Inequalities in MIPS (2/4) Inequalities in MIPS (1/4) Inequalities in MIPS (4/4) Inequalities in MIPS (3/4) UCB CS61C : Machine Structures

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CS61C L8 Introduction to MIPS : Decisions II & Procedures I (1) Instructor Paul Pearce inst.eecs.berkeley.edu/~cs61c UCB CS61C : Machine Structures http://www.xkcd.org/627/ Lecture 8 Decisions & and Introduction to MIPS Procedures 2010-07-01! And in Review Memory is byte-addressable, but lw and sw access one word at a time. A pointer (used by lw and sw) is just a memory address, we can add to it or subtract from it (using offset). A Decision allows us to decide what to execute at runtime rather than compile-time. C Decisions are made using conditional statements within if, while, do while, for. MIPS Decision making instructions are the conditional branches: beq and bne. One can store and load (signed and unsigned) bytes as well as words with lb, lbu Unsigned add/sub don t signal overflow Loops using beq and bne. New Instructions: lw, sw, beq, bne, j, lb, sb, lbu, addu, addiu, subu, srl, sll WOW CS61C L8 Introduction to MIPS : Decisions II & Procedures I (2) Inequalities in MIPS (1/4) Until now, we ve only tested equalities (== and!= in C). General programs need to test < and > as well. Introduce MIPS Inequality Instruction: Set on Less Than Syntax: slt reg1,reg2,reg3 Meaning: reg1 = (reg2 < reg3); if (reg2 < reg3) reg1 = 1; Same thing else reg1 = 0; set means change to 1, reset means change to 0. Inequalities in MIPS (2/4) How do we use this? Compile by hand: if (g < h) goto Less; #g:$s0, h:$s1 Answer: compiled MIPS code slt $t0,$s0,$s1 # $t0 = 1 if g<h # $t0 = 0 if g>=h bne $t0,$0,less # goto Less # if $t0!=0 Register $0 always contains the value 0, so bne and beq often use it for comparison after an slt instruction. A slt bne pair means if( < )goto CS61C L8 Introduction to MIPS : Decisions II & Procedures I (3) CS61C L8 Introduction to MIPS : Decisions II & Procedures I (4) Inequalities in MIPS (3/4) Inequalities in MIPS (4/4) Now we can implement <, but how do we implement >, and? We could add 3 more instructions, but: MIPS goal: Simpler is Better Can we implement in one or more instructions using just slt and branches? What about >? What about? Lets compile this by hand: if (g >= h) goto Less; #g:$s0, h:$s1 Answer: compiled MIPS code slt $t0,$s0,$s1 # $t0 = 1 if g<h # $t0 = 0 if g>=h beq $t0,$0,less # goto less if g>=h Two independent variations possible: Use slt $t0,$s1,$s0 instead of slt $t0,$s0,$s1 Use bne instead of beq CS61C L8 Introduction to MIPS : Decisions II & Procedures I (5) CS61C L8 Introduction to MIPS : Decisions II & Procedures I (6)

CS61C L8 Introduction to MIPS : Decisions II & Procedures I (7) C! M I P S! Immediates in Inequalities There is also an immediate version of slt to test against constants: slti Helpful in for loops Loop:... if (g >= 1) goto Loop; Loop:... slti $t0,$s0,1 # $t0 = 1 if g < i # $t0 = 0 if g>=1) beq $t0,$0,loop # goto Loop if g>=1 An slt beq pair means if( )goto What about unsigned numbers? Also unsigned inequality instructions: sltu, sltiu which sets result to 1 or 0 depending on unsigned comparisons What is value of $t0, $t1? ($s0 = FFFF FFFA hex, $s1 = 0000 FFFA hex ) slt $t0, $s0, $s1 sltu $t1, $s0, $s1 CS61C L8 Introduction to MIPS : Decisions II & Procedures I (8) MIPS Signed vs. Unsigned diff meanings! Peer Instruction MIPS terms Signed/Unsigned overloaded : Do/Don't sign extend (lb, lbu) Do/Don't overflow (add, addi, sub, mult, div) (addu, addiu, subu, multu, divu) Do signed/unsigned compare (slt, slti/sltu, sltiu) Loop:addi $s0,$s0,-1 # i = i - 1 slti $t0,$s1,2 # $t0 = (j < 2) beq $t0,$0,loop # goto Loop if $t0 == 0 slt $t0,$s1,$s0 # $t0 = (j < i) bne $t0,$0,loop # goto Loop if $t0!