part1

1.1-1.26
qufac dikjo wpnry slgxz tbhme v
1.27-1.44
jbfji edgcj dcjbf haa
3.1、1: t0=0
2: t1=1
3: if (a0<t1) t2=1; else t2=0
4: if (t2= =0) goto finish
5: t0=t0+t1
6: t1=t1+2
7: jump to loop
8: v0=t0
3.2、Find the element which appears most times, $v0 stores the the maxium number of the
identical integer in the array, $v1 stores its value.
3.4、addi $t0,$t1,100
3.5、lui $t1,61
addi $t1,$t1,2304
lw $t2,11($t1)
add $t2,$t2,$t0
sw $t2,10($t1)
3.6、Loop: lw $v1, 0($a0)
sw $v1, 0($a1)
addi $a0, $a0, 4
addi $a1, $a1, 4
beq $v1, $zero, End
addi $v0, $v0, 1
j Loop
End:
3.7、1: I format,
3: I format,
4: I format,
5: I format,
6: I format,
3.8、v0=0;
v1=a0;
v0++;
a1=v1;
a0++;
a1++;
if(v1= =0) goto loop;
v0--;
3.9、bne $t0,$s5,exit
loop: add $t1,$s3,$s3
add $t1,$t1,$t1
add $s3,$s3,$s6
lw $t0,0(t1)
beq $t0,$s5,loop exit:
3.11、slti $t5,$t0,1
bne $t0,1,exit
loop: addi $t0,$zero,0
add $t4,$t0,$t0
add $t4,$t4,$t4
lw $t1,0($a0)
add $t2,$t4,$t1
add $t3,$a1,$t1
sw $t2,0($t3)
addi $t0,$t0,1
slti $t5,$t0,101
beq $t5,1,loop
exit:
3.23、loop: addi $t0,$zero,98
lb $t1,0($a0)
addi $a0,$a0,1
beq $t1,$zero,exit
bne $t1,$t0,loop
beq $t1,$t0,next
sub $v0,$a0,1
exit: add $v0,$zero,$zero
next:
3.24、bcount: addi $sp,$sp,-12
sw $ra,8($sp)
sw $a0,4($sp)
sw $s0,0($sp)
add $s0,$zero,$zero
Next: jal bfind
add $a0,$v0,$zero
lb $t0,0($a0)
beq $t0,$zero,End #encounter null
addi $s0,$s0,1
addi $v0,$v0,1
j Next
End: add $v0,$s0,$zero
lw $s0,0($sp)
lw $a0,4($sp)
lw $ra,8($sp)
addi $sp,$sp,12
jr $ra
4.1、00000000000000000000001000000000
4.2、11111111111111111111110000000001
4.4、-500
4.5、-1
4.6、2147483647
4.10、loop: sub $t3,$zero,$t3
slt $t0,$t3,$zero
beq $t0,$zero,loop
add $t2,$t3,$zero how to use 3 instructions?
4.13、If A_lower is positive, A_upper_adjusted equals to A_upper.
If A_lower is negative, A_upper_adjusted equals to A_upper+1.
4.17、addu $t0, $t3,$t4
sltu $t2,$t0,$t3
4.22、srl $t0,$s0,8
andi $s1,$t0,4095
4.23、
when it overflew , set should be converted .
4.40、xor $s0,$s0,$s1
xor $s1,$s1,$s0
xor $s0,$s0,$s1
4.46、ripple carry c1=a0*c0+b0*c0+a0*b0 2T
c2=a1*c1+b1*c1+a1*b1 4T
c3=a2*c2+b2*c2+a2*b2 6T
The worst case : c4=a3*c3+b3*c3+a3*b3 8T
Carry look ahead : gi=ai*bi T pi=ai+bi T
c1=c2=c3=c4=3T
4.53、
4.54、
step1 6.42*101=0.642*102
step2 9.51*102+0.642*102=10.152*102
step3 10.152*102=1.0152*102
step4 1.0152*103 is rounded to three significant digits 1.015*103
4.55、6.42X101=0.642X102
0.642+9.51=10.152
6.42X101+9.51X102=10.152X102=1.0152X103
so the answer is 1.02X103
5.1、
RegDst=0 lw sw and branch
AluSrc=0 R-format beq
MemtoReg=0 R-format sw and beq
Zero=0 R-format sw and lw
5.2、RegDst=1 R-format sw and beq
AluSrc=1 lw and sw
MemtoReg=1 lw sw and beq
Zero=1 R-format lw and sw。