Post on 20-Dec-2015
transcript
Arithmetic IVCPSC 321
Andreas Klappenecker
Any Questions?
Today’s Menu
DivisionFloating Point Numbers
Recall: Multiplier
ControltestWrite
32 bits
64 bits
Shift rightProduct
Multiplicand
32-bit ALU
Done
1. TestProduct0
1a. Add multiplicand to the left half ofthe product and place the result inthe left half of the Product register
2. Shift the Product register right 1 bit
32nd repetition?
Start
Product0 = 0Product0 = 1
No: < 32 repetitions
Yes: 32 repetitions
The Booth Multiplier
Let’s kick it up a notch!
Booth Multiplication
Current and previous bit
00: middle of run of 0s, no action01: end of a run of 1s, add multiplicand10: beginning of a run of 1s, subtract
mcnd11: middle of string of 1s, no action
Negative Numbers
00102 x 11012 = 1111 10102
0) Mcnd 0010 Prod 0000 1101,0
1) Mcnd 0010 Prod 1110 1101,0 sub
1) Mcnd 0010 Prod 1111 0110,1 >>
2) Mcnd 0010 Prod 0001 0110,1 add
2) Mcnd 0010 Prod 0000 1011,0 >>
3) Mcnd 0010 Prod 1110 1011,0 sub
3) Mcnd 0010 Prod 1111 0101,1 >>
4) Mcnd 0010 Prod 1111 0101,1 nop
4) Mcnd 0010 Prod 1111 1010,1 >>
Exercise
• Work through some examples1. Grade school algorithms2. Booth multiplication• Read Chapter 4• Really!
MIPS Assembly Language
Multiplication of two 32 bit operands
x =
mult $a1, $s1 result is stored inmfhi $v0 special registersmflo $v1 high= 32 MSBs
low = 32 LSBs
high low
MIPS Assembly Language
Integer division: divide a by b a = q b + r
quotient q = floor(a/b), remainder r = a % b
div $a0, $a1mflo $v0 # quotientmfhi $v1 # remainder
r qa b/ =
Exponentiation
Suppose you want to raise an integer b to the nth power:
(...((b*b)*b)...*b) = bn
Way too slow for large n!
Use square and multiply algorithm.
Square and Multiply
Power(x,n) =
x if n=1
Power(x2,n/2) if n even
x*Power(x2,(n-1)/2) if n odd
x6= (x2)3=x2 (x2)2
Division
Recall Decimal Division
Grammar school algorithm93:7= 137 2321 (trial multiplication, 3*7 works, 4*7
doesn’t)
2
-
-
Binary division
• Dividend = Quotient x Divisor + Remainder
• See how big a number can be subtracted
• Create a quotient bit (0 or 1) in each step
• Binary division is simpler than decimal division
Binary Division
00000111: 0010 = 11
-00100
0011
-0010
0001
7 = 3x2 + 1
Division Hardware (Version 1)
rem = rem - div
if rem < 0 then
// divisor too big
rem = rem + div
quo <<= 1
LSB(quo) = 0
else
// can divide
quo <<= 1
LSB(quo) = 1
fi
div >>= 1
repeat unless done
iteration
Step Quot Divisor Remainder
0 initial step 0000 0010 0000 0000 0111
1 rem -= div
rem<0 =>+div, Q<<=1, Q0=0
div >>= 1
0000 0000 0000
0010 0000 0010 0000 0001 0000
1110 0111 0000 0111 0000 0111
2 rem -= div
rem<0 =>+div, Q<<=1, Q0=0
div >>= 1
0000 0000 0000
0001 0000 0001 0000 0000 1000
1111 0111 0000 0111 0000 0111
3 rem -= div
rem<0 =>+div, Q<<=1, Q0=0 div >>= 1
0000 0000
0000
0000 1000 0000 1000 0000 0100
1111 1111 0000 0111 0000 0111
4 rem -= div
rem<0 => Q<<=1, Q0=1
div >>= 1
0001 00010001
0000 0100 0000 0100 0000 0010
0000 0011 0000 0011 0000 0011
5 rem -= div
rem<0 => Q<<=1, Q0=1
div >>= 1
0001 0011 0011
0000 0010 0000 0010 0000 0001
0000 0001 0000 0001 0000 0001
• Half of the bits in divisor always 0• Half of adder and divisor register is wasted
• Instead of shifting divisor to the right, shift remainder to the left• Shift left first to save one iteration
Division hardware (Version 2)
ALU and divisor registers are reduced, remainder shifted left
3b. Restore the original value by adding the Divisor register to the left half of the Remainder register, &place the sum in the left half of the Remainder register. Also shift the Quotient register to the left, setting the new least significant bit to 0.
Test Remainder
Remainder < 0Remainder > 0
2. Subtract the Divisor register from the left half of the Remainder register, & place the result in the left half of the Remainder register.
3a. Shift the Quotient register to the left setting the new rightmost bit to 1.
1. Shift the Remainder register left 1 bit.
Done
Yes: n repetitions (n = 4 here)
nthrepetition?
No: < n repetitions
Start: Place Dividend in Remainder
Observations (Divide Version 2)
•Eliminate quotient register => combine with remainder register
•Start by shifting the remainder left, as before. •After that, the loop will contain only two steps
(because remainder register shifts both remainder and quotient)
•Remainder will now be shifted left one time too many.•Thus, a final correction step must shift back only the
remainder in the left half of the register
rem <<= 1
rem -= (div >> 32)
if rem < 0 then
rem += (div >> 32)
rem <<= 1
LSB(rem) = 0
else
rem <<= 1
LSB(rem) = 1
fi
repeat unless done
Correct remainder
Version 3
Floating Point Numbers
IEEE 754 Floating Point Representation
• Float – 1 sign bit, 8 exponent bits, 23 bits for significand.
seeeeeeeefffffffffffffffffffffffvalue = (-1)s x F x 2E-127
with F= 1 + .ffffffff .... fff
• Double – 1 sign bit, 11 exponent bits, 52 bits for significand
Floating Point Representation: double
• 1 bit sign, 11 bits for exponent, 52 bits for significand
• seeeeeeeeeeeffffffffffffffffffff ffffffffffffffffffffffffffffffff
Range of float: 2.0 x 10-38 … 2.0 x 1038
Range of double: 2.0 x 10-308 … 2.0 x 10308
Integer and Floating Point Arithmetic
instructionmemory
PC
integerregister
file
integerALU
integermultiplier
datamemory
flt ptregister
file
flt ptadder
flt ptmultiplier
HILO
Floating Point Registers in MIPS
• $f0 - $f3 function-returned values• $f4 - $f11 temporary values• $f12 - $f15 function arguments• $f16 - $f19 temporary values• $f20 - $f31 saved valuesInterpretation: 32 registers of 32 bits or
16 registers of 64 bits (even indices)
Floating Point in MIPS
.data
Pi: .double 3.14159
Rad: .double 12.34567
.text
main: l.d $f0, Pi # $f0 = Pi; load double
l.d $f4, Rad # $f4 = radius
mul.d $f12, $f4, $f4 # $f12 = radius squared
mul.d $f12, $f12, $f0 # multiply by Pi
li $v0, 3 # print double $f12
syscall # (= print the area)
li $v0, 10 #
syscall # exit
Conclusion
• We learned how to divide• Make sure that you work through some
examples• Read the section on floating point addition and
multiplication in Chapter 4 (What are the challenges here?)• To probe further: Knuth’s “Art of Computer Programming”
contains numerous MIX assembly language programs; you can get involved into the translation to MMIX.