= 0 ($s0=i, $s1=j) What C code properly fills in the blank in loop below? do {i--; while( ); a) j < 2 && j < i b) j 2 && j < i c) j < 2 && j i d) j 2 && j i e) j > 2 && j < i a) j < 2 j < i b) j 2 j < i c) j < 2 j i d) j 2 j i e) j > 2 j < i CS61C L8 Introduction to MIPS : Decisions II & Procedures I (9) CS61C L8 Introduction to MIPS : Decisions II & Procedures I (10) Administrivia Project 1 due Friday! (ok, Saturday, but tell your brain it s Friday!) Projector bulb is dying I tried to adjust the color scheme accordingly. I called ETS. Should be fixed soon. HW 3 and 4 now online Midterm: TWO WEEKS FROM TOMORROW! Friday, July 16 th. 9:30am-12:30pm. Room: 100 Lewis Other administrivia? C functions main() { int i,j,k,m;... i = mult(j,k);... m = mult(i,i);... /* really dumb mult function */ int mult (int mcand, int mlier){ int product = 0; while (mlier > 0) { product = product + mcand; mlier = mlier -1; return product; What information must compiler/programmer keep track of? What instructions can accomplish this? CS61C L8 Introduction to MIPS : Decisions II & Procedures I (12) CS61C L8 Introduction to MIPS : Decisions II & Procedures I (13)

CS61C L8 Introduction to MIPS : Decisions II & Procedures I (14) Function Call Bookkeeping Registers play a major role in keeping track of information for function calls. Register conventions: Return address $ra Arguments $a0, $a1, $a2, $a3 Return value $v0, $v1 Local variables $s0, $s1,, $s7 The stack is also used; more later. C M I P S Instruction Support for Functions (1/6)... sum(a,b);... /* a,b:$s0,$s1 */ int sum(int x, int y) { return x+y; address (shown in decimal) 1000 1004 1008 1012 1016 2000 2004 In MIPS, all instructions are 4 bytes, and stored in memory just like data. So here we show the addresses of where the programs are stored. CS61C L8 Introduction to MIPS : Decisions II & Procedures I (15) C M I P S Instruction Support for Functions (2/6)... sum(a,b);... /* a,b:$s0,$s1 */ int sum(int x, int y) { return x+y; address (shown in decimal) 1000 add $a0,$s0,$zero # x = a 1004 add $a1,$s1,$zero # y = b 1008 addi $ra,$zero,1016 #$ra=1016 1012 j sum #jump to sum 1016 2000 sum: add $v0,$a0,$a1 2004 jr $ra # new instruction C M I P S Instruction Support for Functions (3/6)... sum(a,b);... /* a,b:$s0,$s1 */ int sum(int x, int y) { return x+y; Question: Why use jr here? Why not use j? Answer: sum might be called by many places, so we can t return to a fixed place. The calling proc to sum must be able to say return here somehow. 2000 sum: add $v0,$a0,$a1 2004 jr $ra # new instruction CS61C L8 Introduction to MIPS : Decisions II & Procedures I (16) CS61C L8 Introduction to MIPS : Decisions II & Procedures I (17) Instruction Support for Functions (4/6) Single instruction to jump and save return address: jump and link (jal) Before: 1008 addi $ra,$zero,1016 #$ra=1016 1012 j sum #goto sum After: 1008 jal sum # $ra=1012,goto sum Why have a jal? Make the common case fast: function calls very common. Don t have to know where code is in memory with jal! Instruction Support for Functions (5/6) Syntax for jal (jump and link) is same as for j (jump): jal label jal should really be called laj for link and jump : Step 1 (link): Save address of next instruction into $ra! Why next instruction? Why not current one? Step 2 (jump): Jump to the given label CS61C L8 Introduction to MIPS : Decisions II & Procedures I (18) CS61C L8 Introduction to MIPS : Decisions II & Procedures I (19)

