronics, network/storage equipment, cameras, printers, … � Typical of many modern ISAs Chapter 2 — Instructions: Language of the Computer — 4 Arithmetic Operations � Add and subtract, three operands � Two sources and one destination add a, b, c # a gets b + c � All arithmetic operations have this form � Design Principle 1: Simplicity favours regularity � Regularity makes implementation simpler � Simplicity enables higher performance at lower cost §2.2 Operations of the Computer Hardware Chapter 2 — Instructions: Language of the Computer — 5 Arithmetic Example � C code: f = (g + h) - (i + j); � Compiled MIPS code: add t0, g, h # temp t0 = g + h add t1, i, j # temp t1 = i + j sub f, t0, t1 # f = t0 - t1 Chapter 2 — Instructions: Language of the Computer — 6 Register Operands � Arithmetic instructions use register operands � MIPS has a 32 × 32-bit register file � Use for frequently accessed data � Numbered 0 to 31 � 32-bit data called a “word” � Assembler names � $t0, $t1, …, $t9 for temporary values � $s0, $s1, …, $s7 for saved variables � Design Principle 2: Smaller is faster � c.f. main memory: millions of locations §2.3 Operands of the Computer Hardware Chapter 2 — Instructions: Language of the Computer — 7 Register Operand Example � C code: f = (g + h) - (i + j); � f, …, j in $s0, …, $s4 � Compiled MIPS code: add $t0, $s1, $s2 add $t1, $s3, $s4 sub $s0, $t0, $t1 Chapter 2 — Instructions: Language of the Computer — 8 Memory Operands � Main memory used for composite data � Arrays, structures, dynamic data � To apply arithmetic operations � Load values from memory into registers � Store result from register to memory � Memory is byte addressed � Each address identifies an 8-bit byte � Words are aligned in memory � Address must be a multiple of 4 � MIPS is Big Endian � Most-significant byte at least address of a word � c.f. Little Endian: least-significant byte at least address Chapter 2 — Instructions: Language of the Computer — 9 Memory Operand Example 1 � C code: g = h + A[8]; � g in $s1, h in $s2, base address of A in $s3 � Compiled MIPS code: � Index 8 requires offset of 32 � 4 bytes per word lw $t0, 32($s3) # load word add $s1, $s2, $t0 offset base register Chapter 2 — Instructions: Language of the Computer — 10 Memory Operand Example 2 � C code: A[12] = h + A[8]; � h in $s2, base address of A in $s3 � Compiled MIPS code: � Index 8 requires offset of 32 lw $t0, 32($s3) # load word add $t0, $s2, $t0 sw $t0, 48($s3) # store word Chapter 2 — Instructions: Language of the Computer — 11 Registers vs. Memory � Registers are faster to access than memory � Operating on memory data requires loads and stores � More instructions to be executed � Compiler must use registers for variables as much as possible � Only spill to memory for less frequently used variables � Register optimization is important! Chapter 2 — Instructions: Language of the Computer — 12 Immediate Operands � Constant data specified in an instruction addi $s3, $s3, 4 � No subtract immediate instruction � Just use a negative constant addi $s2, $s1, -1 � Design Principle 3: Make the common case fast � Small constants are common � Immediate operand avoids a load instruction Chapter 2 — Instructions: Language of the Computer — 13 The Constant Zero � MIPS register 0 ($zero) is the constant 0 � Cannot be overwritten � Useful for common operations � E.g., move between registers add $t2, $s1, $zero Chapter 2 — Instructions: Language of the Computer — 14 Unsigned Binary Integers � Given an n-bit number 0 0 1 1 n 2 n 2 n 1 n 1 x 2 x 2 2 x 2 x x + + + + = − − − − L � Range: 0 to +2n – 1 � Example � 0000 0000 0000 0000 0000 0000 0000 10112 = 0 + … + 1×23 + 0×22 +1×21 +1×20 = 0 + … + 8 + 0 + 2 + 1 = 1110 � Using 32 bits � 0 to +4,294,967,295 §2.4 Signed and Unsigned Numbers Chapter 2 — Instructions: Language of the Computer — 15 2s-Complement Signed Integers � Given an n-bit number 0 0 1 1 n 2 n 2 n 1 n 1 x 2 x 2 2 x 2 x x + + + + = − − − − − L � Range: –2n – 1 to +2n – 1 – 1 � Example � 1111 1111 1111 1111 1111 1111 1111 11002 = –1×231 + 1×230 + … + 1×22 +0×21 +0×20 = –2,147,483,648 + 2,147,483,644 = –410 � Using 32 bits � –2,147,483,648 