Basic Arithmetic and the ALU. Basic Arithmetic and the ALU. Background. Background. Integer multiplication, division

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1 Basic Arithmetic and the ALU Forecast Representing numbs, 2 s Complement, unsigned ition and subtraction /Sub ALU full add, ripple carry, subtraction, togeth Carry-Lookahead addition, etc. Logical opations and, or, xor, nor, shifts - barrel shift Ovflow, MMX Basic Arithmetic and the ALU Integ multiplication, division floating point arithmetic lat not crucial for the project CS/ECE 552 Lecture Notes: Chapt 4 1 CS/ECE 552 Lecture Notes: Chapt 4 2 Background Background Recall: n bits give rise to 2 n combinations let us call a string of 32 bits as b31 b30... b3 b2 b1 b0 No inhent meaning one intpretation f(b31... b4 b3 b2 b1 b0) -> value anoth f(b31... b4 b3 b2 b1 b0) -> control signals 32-bit types include unsigned integs singed integs single-precision floating point MIPS instructions (A.10) CS/ECE 552 Lecture Notes: Chapt 4 3 CS/ECE 552 Lecture Notes: Chapt 4 4

2 Unsigned integs Signed Integs f(b31... b0) = b31 x b1 x 2 + b0 x 2 0 Treat as normal binary numb e.g., =1x2 7 +1x2 6 +0x2 5 +1x2 4 +0x2 3 +1x2 2 +0x2 1 +1x2 0 = = 213 max f ( ) = = 4, 294, 967, 295 min f( ) = 0 2 s Complement f(b31 b30... b1 b0) = -b31 x b1 x 2 + b0 x 2 0 max f( ) = = min f( ) = = (asymmetric) range [-2 31, ] => #values ( ) = 2 32 E.g., > > range [0, ] => # values (2 32-1) = 2 32 CS/ECE 552 Lecture Notes: Chapt 4 5 CS/ECE 552 Lecture Notes: Chapt 4 6 Why 2 s Complement ition and Subtraction why not use signed magnitude 2 s complement makes comput arithmetic simpl just like humans don t work with Roman Numals Representation affects ease of calculation 4-bit unsigned example not answ bit 2 s Complement - ignoring ovflow CS/ECE 552 Lecture Notes: Chapt 4 7 CS/ECE 552 Lecture Notes: Chapt 4 8

3 Subtraction A - B = A + 2 s complement of B E.g., full add (a, b, c in )--> (c out, s) c out = two of more of (a, b, c in ) s = exactly one or three of (a, b, c in ) CS/ECE 552 Lecture Notes: Chapt 4 9 CS/ECE 552 Lecture Notes: Chapt 4 10 Ripple-carry Ripple-carry Subtractor Just concatenate the full adds A - B = A + (-B) => invt B and set C in0 to 1 c in C out 1 C out a 0 b 0 a 1 b 1 a 2 b 2 a 31b 31 a 0 b 0 a 1 b 1 a 2 b 2 a 3 b 3 CS/ECE 552 Lecture Notes: Chapt 4 11 CS/ECE 552 Lecture Notes: Chapt 4 12

4 Combined Ripple-carry /Subtractor Carry Lookahead control = 1 => subtract XOR B with control and set C in0 to control a 0 b 0 a 1 b 1 a 2 b 2 C out a 31 b 31 opation The above ALU is too slow - gate delays for add = 32 x FA + XOR ~= 64 - too slow Theoretically: In parallel sum 0 = f(c in, a 0, b 0 ) sum i = f(c in, a i... a 0, b i... b 0 ) sum 31 = f(c in, a a 0, b b 0 ) CS/ECE 552 Lecture Notes: Chapt 4 13 CS/ECE 552 Lecture Notes: Chapt 4 14 Carry Lookahead Carry Lookahead Need compromise build tree so delay is O(log 2 n) for n bits E.g., 2 x 5 gate delays for 32-bits Need both to genate and at least one to propagate Define: g i = a i * b i ## carry genate We will give the basic idea with (a) 4-bit then (b) 16-bit add p i = a i + b i ## carry propagate Recall: c i+1 = a i * b i + a i * c i + b i * c i A little convoluted! = a i * b i + (a i + b i ) * c i = g i + p i * c i CS/ECE 552 Lecture Notes: Chapt 4 15 CS/ECE 552 Lecture Notes: Chapt 4 16