CS61C L8 Introduction to MIPS : Decisions II & Procedures I (20) Instruction Support for Functions (6/6) Syntax for jr (jump register): jr register Instead of providing a label to jump to, the jr instruction provides a register which contains an address to jump to. Very useful for function calls: jal stores return address in register ($ra) jr $ra jumps back to that address Nested Procedures (1/2) int sumsquare(int x, int y) { return mult(x,x)+ y; Something called sumsquare, now sumsquare is calling mult. So there s a value in $ra that sumsquare wants to jump back to, but this will be overwritten by the call to mult. Need to save sumsquare return address before call to mult. CS61C L8 Introduction to MIPS : Decisions II & Procedures I (21) Nested Procedures (2/2) In general, may need to save some other info in addition to $ra. When a C program is run, there are 3 important memory areas allocated: Static: Variables declared once per program, cease to exist only after execution completes. E.g., C globals Heap: Variables declared dynamically via malloc Stack: Space to be used by procedure during execution; this is where we can save register values CS61C L8 Introduction to MIPS : Decisions II & Procedures I (22) C memory Allocation review Address-> $sp stack pointer 0 Stack Heap Static Code Program CS61C L8 Introduction to MIPS : Decisions II & Procedures I (23) Space for saved procedure information Explicitly created space, i.e., malloc() Variables declared once per program; e.g., globals Using the Stack (1/2) So we have a register $sp which always points to the last used space in the stack. To use stack, we decrement this pointer by the amount of space we need and then fill it with info. So, how do we compile this? int sumsquare(int x, int y) { return mult(x,x)+ y; CS61C L8 Introduction to MIPS : Decisions II & Procedures I (24) Using the Stack (2/2) Hand-compile int sumsquare(int x, int y) { sumsquare: return mult(x,x)+ y; addi $sp,$sp,-8 # space on stack sw $ra, 4($sp) # save ret addr push sw $a1, 0($sp) # save y add $a1,$a0,$zero # mult(x,x) jal mult # call mult lw $a1, 0($sp) # restore y add $v0,$v0,$a1 # mult()+y pop lw $ra, 4($sp) # get ret addr addi $sp,$sp,8 # restore stack jr $ra mult:... CS61C L8 Introduction to MIPS : Decisions II & Procedures I (25)

CS61C L8 Introduction to MIPS : Decisions II & Procedures I (26) Steps for Making a Procedure Call 1. Save necessary values onto stack. 2. Assign argument(s), if any. 3. jal call 4. Restore values from stack. Rules for Procedures Called with a jal instruction, returns with a jr $ra Accepts up to 4 arguments in $a0, $a1, $a2 and $a3 Return value is always in $v0 (and if necessary in $v1) Must follow register conventions So what are they? NEXT TIME CS61C L8 Introduction to MIPS : Decisions II & Procedures I (27) Basic Structure of a Function Prologue entry_label: addi $sp,$sp, -framesize sw $ra, framesize-4($sp) # save $ra save other regs if need be Body... (call other functions ) memory Epilogue restore other regs if need be lw $ra, framesize-4($sp) # restore $ra addi $sp,$sp, framesize