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1、Patterson-1610874 978-0-12-407726-3PII Solutions 2 Chapter 2 Solutions S-3 2.1 addi f, h, -5 (note, no subi) add f, f, g 2.2 f = g + h + i 2.3 sub $t0, $s3, $s4 add $t0, $s6, $t0 lw $t1, 16($t0) sw $t1, 32($s7) 2.4 Bg = Af + A1+f; 2.5 add $t0, $s6, $s0 add $t1, $s7, $s1 lw $s0, 0($t0) lw $t0, 4($t0)
2、 add $t0, $t0, $s0 sw $t0, 0($t1) 2.6 2.6.1 temp = Array0; temp2 = Array1; Array0 = Array4; Array1 = temp; Array4 = Array3; Array3 = temp2; 2.6.2 lw $t0, 0($s6) lw $t1, 4($s6) lw $t2, 16($s6) sw $t2, 0($s6) sw $t0, 4($s6) lw $t0, 12($s6) sw $t0, 16($s6) sw $t1, 12($s6) S-4 Chapter 2 Solutions 2.7 Li
3、ttle-EndianBig-Endian AddressDataAddressData 12ab1212 8cd 8ef 4ef 4cd 012 0ab 2.8 2882400018 2.9 sll $t0, $s1, 2 # $t0 0) 2.26.3 5*N 2.27 addi $t0, $0, 0 beq $0, $0, TEST1 LOOP1: addi $t1, $0, 0 beq $0, $0, TEST2 LOOP2: add $t3, $t0, $t1 sll $t2, $t1, 4 add $t2, $t2, $s2 sw $t3, ($t2) addi $t1, $t1,
4、 1 TEST2: slt $t2, $t1, $s1 bne $t2, $0, LOOP2 addi $t0, $t0, 1 TEST1: slt $t2, $t0, $s0 bne $t2, $0, LOOP1 2.28 14 instructions to implement and 158 instructions executed 2.29 for (i=0; i0, test if n=1 add $v0, $0, $0 # else fib(0) = 0 j rtn # test2: addi $t0, $0, 1 # bne $t0, $a0, gen # if n1, gen
5、 add $v0, $0, $t0 # else fib(1) = 1 j rtn gen: subi $a0, $a0,1 # n-1 jal fib # call fib(n-1) add $s0, $v0, $0 # copy fib(n-1) sub $a0, $a0,1 # n-2 jal fib # call fib(n-2) add $v0, $v0, $s0 # fib(n-1)+fib(n-2) rtn: lw $a0, 0($sp) # pop $a0 lw $s0, 4($sp) # pop $s0 lw $ra, 8($sp) # pop $ra addi $sp, $
6、sp, 12 # restore sp jr $ra # fib(0) = 12 instructions, fib(1) = 14 instructions, # fib(N) = 26 + 18N instructions for N =2 2.32 Due to the recursive nature of the code, it is not possible for the compiler to in-line the function call. 2.33 after calling function fib: old $sp - 0x7ffffffc ? -4 conten
7、ts of register $ra for fib(N) -8 contents of register $s0 for fib(N) $sp- -12 contents of register $a0 for fib(N) there will be N-1 copies of $ra, $s0 and $a0 S-8 Chapter 2 Solutions 2.34 f: addi $sp,$sp,-12 sw $ra,8($sp) sw $s1,4($sp) sw $s0,0($sp) move $s1,$a2 move $s0,$a3 jal func move $a0,$v0 ad
8、d $a1,$s0,$s1 jal func lw $ra,8($sp) lw $s1,4($sp) lw $s0,0($sp) addi $sp,$sp,12 jr $ra 2.35 We can use the tail-call optimization for the second call to func, but then we must restore $ra, $s0, $s1, and $sp before that call. We save only one instruction (jr $ra). 2.36 Register $ra is equal to the r
9、eturn address in the caller function, registers $sp and $s3 have the same values they had when function f was called, and register $t5 can have an arbitrary value. For register $t5, note that although our function f does not modify it, function func is allowed to modify it so we cannot assume anythi
10、ng about the of $t5 aft er function func has been called. 2.37 MAIN: addi $sp, $sp, -4 sw $ra, ($sp) add $t6, $0, 0x30 # 0 add $t7, $0, 0x39 # 9 add $s0, $0, $0 add $t0, $a0, $0 LOOP: lb $t1, ($t0) slt $t2, $t1, $t6 bne $t2, $0, DONE slt $t2, $t7, $t1 bne $t2, $0, DONE sub $t1, $t1, $t6 beq $s0, $0,
11、 FIRST mul $s0, $s0, 10 FIRST: add $s0, $s0, $t1 addi $t0, $t0, 1 j LOOP Chapter 2 Solutions S-9 DONE: add $v0, $s0, $0 lw $ra, ($sp) addi $sp, $sp, 4 jr $ra 2.38 0x00000011 2.39 Generally, all solutions are similar: lui $t1, top_16_bits ori $t1, $t1, bottom_16_bits 2.40 No, jump can go up to 0x0FFF
12、FFFC. 2.41 No, range is 0x604 + 0x1FFFC = 0x0002 0600 to 0x604 0x20000 = 0xFFFE 0604. 2.42 Yes, range is 0x1FFFF004 + 0x1FFFC = 0x2001F000 to 0x1FFFF004 - 0x20000 = 1FFDF004 2.43 trylk: li $t1,1 ll $t0,0($a0) bnez $t0,trylk sc $t1,0($a0) beqz $t1,trylk lw $t2,0($a1) slt $t3,$t2,$a2 bnez $t3,skip sw
13、$a2,0($a1) skip: sw $0,0($a0) 2.44 try: ll $t0,0($a1) slt $t1,$t0,$a2 bnez $t1,skip mov $t0,$a2 sc $t0,0($a1) beqz $t0,try skip: 2.45 It is possible for one or both processors to complete this code without ever reaching the SC instruction. If only one executes SC, it completes successfully. If both
14、reach SC, they do so in the same cycle, but one SC completes fi rst and then the other detects this and fails. S-10 Chapter 2 Solutions 2.46 2.46.1 Answer is no in all cases. Slows down the computer. CCT ? clock cycle time ICa ? instruction count (arithmetic) ICls ? instruction count (load/store) IC
15、b ? instruction count (branch) new CPU time ? 0.75*old ICa*CPIa*1.1*oldCCT ? oldICls*CPIls*1.1*oldCCT ? oldICb*CPIb*1.1*oldCCT Th e extra clock cycle time adds suffi ciently to the new CPU time such that it is not quicker than the old execution time in all cases. 2.46.2 107.04%, 113.43% 2.47 2.47.1 2.6 2.47.2 0.88 2.47.3 0.533333333
