Transcript Answers to Test 1, Question 1

Answers to Test 2, Question 1
• The functions receives the radius and height of a cylinder
(‫ )גליל‬and returns its volume (‫)נפח‬.
const float PI=3.14
float cyl_vol(float r, float h)
{
return(r*r*h*PI);
}
1
(CarryIn (0 if addition 1 if subtraction
Question 2
Operation
a
b
0
Result
1
CarryIn
Operation
a
b
CarryOut
0
Result
1
CarryOut
2
Question 3
•
•
•
•
ETA = 324M * 2.5 * 2ns = 1620M ns = 1.62 sec
ETB = 290M *3.2 *(1/550M)ns = 1687M ns = 1.687s
A is faster than B by 4% (1687/1620)
Computer A: FE=1.00 (all the instructions are
effected), SE = 2/1.8 = 1.11.
(1.00/1.11 + 1 -1)*1.62 = 1.458 sec
• Computer B: FE=0.50 , SE=3.2/2.1 = 1.52
(0.50/1.52 + 0.50)*1.687 = 1.398 sec
Now B is faster than A by 4%
3
Question 4
• mov $t0,$t1 -> add $t0,$t1,$zero
• bgt $t0,$t1,L1 -> slt $t2,$t1,$t0
bne $t2,$zero,L1
• seq $t0,$s1,$s2 ->
bne $s1,$s2,L1
addi $t0,$zero,1
j L2
L1: addi $t0,$zero,0
L2:
• div $t0,$t1,$t2 -> div $t1,$t2
mflo $t0
4
Question 5
• Register $s0 contains
0111 1111 1111 1111 1111 1111 1111 1111
Register $s1 contains
1000 0000 0000 0000 0000 0000 0000 0001
slt $t0,$s0,$s1 compares 2 signed numbers.
$s0 contains 231-1 and $s1 contains -231-1. 231-1 > 231-1 so $t0 equals 0.
sltu $t1,$s0,$s1 compares 2 unsigned numbers.
$s0 contains 231-1 and $s1 contains 231+1 . 232-1 <
231+1 so $t1 equals 1.
• 0x80000000 = 2147483648 so both $t0 and $t1
will contain 0.
5
Question 6
• 16KB=214 , thus 14 address lines are needed. The
width of the SRAM is 4 bits, so 4 input lines and 4
output lines are needed.
• 2MB = 221, thus 21 address lines, 1 data input line,
and 1 data output line are needed.
• To read 4 words takes 1 + 4*12 + 4*2 = 57 cycles
• If memory is interleaved in 2 banks we can read
each 2 words in parallel, although we will have to
still send them one at a time. So to read 4 words
now takes 1 + 2*12 + 4*2 = 33 cycles.
6
Question 7
sum:
subi $sp,$sp,8 # make room for 2 items
sw
$ra,4($sp)# push the return address
sw
$a0,0($sp)# push the argument n
slt $t0,$a0,1 # test for n<1
beq $t0,$zero,L1 # if n>=1 goto L1
li
$v0,0 # pseudoinstruction $v0=0
addi $sp,$sp,8 # pop 2 items off stack
jr
$ra
The following is the recursive call to sum(n-1)
L1:
subi $a0,$a0,1 # n-jal sum # call sum(n-1)
lw
$a0,0($sp) # return from fact(n-1)
lw
$ra,4($sp) # pop n and return address
addi $sp,$sp,8 # pop 2 items off stack
add $v0,$a0,$v0 # return n + sum(n-1)
jr
$ra
7
Question 8
• What is reduced is the number of bits for the
opcode, this reduces the number of possible
instructions.
• RISC processors have instructions that are the
same length (1 word) as opposed to GPRs that
have variable length instructions.
RISC processors have 3 operand instructions as
opposed to 2 operand instructions.
RISC processors access memory only through
Load/Store instructions, as opposed to GPRs
where all instructions (even ALU instructions) can
have memory operands.
• RISC programs might have more instructions. 8