Download presentation

Presentation is loading. Please wait.

Published byWinston Wiswell Modified over 2 years ago

1
ECE 232 L6.Assemb.1 Adapted from Patterson 97 ©UCBCopyright 1998 Morgan Kaufmann Publishers ECE 232 Hardware Organization and Design Lecture 6 MIPS Assembly Instructions, cont’d Maciej Ciesielski www.ecs.umass.edu/ece/labs/vlsicad/ece232/sp$r2002/index_232.html

2
ECE 232 L6.Assemb.2 Adapted from Patterson 97 ©UCBCopyright 1998 Morgan Kaufmann Publishers Administrative issues °Homework #2 assigned today, due Th., Feb. 21 in class. °Excuses that cannot be tolerated Forgot to buy the textbook Forgot to do the homework Dog ate the homework, etc … (you are pretty creative at those) °Solving the textbook problem Putting one copy of text on reserve in Physical Science Library °Change in late homework policy 25% per day – to allow to post solutions in 3-4 days °Midterm exam I – Th., March 7, 6:30 pm Morrill Science Center (Bldg II) room 131 No excuses (forgot, overslept, overate, etc) Only legitimate reasons (in writing) can be accepted

3
ECE 232 L6.Assemb.3 Adapted from Patterson 97 ©UCBCopyright 1998 Morgan Kaufmann Publishers Outline °Review: MIPS instructions, formats MIPS instructions: data transfers, arithmetic, logical Pseudo-instruction example: loading large constant MIPS register organization °Implementing loops °Implementing switch/case statement °Procedures and subroutines Stacks and pointers °Running a program Compiler, Assembler, Linker, Loader

4
ECE 232 L6.Assemb.4 Adapted from Patterson 97 ©UCBCopyright 1998 Morgan Kaufmann Publishers MIPS Addressing Modes/Instruction Formats immedoprsrt Immediate All instructions are 32-bit wide Base+index immedoprsrt register + Memory PC-relative immedoprsrt PC Memory + register Register (direct ) oprsrtrdfnc...

5
ECE 232 L6.Assemb.5 Adapted from Patterson 97 ©UCBCopyright 1998 Morgan Kaufmann Publishers R0$zero constant 0 R1$at reserved for assembler R2$v0 value registers & R3$v1 function results R4$a0 arguments R5$a1 R6$a2 R7$a3 R8$t0 temporary: caller saves... (callee can clobber) R15 $t7 MIPS: Software conventions for registers R16 $s0callee saves... (caller can clobber) R23 $s7 R24 $t8 temporary (cont’d) R25 $t9 R26 $k0 reserved for OS kernel R27 $k1 R28 $gp pointer to global area R29 $spStack pointer R30 $fpFrame pointer R31 $ra return Address

6
ECE 232 L6.Assemb.6 Adapted from Patterson 97 ©UCBCopyright 1998 Morgan Kaufmann Publishers MIPS data transfer instructions Instruction Comment sw 500($r4), $r3Store word sh 502($r2), $r3Store half sb 41($r3), $r2Store byte lw $r1, 30($r2)Load word lh $r1, 40($r3)Load halfword lhu $r1, 40($r3)Load halfword unsigned lb $r1, 40($r3)Load byte lbu $r1, 40($r3)Load byte unsigned lui $r1, 40Load Upper Immediate (16 bits shifted left by 16) 0000 … 0000 LUI $r5 $r5

7
ECE 232 L6.Assemb.7 Adapted from Patterson 97 ©UCBCopyright 1998 Morgan Kaufmann Publishers Loading large numbers °Pseudo-instruction li $t0, big: load 32-bit constant lui $t0, upper# $t0 [31:16] upper ori $t0, $t0, lower# $t0 ($t0 Or [0ext.lower]) $to: upper lower upper 0 163115 0000 0000 lower0000 0000 Or 32-bit constant

8
ECE 232 L6.Assemb.8 Adapted from Patterson 97 ©UCBCopyright 1998 Morgan Kaufmann Publishers Loop with variable array index °Compile the following loop, with A[ ] = array with base address in $s5; variables g, h, i, j associated with registers $s1, $s2, $s3, $s4. Loop:g = g + A[i]; i = i + j; if (i h) go to Loop; MIPS instructions: Loop:add $t1, $s3, $s3 # $t1 i+i = 2i add $t1, $t1, $t1 # $t1 2i+2i = 4i add $t1, $t1, $s5 # $t1 address of A[i] lw $t0, 0 ($t1) # $t0 A[i] add $s1, $s1, $t0 # $s1 g + A[i] add $s3, $s3, $s4 # $s3 i + j bne $s3, $s2, Loop # if (i h) go to Loop

9
ECE 232 L6.Assemb.9 Adapted from Patterson 97 ©UCBCopyright 1998 Morgan Kaufmann Publishers While loop °Base address of A[i] is in $s6; variables i, j, k are in $s3, $s4, $s5. °Compile the following while loop while (A[i] == k) i = i + j; into MIPS instructions: Loop:add $t1, $s3, $s3 # $t1 i+i = 2i add $t1, $t1, $t1 # $t1 2i+2i = 4i add $t1, $t1, $s6 # $t1 address of A[i] lw $t0, 0 ($t1) # $t0 A[i] bne $t0, $s5, Exit # if (A[I] k) go to Exit add $s3, $s3, $s4 # $s3 i + j j Loop # go to Loop Exit:

