Organization Lecture-4 Code Translation-2 Memory, Data transfer - PowerPoint PPT Presentation

CSE 2021: Computer Organization Lecture-4 Code Translation-2 Memory, Data transfer instructions, Data segment, ASCII, Procedures, Stack Shakil M. Khan (adapted from Prof. Roumani)

Previously… Registers • – $s0 - $s7, $t0 - $t9, $zero, $ra, $v0, $a0 Arithmetic and logical instructions • – add, sub, slt, mul, div – and, or, xor, nor, sll, srl, sra – the i and u suffix lui • Jump and branch • – j, jr – beq, bne I/O (syscall) • MIPS instruction formats • – R-type (register: e.g. add) – I-type (immediate: e.g. addi) – J-type (jump: e.g. j) CSE-2021 2

Memory Organization Memory: large one dimensional • array of cells Addresses are indices to the array • 12 100 Each cell is 1 word (4 bytes) long • 8 10 4 101 Each word in a memory has an • address, which is a multiple of 4 0 1 – min value of byte address = 0 Address Data – max value = 2 32 – 1 Processor Memory Data is transferred from memory • into registers using data transfer instructions CSE-2021 3

The Load/Store Family (1) • Used to transfer data to/from DRAM – to read/write .data, the heap, and the stack • lw and sw – what are their algorithms? • lb and sb – what does the u suffix do to lb? CSE-2021 4

The Load/Store Family (2) Syntax Meaning Comments Example Memory to lw rt, imm(rs) rt = M[rs + SignExtImm] lw $s1, 100($s2) Register Register to sw rt, imm(rs) M[rs + SignExtImm] = rt sw $s1, 100($s2) memory Memory to lb rt, imm(rs) rt(7:0) = M[rs + SignExtImm](7:0) lb $s1, 100($s2) Register Register to sb rt, imm(rs) M[rs + SignExtImm](7:0) = rt(7:0) sb $s1, 100($s2) memory CSE-2021 5

The Load/Store Family (Example) C instruction: g = h + A[k] • Register allocation: • A[k] 1 0 0 Address + 4xk $s1 => g $s2 => h – … 1 0 … $s3 contains base address – A[1] 1 0 1 address + 4 of array (i.e. address of A[0]) A[0] 1 address $s4 contains value of k – Array D a t a MIPS Code: sll $t1,$s4,2 # $t1 = 4*k add $t1,$t1,$s3 # $t1 = address of A[0] + 4*k lw $t0,0($t1) # $t0 = A[k] add $s1,$s2,$t0 # $s3 = h + A[k] CSE-2021 6

MIPS Fields for Data Transfer Operations (I-type) op rs rt address 6 bits 5 bits 5 bits 16 bits 1 st operand 2 nd operand opcode Memory address (offset) Example: lw $t0,32($s3) • opcode = 35 10 = (100011) 2 – rs = $s3 = 19 10 = (10011) 2 – rt = $t0 = 8 10 = (01000) 2 – address = 32 10 = (0000 0000 0010 0000) 2 – leads to the binary machine language code: – 100011 10011 01000 0000000000100000 CSE-2021 7

Data Transfer Instructions (Exercise) • Write the MIPS assembly code for the following C assignment instruction – A[12] = h + A[8] – assuming that the variable h is stored in $s2 and the base address of the array A is in $s3 CSE-2021 8

Assumptions for Phase II A Utility Class • A single class plus its client – hence, two linked classes but no objects • Static Attributes only – hence .data is needed but no heap • Can have several static methods – hence stack is needed CSE-2021 9

Static Attributes and .data • To allocate static int x = 5 in .data: x: .word* 5 • To transfer the value of x to register s0: lb/lh/lw $s0, x($0) • To transfer the value of register s0 to x: sb/sh/sw $s0, x($0) * can use byte/half/word, ascii/asciiz, or space CSE-2021 10

ASCII Characters • American Standard Code for Information Interchange – 8 bits per char (in contrast to 16 bits unicode in Java) – e.g.: ASCII code for char C is (67) 10 or (01000011) 2 • Strings are sequences of characters + something to represent its length • Three possible choices: – end a string with a null character `\0 ’ ( ANSI C) – use first byte to represent length (JAVA) – use accompanying variable to indicate length • How to load a 8 or 16 bits at a time? – lb, sb, lh, sh CSE-2021 11

ASCII ASCII ASCII ASCII ASCII ASCII Char Char Char Char Char Char value value value value value value space 0 @ P ` p 32 48 64 80 96 112 ! 1 A Q a q 33 49 65 81 97 113 “ 2 B R b r 34 50 66 82 98 114 # 3 C S c s 35 51 67 83 99 115 $ 4 D T d t 36 52 68 84 100 116 % 5 E U e u 37 53 69 85 101 117 & 6 F V f v 38 54 70 86 102 118 ‘ 7 G W g w 39 55 71 87 103 119 ( 8 H X h x 40 56 72 88 104 120 ) 9 I Y i y 41 57 73 89 105 121 * : J Z j z 42 58 74 90 106 122 + ; K [ k { 43 59 75 91 107 123 ‘ < L \ l | 44 60 76 92 108 124 - = M ] m } 45 61 77 93 109 125 . > N ^ n ~ 46 62 78 94 110 126 / ? O _ o DEL 47 63 79 95 111 127 CSE-2021 12