to +2,147,483,647 Chapter 2 — Instructions: Language of the Computer — 16 2s-Complement Signed Integers � Bit 31 is sign bit � 1 for negative numbers � 0 for non-negative numbers � –(–2n – 1) can’t be represented � Non-negative numbers have the same unsigned and 2s-complement representation � Some specific numbers � 0: 0000 0000 … 0000 � –1: 1111 1111 … 1111 � Most-negative: 1000 0000 … 0000 � Most-positive: 0111 1111 … 1111 Chapter 2 — Instructions: Language of the Computer — 17 Signed Negation � Complement and add 1 � Complement means 1 → 0, 0 → 1 x 1 x 1 1111...111 x x 2 = − + = − = + � Example: negate +2 � +2 = 0000 0000 … 00102 � –2 = 1111 1111 … 11012 + 1 = 1111 1111 … 11102 Chapter 2 — Instructions: Language of the Computer — 18 Sign Extension � Representing a number using more bits � Preserve the numeric value � In MIPS instruction set � addi: extend immediate value � lb, lh: extend loaded byte/halfword � beq, bne: extend the displacement � Replicate the sign bit to the left � c.f. unsigned values: extend with 0s � Examples: 8-bit to 16-bit � +2: 0000 0010 => 0000 0000 0000 0010 � –2: 1111 1110 => 1111 1111 1111 1110 Chapter 2 — Instructions: Language of the Computer — 19 Representing Instructions � Instructions are encoded in binary � Called machine code � MIPS instructions � Encoded as 32-bit instruction words � Small number of formats encoding operation code (opcode), register numbers, … � Regularity! � Register numbers � $t0 – $t7 are reg’s 8 – 15 � $t8 – $t9 are reg’s 24 – 25 � $s0 – $s7 are reg’s 16 – 23 §2.5 Representing Instructions in the Computer Chapter 2 — Instructions: Language of the Computer — 20 MIPS R-format Instructions � Instruction fields � op: operation code (opcode) � rs: first source register number � rt: second source register number � rd: destination register number � shamt: shift amount (00000 for now) � funct: function code (extends opcode) op rs rt rd shamt funct 6 bits 6 bits 5 bits 5 bits 5 bits 5 bits Chapter 2 — Instructions: Language of the Computer — 21 R-format Example add $t0, $s1, $s2 special $s1 $s2 $t0 0 add 0 17 18 8 0 32 000000 10001 10010 01000 00000 100000 000000100011001001000000001000002 = 0232402016 op rs rt rd shamt funct 6 bits 6 bits 5 bits 5 bits 5 bits 5 bits Chapter 2 — Instructions: Language of the Computer — 22 Hexadecimal � Base 16 � Compact representation of bit strings � 4 bits per hex digit 0 0000 4 0100 8 1000 c 1100 1 0001 5 0101 9 1001 d 1101 2 0010 6 0110 a 1010 e 1110 3 0011 7 0111 b 1011 f 1111 � Example: eca8 6420 � 1110 1100 1010 1000 0110 0100 0010 0000 Chapter 2 — Instructions: Language of the Computer — 23 MIPS I-format Instructions � Immediate arithmetic and load/store instructions � rt: destination or source register number � Constant: –215 to +215 – 1 � Address: offset added to base address in rs � Design Principle 4: Good design demands good compromises � Different formats complicate decoding, but allow 32-bit instructions uniformly � Keep formats as similar as possible op rs rt constant or address 6 bits 5 bits 5 bits 16 bits Machine Code � What is the machine code of the following instruction: � sub $t0, t1, $t2 � What is the assembly language statement to this machine instruction: � 00af802016 Chapter 2 — Instructions: Language of the Computer — 24 Chapter 2 — Instructions: Language of the Computer — 26 Logical Operations � Instructions for bitwise manipulation Operation C Java MIPS Shift left << << sll Shift right >> >> srl Bitwise AND & & and, andi Bitwise OR | | or, ori Bitwise NOT ~ ~ nor � Useful for extracting and inserting groups of bits in a word §2.6 Logical Operations Chapter 2 — Instructions: Language of the Computer — 27 Shift Operations � shamt: how many positions to shift � Shift left logical � Shift left and fill with 0 bits � sll by i bits multiplies by 2i � Shift right logical � Shift right and fill with 0 bits � srl by i bits divides by 2i (unsigned only) op rs rt rd shamt funct 6 bits 6 bits 5 bits 5 bits 5 bits 5 bits Chapter 2 — Instructions: Language of the Computer — 28 AND Operations � Useful to mask bits in a word � Select some bits, clear others to 0 and $t0, $t1, $t2 0000 0000 0000 0000 0000 1101 1100 0000 0000 0000 0000 0000 0011 1100 0000 0000 $t2 $t1 0000 0000 0000 0000 0000 1100 0000 0000 $t0 Chapter 2 — Instructions: Language of the Computer — 29 OR Operations � Useful to include bits in a word � Set some bits to 1, leave others unchanged or $t0, $t1, $t2 0000 0000 0000 0000 0000 1101 1100 0000 0000 0000 0000 0000 0011 1100 0000 0000 $t2 $t1 0000 0000 0000 0000 0011 1101 1100 0000 $t0 Chapter 2 — Instructions: Language of the Computer — 30 NOT Operations � Useful to invert bits in a word � Change 0 to 1, and 1 to 0 � MIPS has NOR 3-operand instruction � a NOR b == NOT ( a OR b ) nor $t0, $t1, $zero 0000 0000 0000 0000 0011 1100 0000 0000 $t1 1111 1111 1111 1111 1100 0011 1111 1111 $t0 Register 0: always read as zero Chapter 2 — Instructions: Language of the Computer — 31 Conditional Operations � Branch to a labeled instruction if a condition is true � Otherwise, continue sequentially � beq rs, rt, L1 � if (rs == rt) branch to instruction labeled L1; � bne rs, rt, L1 � if (rs != rt) branch to instruction labeled L1; � j L1 � unconditional jump to instruction labeled L1 §2.7 Instructions for Making Decisions Chapter 2 — Instructions: Language of the Computer — 32 Compiling If Statements � C code: if (i==j) f = g+h; else f = g-h; � f, g, … in $s0, $s1, … � Compiled MIPS code: bne $s3, $s4, Else add $s0, $s1, $s2 j Exit Else: sub $s0, $s1, $s2 Exit: … Assembler calculates addresses Chapter 2 — Instructions: Language of the Computer — 33 Compiling Loop Statements � C code: while (save[i] == k) i += 1; � i in $s3, k in $s5, address of save in $s6 � Compiled MIPS code: Loop: sll $t1, $s3, 2 add $t1, $t1, $s6 lw $t0, 0($t1) bne $t0, $s5, Exit addi $s3, $s3, 1 j Loop Exit: … Chapter 2 — Instructions: Language of the Computer — 35 More Conditional Operations � Set result to 1 if a condition is true � Otherwise, set to 0 � slt rd, rs, rt � if (rs < rt) rd = 1; else rd = 0; � slti rt, rs, constant � if (rs < constant) rt = 1; else rt = 0; � Use in combination with beq, bne slt $t0, $s1, $s2 # if ($s1 < $s2) bne $t0, $zero, L # branch to L Chapter 2 — Instructions: Language of the Computer — 36 Branch Instruction Design � Why not blt, bge, etc? � Hardware for <, ≥, … slower than =, ≠ � Combining with branch involves more work per instruction, requiring a slower clock � All instructions penalized! � beq and bne are the common case � This is a good design compromise Chapter 2 — Instructions: Language of the Computer — 37 Signed vs. Unsigned � Signed comparison: slt, slti � Unsigned comparison: sltu, sltui � Example � $s0 = 1111 1111 1111 1111 1111 1111 1111 1111 � $s1 = 0000 0000 0000 0000 0000 0000 0000 0001 � slt $t0, $s0, $s1 # signed � –1 < +1 ⇒ $t0 = 1 � sltu $t0, $s0, $s1 # unsigned � +4,294,967,295 > +1 ⇒ $t0 = 0 Chapter 2 — Instructions: Language of the Computer — 38 Procedure Calling � Steps required 1. Place parameters in registers 2. Transfer control to procedure 3. Acquire storage for procedure 4. Perform procedure’s operations 5. Place result in register for caller 6. Return to place of call §2.8 Supporting Procedures in Computer Hardware Chapter 2 — Instructions: Language of the Computer — 39 Register Usage � $a0 – $a3: arguments (reg’s 4 – 7) � $v0, $v1: result values (reg’s 2 and 3) � $t0 – $t9: temporaries � Can be overwritten by callee � $s0 – $s7: saved � Must be saved/restored by callee � $gp: global pointer for static data (reg 28) � $sp: stack pointer (reg 29) � $fp: frame pointer (reg 30) � $ra: return address (reg 31) Chapter 2 — Instructions: Language of the Computer — 40 Procedure Call Instructions � Procedure call: jump and link jal ProcedureLabel � Address of following instruction put in $ra � Jumps to target address � Procedure return: jump register jr $ra � Copies $ra to program counter � Can also be used for computed jumps � e.g., for case/switch statements Chapter 2 — Instructions: Language of the Computer — 41 Leaf Procedure Example � C code: int leaf_example (int g, h, i, j) { int f; f = (g + h) - (i + j); return f; } � Arguments g, …, j in $a0, …, $a3 � f in $s0 (hence, need to save $s0 on stack) � Result in $v0 Chapter 2 — Instructions: Language of the Computer — 42 Leaf Procedure Example � MIPS code: leaf_example: addi $sp, $sp, -4 sw $s0, 0($sp) add $t0, $a0, $a1 add $t1, $a2, $a3 sub $s0, $t0, $t1 add $v0, $s0, $zero lw $s0, 0($sp) addi $sp, $sp, 4 jr $ra Save $s0 on stack Procedure body Restore $s0 Result Return Chapter 2 — Instructions: Language of the Computer — 46 Local Data on the Stack � Local data allocated by callee � e.g., C automatic variables � Procedure frame (activation record) � Used by some compilers to manage stack storage Chapter 2 — Instructions: Language of the Computer — 47 Memory Layout � Text: program code � Static data: global variables � e.g., static variables in C, constant arrays and strings � $gp initialized to address allowing ±offsets into this segment � Dynamic data: heap � E.g., malloc in C, new in Java � Stack: automatic storage Chapter 2 — Instructions: Language of the Computer — 48 Character Data � Byte-encoded character sets � ASCII: 128 characters � 95 graphic, 33 control � Latin-1: 256 characters � ASCII, +96 more graphic characters � Unicode: 32-bit character set � Used in Java, C++ wide characters, … � Most of the world’s alphabets, plus symbols §2.9 Communicating with People Chapter 2 — Instructions: Language of the Computer — 49 Byte/Halfword Operations � Could use bitwise operations � MIPS byte/halfword load/store � String processing is a common case lb rt, offset(rs) lh rt, offset(rs) � Sign extend to 32 bits in rt lbu rt, offset(rs) lhu rt, offset(rs) � Zero extend to 32 bits in rt sb rt, offset(rs) sh rt, offset(rs) � Store just rightmost byte/halfword Chapter 2 — Instructions: Language of the Computer — 50 String Copy Example � C code (naïve): � Null-terminated string void strcpy (char x[], char y[]) { int i; i = 0; while ((x[i]=y[i])!='\0') i += 1; } � Addresses of x, y in $a0, $a1 � i in $s0 Chapter 2 — Instructions: Language of the Computer — 51 String Copy Example � MIPS code: strcpy: addi $sp, $sp, -4 # adjust stack for 1 item sw $s0, 0($sp) # save $s0 add $s0, $zero, $zero # i = 0 L1: add $t1, $s0, $a1 # addr of y[i] in $t1 lbu $t2, 0($t1) # $t2 = y[i] add $t3, $s0, $a0 # addr of x[i] in $t3 sb $t2, 0($t3) # x[i] = y[i] beq $t2, $zero, L2 # exit loop if y[i] == 0 addi $s0, $s0, 1 # i = i + 1 j L1 # next iteration of loop L2: lw $s0, 0($sp) # restore saved $s0 addi $sp, $sp, 4 # pop 1 item from stack jr $ra # and return Chapter 2 — Instructions: Language of the Computer — 52 0000 0000 0111 1101 0000 0000 0000 0000 32-bit Constants � Most constants are small � 16-bit immediate is sufficient � For the occasional 32-bit constant lui rt, constant � Copies 16-bit constant to left 16 bits of rt � Clears right 16 bits of rt to 0 lui $s0, 61 0000 0000 0111 1101 0000 1001 0000 0000 ori $s0, $s0, 2304 §2.10 MIPS Addressing for 32-Bit Immediates and Addresses Chapter 2 — Instructions: Language of the Computer — 53 Branch Addressing � Branch instructions specify � Opcode, two registers, target address � Most branch targets are near branch � Forward or backward op rs rt constant or address 6 bits 5 bits 5 bits 16 bits � PC-relative addressing � Target address = PC + offset × 4 � PC already incremented by 4 by this time Chapter 2 — Instructions: Language of the Computer — 54 Jump Addressing � Jump (j and jal) targets could be anywhere in text segment � Encode full address in instruction op address 6 bits 26 bits � (Pseudo)Direct jump addressing � Target address = PC31…28 : (address × 4) Chapter 2 — Instructions: Language of the Computer — 55 Target Addressing Example � Loop code from earlier example � Assume Loop at location 80000 Loop: sll $t1, $s3, 2 80000 0 0 19 9 2 0 add $t1, $t1, $s6 80004 0 9 22 9 0 32 lw $t0, 0($t1) 80008 35 9 8 0 bne $t0, $s5, Exit 80012 5 8 21 2 addi $s3, $s3, 1 80016 8 19 19 1 j Loop 80020 2 20000 Exit: … 80024 Chapter 2 — Instructions: Language of the Computer — 56 Branching Far Away � If branch target is too far to encode with 16-bit offset, assembler rewrites the code � Example beq $s0,$s1, L1 ↓ bne $s0,$s1, L2 j L1 L2: … Chapter 2 — Instructions: Language of the Computer — 57 Addressing Mode Summary Chapter 2 — Instructions: Language of the Computer — 60 Translation and Startup Many compilers produce object modules directly Static linking §2.12 Translating and Starting a Program Chapter 2 — Instructions: Language of the Computer — 92 Concluding Remarks � Design principles 1. Simplicity favors regularity 2. Smaller is faster 3. Make the common case fast 4. Good design demands good compromises � Layers of software/hardware � Compiler, assembler, hardware � MIPS: typical of RISC ISAs � c.f. x86 §2.19 Concluding Remarks REVISION Chapter 2 — Instructions: Language of the Computer — 94 95 Logical Operators � Logical operators � Allows for forming more complex conditions � Combines simple conditions � Java logical operators � && (conditional AND) � || (conditional OR) � & (boolean logical AND) � | (boolean logical inclusive OR) � ^ (boolean logical exclusive OR) � ! (logical NOT) 96 Logical Operators (Cont.) � Conditional AND (&&) Operator � Consider the following if statement � if ( gender == FEMALE && age >= 65 ) � ++seniorFemales; � Combined condition is true � if and only if both simple conditions are true � Combined condition is false � if either or both of the simple conditions are false 97 && && && && (conditional AND) operator truth table. expression1 expression2 expression1 && && && && expression2 false false false false false false false false FFFFalse alse alse alse false false false false true true true true FFFFalse alse alse alse true true true true false false false false FFFFalse alse alse alse true true true true true true true true TTTTrue rue rue rue 98 Logical Operators (Cont.) � Conditional OR (||) Operator � Consider the following if statement � if ( ( semesterAverage >= 90 ) || ( finalExam >= 90 )) � System.out.println( “Student grade is A” ); � Combined condition is true � if either or both of the simple condition are true � Combined condition is false � if both of the simple conditions are false 99 || || || || (conditional OR) operator truth table. expression1 expression2 expression1 || || || || expression2 false false false false false false false false false false false false false false false false true true true true true true true true true true true true false false false false true true true true true true true true true true true true true true true true 100 Logical Operators (Cont.) � Short-Circuit Evaluation of Complex Conditions � Parts of an expression containing && or || operators are evaluated only until it is known whether the condition is true or false � E.g., ( gender == FEMALE ) && ( age >= 65 ) � Stops immediately if gender is not equal to FEMALE 101 Common Programming Error � In expressions using operator &&, a condition- we will call this the dependent condition-may require another condition to be true for the evaluation of the dependent condition to be meaningful. In this case, the dependent condition should be placed after the other condition, or an error might occur. For example, in the expression ( i != 0 ) && ( 10 / i == 2 ), the second condition must appear after the first condition, or a divide-by-zero error might occur. 102 Logical Operators (Cont.) � Boolean Logical AND (&) Operator � Works identically to && � Except & always evaluate both operands � Boolean Logical OR (|) Operator � Works identidally to || � Except | always evaluate both operands 103 Logical Operators (Cont.) � Boolean Logical Exclusive OR (^) � One of its operands is true and the other is false � Evaluates to true � Both operands are true or both are false � Evaluates to false � Logical Negation (!) Operator � Unary operator 104 ^^^^ (boolean logical exclusive OR) operator truth table. expression1 expression2 expression1 ^^^^ expression2 false false false false false false false false false false false false false false false false true true true true true true true true true true true true false false false false true true true true true true true true true true true true false false false false 105 (logical negation, or logical NOT) operator truth table. expression !!!!expression false false false false true true true true true true true true false false false false