jr $ra CS61C L8 Introduction to MIPS : Decisions II & Procedures I (28) ra MIPS Registers The constant 0 $0 $zero Reserved for Assembler $1 $at Return Values $2-$3 $v0-$v1 Arguments $4-$7 $a0-$a3 Temporary $8-$15 $t0-$t7 Saved $16-$23 $s0-$s7 More Temporary $24-$25 $t8-$t9 Used by Kernel $26-27 $k0-$k1 Global Pointer $28 $gp Stack Pointer $29 $sp Frame Pointer $30 $fp Return Address $31 $ra Already discussed New Not yet (From MIPS green sheet) Use names for registers -- code is clearer! CS61C L8 Introduction to MIPS : Decisions II & Procedures I (29) Other Registers $at: may be used by the assembler at any time; unsafe to use $k0-$k1: may be used by the OS at any time; unsafe to use $gp, $fp: don t worry about them Note: Feel free to read up on $gp and $fp in Appendix A, but you can write perfectly good MIPS code without them. CS61C L8 Introduction to MIPS : Decisions II & Procedures I (30) Peer Instruction int fact(int n){ if(n == 0) return 1; else return(n*fact(n-1)); When translating this to MIPS 1) We COULD copy $a0 to $a1 (& then not store $a0 or $a1 on the stack) to store n across recursive calls. 2) We MUST save $a0 on the stack since it gets changed. 3) We MUST save $ra on the stack since we need to know where to return to CS61C L8 Introduction to MIPS : Decisions II & Procedures I (31) 123 a) FFF b) FFT c) FTF c) FTT d) TFF d) TFT e) TTF e) TTT

CS61C L8 Introduction to MIPS : Decisions II & Procedures I (33) And in Conclusion In order to help the conditional branches make decisions concerning inequalities, we introduce a single instruction: Set on Less Than called slt, slti, sltu, sltiu Functions called with jal, return with jr $ra. The stack is your friend: Use it to save anything you need. Just leave it the way you found it! Instructions we know so far Arithmetic: add, addi, sub, addu, addiu, subu Memory: lw, sw, lb, sb Decision: beq, bne, slt, slti, sltu, sltiu Unconditional Branches (Jumps): j, jal, jr Registers we know so far All of them! And in Conclusion to the conclusion We are 1/4 th of the way done! CS61C L8 Introduction to MIPS : Decisions II & Procedures I (34) Bonus Slides Example: The C Switch Statement (1/3) Choose among four alternatives depending on whether k has the value 0, 1, 2 or 3. Compile this C code: switch (k) { case 0: f=i+j; break; /* k=0 */ case 1: f=g+h; break; /* k=1 */ case 2: f=g h; break; /* k=2 */ case 3: f=i j; break; /* k=3 */ CS61C L8 Introduction to MIPS : Decisions II & Procedures I (35) CS61C L8 Introduction to MIPS : Decisions II & Procedures I (36) Example: The C Switch Statement (2/3) This is complicated, so simplify. Rewrite it as a chain of if-else statements, which we already know how to compile: if(k==0) f=i+j; else if(k==1) f=g+h; else if(k==2) f=g h; else if(k==3) f=i j; Use this mapping: f:$s0, g:$s1, h:$s2, i:$s3, j:$s4, k:$s5 Example: The C Switch Statement (3/3) Final compiled MIPS code: bne $s5,$0,l1 # branch k!=0 add $s0,$s3,$s4 #k==0 so f=i+j j Exit # end of case so Exit L1: addi $t0,$s5,-1 # $t0=k-1 bne $t0,$0,l2 # branch k!=1 add $s0,$s1,$s2 #k==1 so f=g+h j Exit # end of case so Exit L2: addi $t0,$s5,-2 # $t0=k-2 bne $t0,$0,l3 # branch k!=2 sub $s0,$s1,$s2 #k==2 so f=g-h j Exit # end of case so Exit L3: addi $t0,$s5,-3 # $t0=k-3 bne $t0,$0,exit # branch k!=3 sub $s0,$s3,$s4 # k==3 so f=i-j Exit: CS61C L8 Introduction to MIPS : Decisions II & Procedures I (37) CS61C L8 Introduction to MIPS : Decisions II & Procedures I (38)