.data Example #1 .data x: .byte 65 y: .word 123 z: .byte -30 u: .half 120 v: .ascii “York” s: .asciiz “York” t: .float 2.45 .text main: lb $t0, x($0) lw $t0, y($0) lb $t0, z($0) lbu $t0, z($0) lhu $t0, u($0) lbu $t0, v($0) CSE-2021 13

.data Example #2 .data y: .word 123, 150, 22 z: .byte -30, 12 p: .word y .text main: addi $t1, $0, 8 lw $t0, y($t1) # index-like addi $t1, $0, 1 lb $t0, z($t1) # index-like la $t1, y lw $t0, 0($t1) # pointer-like lw $t0, 4($t1) # pointer-like lw $t1, p($0) addi $t1, $t1, 8 lw $t0, 0($t1) # pointer-like CSE-2021 14

.data Example #3 .data x: .byte 65 y: .word 123, 150, 22 z: .byte -30, 12 u: .half 120 v: .ascii “York” s: .asciiz “York” p: .word y .text main: la $t1, x addi $t1, $t1, 8 lw $t0, 0($t1) # alignment! lw $t1, p($0) addi $t1, $t1, 16 # endianness lw $t0, 0($t1) CSE-2021 15

Procedures (1) Input Parameters $a0 - $a3 Calling Procedure Program Output Parameters $v0 - $v1 • Calling program: – places parameters in registers $a0 - $a3 where the called procedure can access them – transfers control to the procedure (recall PC)  address of the next instruction in the calling program is automatically placed in register $ra  jal proc CSE-2021 16

Procedures (2) Input Parameters $a0 - $a3 Calling Procedure Program Output Parameters $v0 - $v1 Called Procedure: • – acquires additional space needed to perform the task  saves values of required registers in a stack $sp – performs the desired task – restores the values of registers that it used – saves the result in registers $v0 - $v1 – returns control to the calling program by returning to instruction whose address is saved in $ra (jr $ra) CSE-2021 17

Procedures (Stack) H i g h a d d r e s s $fp $fp $ s p $ s p $fp C o n t e n t s o f r e g i s t e r $ t 1 C o n t e n t s o f r e g i s t e r $ t 0 $ s p C o n t e n t s o f r e g i s t e r $ s 0 L o w a d d r e s s a . b . c . • Where are values of the calling function stored? – a stack (assigned space) in the memory is used – register $sp contains the address of the stack pointer CSE-2021 18

Stack Usage • To push the content of register $s0 on the stack: addi $sp, $sp, -4 sw $s0, 0($sp) • To pop the word at the top of the stack into $s0: lw $s0, 0($sp) addi $sp, $sp, 4 • Every method must preserve $sp, $ra, and $sx plus any other register it needs after a call CSE-2021 19

Proc. Example #1 (1) Example: • int leaf_example (int g, int h, int i, int j) { int f; f = (g + h) – (i + j); return f; } Assume: g, h, i, j are stored in registers $a0 - $a3 • Calling function in MIPS: • main: … jal leaf_example … CSE-2021 20

Proc. Example #1 (2) leaf_example: # save registers $t0, $t1, $s1 addi $sp, $sp, -12 sw $t1, 8($sp) sw $t0, 4($sp) sw $s0, 0 ($sp) # perform the required operation add $t0, $a0, $a1 add $t1, $a2, $a3 sub $s0, $t0, $t1 add $v0, $s0,$zero # restore registers $t0, $t1, $s1 lw $s0, 0 ($sp) lw $t0, 4($sp) lw $t1, 8($sp) addi $sp, $sp, 12 # return control jr $ra CSE-2021 21

Proc. Example #2 (1) • Nested Procedure Example: int fact (int n) { if (n < 1) return (1) else return (n * fact(n-1)); } • Assume variable n is stored in $a0 CSE-2021 22

Proc. Example #2 (2) • Before each function call – value of $ra must be saved in the stack – value of $a0 must be saved in the stack • After returning from a function – value of $a0 must be recalled for ($v0 × $a0) – value of $ra must be restored to return control to the calling function CSE-2021 23

Proc. Example #2 (3) Flow Diagram main $v0 = 2 x ($a0) 1 ($ra) 1 = return address in main = 6 ($a0) 1 = 3 fact # 1 ($ra) 2 = return address in fact # 1 $v0 = 1 x ($a0) 2 ($a0) 2 = 2 = 2 fact # 2 ($ra) 3 = return address in fact # 2 $v0 = 1 x ($a0) 3 ($a0) 3 = 1 = 1 fact # 3 CSE-2021 24

Proc. Example #2 (4): Partial MIPS Code fact: # input value is stored in $a0 and output value in $v0 # add commands for saving $ra, $a0 addi $t1, $zero, 1 # initialize $t1 = 1 slt $t0, $a0, $t1 # if (n < 1), $t0 = 1 beq $t0, $zero, L1 # if $t0 == 0, go to L1 # if n < 1 addi $v0, $zero,1 # return 1 jr $ra # if n >= 1 L1: addi $a0, $a0, -1 # $a0 = $a0 -1 jal fact # add commands to restore $ra, $a0 # process & return answer in $v0 CSE-2021 25