10
ECE 232 L6.Assemb.10 Adapted from Patterson 97 ©UCBCopyright 1998 Morgan Kaufmann Publishers Switch/case statement °Variables f - k are in $s0 - $s5. Register $t2 contains constant 4. °Compile the following switch statement into MIPS instructions switch (k) { case 0: f = i + j; break; /* k=0 */ case 1: f = g + h; break; /* k=1 */ case 2: f = g - h; break; /* k=2 */ case 3: f = i - j; break; /* k=3 */ } °Use the switch variable k to index the jump address table. First, test for k, if in correct range (0-3). slt $t3, $s5, $zero # test if k, 0 bne $t3, $zero, Exit # if k < 0, go to Exit slt $t3, $s5, $t2 # test if k < 4 beq $t3, $zero, Exit # if k 4, go to Exit

11
ECE 232 L6.Assemb.11 Adapted from Patterson 97 ©UCBCopyright 1998 Morgan Kaufmann Publishers Switch statement, cont’d °Access jump table T [ ] with addresses L0,L1,L2,L3: add $t1, $s5, $s5 # $t1 2k add $t1, $t1, $t1 # $t1 4k add $t1, $t1, $t4 # $t1 address of T [k] lw $t0, 0 ($t1) # $t0 T [k] °Use jump register instruction to jump via $t1 to the right address jr $to # jump based on register $t0 L0:add $s0, $s3, $s4 # k = 0, so f=$s0 i + j j Exit # go to Exit L1:add $s0, $s1, $s2 # k = 1, so f=$s0 g + h j Exit # go to Exit L2:sub $s0, $s1, $s2 # k = 2, so f=$s0 g + h j Exit # go to Exit L3:sub $s0, $s3, $s4 # k = 3, so f=$s0 i - j Exit:

12
ECE 232 L6.Assemb.12 Adapted from Patterson 97 ©UCBCopyright 1998 Morgan Kaufmann Publishers Stacks in Procedure Calls Stacking of Subroutine Calls and Returns: A: CALL B CALL C C: RET B: A AB ABC AB A Some machines provide a memory stack as part of the architecture (e.g., VAX) Sometimes stacks are implemented via software convention (e.g., MIPS)

13
ECE 232 L6.Assemb.13 Adapted from Patterson 97 ©UCBCopyright 1998 Morgan Kaufmann Publishers Memory Stacks Useful for stacked environments/subroutine call and return Stacks that Grow Up vs. Stacks that Grow Down: a b c 0 Little inf. Big 0 Little inf. Big Memory Addresses SP Next Empty? Last Full? How is empty stack represented? Little --> Big/Last Full POP: Read from Mem(SP) Decrement SP PUSH: Increment SP Write to Mem(SP) grows up grows down Little --> Big/Next Empty POP: Decrement SP Read from Mem(SP) PUSH: Write to Mem(SP) Increment SP

14
ECE 232 L6.Assemb.14 Adapted from Patterson 97 ©UCBCopyright 1998 Morgan Kaufmann Publishers Call-Return Linkage: Stack Frames Reference args and local variables at fixed (positive) offset from FP Grows and shrinks during expression evaluation FP Arguments Callee save registers Local variables SP (old FP, RA) °Many variations on stacks possible (up/down, last pushed / next ) °Block structured languages contain link to lexically enclosing frame °Compilers normally keep scalar variables in registers, not memory! High Mem Low Mem

15
ECE 232 L6.Assemb.15 Adapted from Patterson 97 ©UCBCopyright 1998 Morgan Kaufmann Publishers Compiling a leaf procedure (not nested) °int leaf_example (int g, int h, int i, int j) {int f; f = (g + h) – (i + j); return f; } °Let parameter variables g, h, i, j, correspond to the argument registers $a0, $a1, $a2, $a3. Will use temp. registers $t0= (g + h) and $t1=(i + j). Function f will be stored in $s0. °Steps: °Save the old values of registers ($s0, $t0, $t1) on stack (push) °Issue a jal sub_address instruction ($ra ret_addr, j sub_address) °Perform the computation for $t0, $t1, $s0 using argument registers °Copy the value of f into a return value register $v0 °Restore the old values of the saved registers from stack (pop) °Finally, jump back to the calling routine, jr $ra (PC return_address=PC+4)

16
ECE 232 L6.Assemb.16 Adapted from Patterson 97 ©UCBCopyright 1998 Morgan Kaufmann Publishers Compiling a leaf procedure, cont’d Leaf_example:# label of the procedure Save the old values of registers ($s0, $t0, $t1) on stack (push) sub $sp, $sp, 12# adjust stack to make room for 3 items sw $t1, 8 ($sp)# save reg $t1 on stack …….# repeat for $t0, $s0 Perform the computation for $t0, $t1, $s0 using argument registers add $t0, $a0, $a1# $t0 g + h add $t1, $a2, $a3# $t1 i + j sub $s0, $t0, $t1# $s0 (g + h) – (i + j) Copy the value of f into a return value register $v0 add $v0, $s0, $zero# returns f ($v0 $s0 + 0) Restore the old values of the saved registers from stack (pop ) lw $s0, 0 ($sp)# restore reg. $s0 for the caller ……. # repeat for $t0, $t1 … add $sp, $sp, 12# adjust the stack to delete 3 items Finally, jump back to the calling routine (PC return address) jr $ra# PC $ra

Similar presentations

OK

1 Lecture 4: Procedure Calls Today’s topics: Procedure calls Large constants The compilation process Reminder: Assignment 1 is due on Thursday.

1 Lecture 4: Procedure Calls Today’s topics: Procedure calls Large constants The compilation process Reminder: Assignment 1 is due on Thursday.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

5 components of reading ppt on ipad Ppt on mars one news Ppt on obesity management association Ppt on art of war audio Ppt on acid-base indicators images Ppt on job rotation evaluation Ppt on anti rigging voting system Download ppt on oxidation and reduction for dummies Ppt on shell scripting in linux Ppt on writing effective research questions