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Lecture 6 Sept 16 Chapter 2 continue MIPS

Lecture 6 Sept 16 Chapter 2 continue MIPS translating c into MIPS – examples MIPS simulator (MARS and SPIM). Logical Operations. Instructions for bitwise manipulation. §2.6 Logical Operations. Useful for extracting and inserting groups of bits in a word.

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Lecture 6 Sept 16 Chapter 2 continue MIPS

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  1. Lecture 6 Sept 16 • Chapter 2 • continue MIPS • translating c into MIPS – examples • MIPS simulator (MARS and SPIM)

  2. Logical Operations • Instructions for bitwise manipulation §2.6 Logical Operations • Useful for extracting and inserting groups of bits in a word

  3. op rs rt rd shamt funct 6 bits 5 bits 5 bits 5 bits 5 bits 6 bits 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 (arithmetic) • Shift right and fill with 0 bits (sign bit) • srl, sra by i bits divides by 2i

  4. AND Operations • Useful to mask bits in a word • Select some bits, clear others to 0 and $t0, $t1, $t2 $t2 0000 0000 0000 0000 0000 1101 1100 0000 $t1 0000 0000 0000 0000 0011 1100 0000 0000 $t0 0000 0000 0000 0000 0000 1100 0000 0000

  5. OR, XOR Operations • Useful to include bits in a word • Set some bits to 1, leave others unchanged or $t0, $t1, $t2 $t2 0000 0000 0000 0000 0000 1101 1100 0000 $t1 0000 0000 0000 0000 0011 1100 0000 0000 $t0 0000 0000 0000 0000 0011 1101 1100 0000 • MIPS also has a bitwise exclusive-or instruction • xor $t0, $t1, $t2

  6. NOT Operations • Useful to invert bits in a word • change 0 to 1, and 1 to 0 • MIPS has 3-operand NOR instruction • a NOR b == NOT ( a OR b ) nor $t0, $t1, $zero Register 0: always read as zero $t1 0000 0000 0000 0000 0011 1100 0000 0000 $t0 1111 1111 1111 1111 1100 0011 1111 1111

  7. Initializing a Register Show how each of these bit patterns can be loaded into $s0: 0010 0001 0001 0000 0000 0000 0011 1101 1111 1111 1111 1111 1111 1111 1111 1111 We can use a MIPS instruction load upper immediate. This loads the given 16 bit number as the upper 16 bits of a register, rest filled with 0. lui R, 0x ab98 # copy ab980000 to R.

  8. Initializing a Register Show how each of these bit patterns can be loaded into $s0: 0010 0001 0001 0000 0000 0000 0011 1101 1111 1111 1111 1111 1111 1111 1111 1111 Solution The first bit pattern has the hex representation: 0x2110003d lui $s0,0x2110 # put the upper half in $s0 ori $s0,$s0,0x003d # put the lower half in $s0 Same can be done, with immediate values changed to 0xffff for the second bit pattern. But, the following is simpler and faster: nor $s0,$zero,$zero

  9. Loading data via memory • It is possible to load a value into register by writing it into memory and reading it from there usinglw. • This is done in two steps in MIPS: • Load the memory address using la. • Then use lw to read the content.

  10. 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

  11. Exercise: Write a sequence of MIPS instructions that adds two integers stored in $s1 and $s2, stores the sum (mod 2^32) in register $s3, and sets the register $t0 to 1 (0) if there is an overflow (if there is no overflow). Claim: The sum A + B (where A, B and the result C are 32 bit two’s complement integers) causes overflow if and only if (a) both are of same sign and (b) both A and B are positive (negative) and C < A (C > A). Proof:

  12. 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 ExitElse: sub $s0, $s1, $s2Exit: … Assembler calculates addresses

  13. 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

  14. Branch Instruction Design Why not blt, bge, etc? Hardware for <, ≥, … slower than =, ≠ Combining with branch involves more work per instruction, requiring a slower clock beq and bne are the common case This is a good design compromise

  15. op rs rt constant or address 6 bits 5 bits 5 bits 16 bits Branch Addressing • Branch instructions specify • Opcode, two registers, target address • Most branch targets are near branch • Forward or backward • PC-relative addressing • Target address = PC + offset × 4 • PC already incremented by 4 by this time

  16. 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

  17. Jump Instruction Unconditional jump instruction j verify # go to mem loc named “verify”

  18. op address 26 bits 6 bits Jump Addressing • Jump (j and jal) targets could be anywhere in text segment • Encode full address in instruction • (Pseudo)Direct jump addressing • Target address = PC31…28 : (address × 4)

  19. Examples for Conditional Branching If the branch target is too far to be reachable with a 16-bit offset (rare occurrence), the assembler replaces the branch instructionbeq $s0,$s1,L1with: bne $s1,$s2,L2 # skip jump if (s1)(s2) j L1 # goto L1 if (s1)=(s2) L2: ... Forming if-then constructs; e.g.,if (i == j) x = x + y bne $s1,$s2,endif # branch on ij add $t1,$t1,$t2 # execute the “then” part endif: ... If the condition were (i < j), we would change the first line to: slt $t0,$s1,$s2 # set $t0 to 1 if i<j beq $t0,$0,endif # branch if ($t0)=0; # i.e., i not< j or ij

  20. Compiling if-then-else Statements Show a sequence of MIPS instructions corresponding to: if (i<=j) x = x+1; z = 1; else y = y–1; z = 2*z Solution Similar to the “if-then” statement, but we need instructions for the “else” part and a way of skipping the “else” part after the “then” part. slt $t0,$s2,$s1 # j<i? (inverse condition) bne $t0,$zero,else # if j<i goto else part addi $t1,$t1,1 # begin then part: x = x+1 addi $t3,$zero,1 # z = 1 j endif # skip the else part else: addi $t2,$t2,-1 # begin else part: y = y–1 add $t3,$t3,$t3 # z = z+z endif:...

  21. compiling loop statements • c code: while (save[i] == k) i += 1; • i in $s3, k in $s5, base 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 LoopExit: …

  22. MIPS Instructions Covered So Far op 15 0 0 0 8 10 0 0 0 0 12 13 14 35 43 2 0 1 4 5 fn 32 34 42 36 37 38 39 8 Copy Arithmetic Logic Memory access Control transfer

  23. Finding the Maximum in an array Array A is stored in memory beginning at the address given in $s1. Array length is given in $s2. Find the largest integer in the list and copy it into $t0. Solution Scan the list, holding the largest element identified thus far in $t0.

  24. Finding the Maximum in an array lw $t0,0($s1) # initialize maximum to A[0] addi $t1,$zero,0 # initialize index i to 0 loop: add $t1,$t1,1 # increment index i by 1 beq $t1,$s2,done # if all elements examined, quit sll $t2,$t2,2 # compute 4i in $t2 add $t2,$t2,$s1 # form address of A[i] in $t2 lw $t3,0($t2) # load value of A[i] into $t3 slt $t4,$t0,$t3 # maximum < A[i]? beq $t4,$zero,loop # if not, repeat with no change addi $t0,$t3,0 # if so, A[i] is the new maximum j loop # change completed; now repeat done: # continuation of the program

  25. SPIM – A simulator for MIPS Assembly language • Home page of SPIM: • http://www.cs.wisc.edu/~larus/spim.html • A useful document: • http://www.cs.wisc.edu/HP_AppA.pdf

  26. SPIM simulator windows

  27. Tutorials on SPIM There are many good tutorials on SPIM. Here are two introductory ones: http://www.cs.umd.edu/class/fall2001/cmsc411/projects/spim/ http://users.ece.gatech.edu/~sudha/2030/temp/spim/spim-tutorial.html Please read (watch) them once and refer to them when you need help.

  28. SPIM simulator – how to run? We will implement the code written earlier (finding the max element in an array) using the SPIM simulator. Code is shown below:

  29. Creating the code and the text segments in SPIM • .text • .globl __start • __start: • la $s1, array # initialize array • lw $t0, ($s1) • la $t6, count • lw $s2, ($t6) • la is a pseudo-instruction (load address). • We use this to input the starting address and the size of the array into registers $s1 and $t6.

  30. We use the code for finding the maximum addi $t1,$zero, 0 # initialize index i to 0 loop: add $t1,$t1,1 # increment index i by 1 beq $t1,$s2,done # if all elements examined, quit add $t2,$t1,$t1 # compute 2i in $t2 add $t2,$t2,$t2 # compute 4i in $t2 add $t2,$t2,$s1 # form address of A[i] in $t2 lw $t3,0($t2) # load value of A[i] into $t3 slt $t4,$t0,$t3 # maximum < A[i]? beq $t4,$zero,loop # if not, repeat with no change addi $t0,$t3,0 # if so, A[i] is the new maximum j loop # change completed; now repeat done:

  31. Input to the program – Data Segment .data array: .word 3,4,2,6,12,7,18,26,2,14,19,7,8,12,13 count: .word 15 endl: .asciiz "\n" ans2: .asciiz "\nmax = "

  32. Output the result Input array: 3,4,2,6,12,7,18,26,2,14,19,7,8,12,13

  33. System calls for output la $a0,ans2 li $v0,4 syscall # print "\nmax = " move $a0,$t0 li $v0,1 syscall # print max la $a0,endl # system call to print li $v0,4 # out a newline syscall li $v0,10 syscall # end

  34. Output read from simulators SPIM

  35. Output read from simulators MARS

  36. Procedure Calling Steps required • Place parameters (arguments) in registers • Transfer control to procedure • Acquire storage for procedure • Perform procedure’s operations • Place result in register for caller • Return to place of call §2.8 Supporting Procedures in Computer Hardware

  37. 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)

  38. 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

  39. 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

  40. 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 Result Restore $s0 Return

  41. Non-Leaf Procedures • Procedures that call other procedures are known as non-leaf procedures. • For nested call, caller needs to save on the stack: • Its return address • Any arguments and temporaries needed after the call • Restore from the stack after the call

  42. Non-Leaf Procedure Example • C code: int fact (int n){ if (n < 1) return f; else return n * fact(n-1);} • Argument n in $a0 • Result in $v0

  43. Non-Leaf Procedure Example • MIPS code: fact: addi $sp, $sp, -8 # adjust stack for 2 items sw $ra, 4($sp) # save return address sw $a0, 0($sp) # save argument slti $t0, $a0, 1 # test for n < 1 beq $t0, $zero, L1 addi $v0, $zero, 1 # if so, result is 1 addi $sp, $sp, 8 # pop 2 items from stack jr $ra # and returnL1: addi $a0, $a0, -1 # else decrement n jal fact # recursive call lw $a0, 0($sp) # restore original n lw $ra, 4($sp) # and return address addi $sp, $sp, 8 # pop 2 items from stack mul $v0, $a0, $v0 # multiply to get result jr $ra # and return

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