5 Carry Lookahead 4-bit Carry Lookahead Thefore c 1 = g 0 + p 0 * c 2 = g 1 + p 1 * c 1 = g 1 + p 1 * (g 0 + p 0 * ) c 4 Carry Lookahead Block = g 1 + p 1 * g 0 + p 1 * p 0 * g 3 p 3 a 3 b 3 g 2 p 2 a 2 b 2 g 1 p 1 a 1 b 1 g 0 p 0 a 0 b 0 c 3 = g 2 + p 2 * g 1 + p 2 * p 1 * g 0 + p 2 * p 1 * p 0 * c 4 = g 3 + p 3 *g 2 + p 3 *p 2 *g 1 + p 3 *p 2 *p 1 *g 0 + p 3 *p 2 *p 1 *p 0 * c 3 c 2 c 1 Uses one level to form p i and g i, two levels for carry But, this needs n+1 fanin at the OR and the rightmost AND s 3 s 2 s 1 s 0 CS/ECE 552 Lecture Notes: Chapt 4 17 CS/ECE 552 Lecture Notes: Chapt 4 18 Hiachical Carry Lookahead for 16 bits Hiachical Carry Lookahead for 16 bits Build 16-bit add from four 4-bit adds c 15 Carry Lookahead Block Figure out Genate and Propagate for 4-bits togeth G 0,3 = g 3 + p 3 * g 2 + p 3 * p 2 * g 1 + p 3 * p 2 * p 1 * g 0 P G a,b P G a,b 8-11 P G a 4-7 b 4-7 P G a 0-3 b 0-3 P 0,3 = p 3 * p 2 * p 1 * p 0 (Notation a little diffent from the book) G 4,7 = g 7 + p 7 * g 6 + p 7 * p 6 * g 5 + p 7 * p 6 * p 5 * g 4 c 12 c 8 c 4 P 4,7 = p 7 * p 6 * p 5 * p 4 G 12,15 = g 15 + p 15 * g 14 + p 15 * p 14 * g 13 + p 15 * p 14 * p 13 * g 12 s s 8-11 s 4-7 s 0-3 P 12,15 = p 15 * p 14 * p 13 * p 12 CS/ECE 552 Lecture Notes: Chapt 4 19 CS/ECE 552 Lecture Notes: Chapt 4 20

6 Carry Lookahead Basics Carry Lookahead: Compute G s and P s Fill in the holes in G s and P s: G i, k = G j+1,k + P j+1, k * G i,j (assume i < j +1 < k ) P i,k = P i,j * P j+1, k G 12,15 P 12,15 G 8,11 P 8,11 G 4,7 P 4,7 G 0,3 P 0,3 G 0,7 = G 4,7 + P 4,7 * G 0,3 P 0,7 = P 0,3 * P 4,7 G 8,15 = G 12,15 + P 12,15 * G 8,11 P 8,15 = P 8,11 * P 12, 15 G 0,15 = G 8,15 + P 8,15 * G 0,7 P 0,15 = P 0,7 * P 8, 15 G 8,15 P 8,15 G 0,7 P 0,7 G 0,15 P 0,15 CS/ECE 552 Lecture Notes: Chapt 4 21 CS/ECE 552 Lecture Notes: Chapt 4 22 c 12 Carry Lookahead: Compute c s g 12 - g 15 g 8 - g 11 g 4 - g 7 g 0 - g 3 p 12 - p 15 p 8 - p 11 p 4 - p 7 p 0 - p 3 c4 c0 G 8,11 P 8,11 c 8 G 0,3 P 0,3 Oth s: Carry Select Two adds in parallel - one with C in 0 and the oth c in 1 When C in is done, select the right result 0 1 c 8 G 0,7 P 0,7 next select 2-1 Mux CS/ECE 552 Lecture Notes: Chapt 4 24 select CS/ECE 552 Lecture Notes: Chapt 4 23

7 Oth s: Carry Save Wallace Tree A + B -> S f e d c b a Save carries A + B -> S, C out CSA CSA Use C in A + B + C -> S1, S2 (3# to 2# in parallel) Used in combinational multiplis by building a Wallace Tree c b a CSA CSA CSA c s CS/ECE 552 Lecture Notes: Chapt 4 25 CS/ECE 552 Lecture Notes: Chapt 4 26 Logical Opations Shift Bitwise AND, OR, XOR, NOR Implement with 32 gates in parallel Shifts and rotates rol -> rotate left (MSB --> LSB) ror -> rotate right (LSB --> MSB) sll -> shift left logical (0 --> LSB) srl -> shift right logical (0 --> MSB) srl -> shift right arithmetic (old MSB --> new MSB) E.g., Shift left logical for d<7:0> and shamt<2:0> Using 2-1 Muxes called Mux(select, in 0, in 1 ) stage0<7:0> = Mux(shamt<0>,d<7:0>, 0 d<7:1>) stage1<7:0> = Mux(shamt<1>, stage0<7:0>, 00 stage0<6:2>) dout<7:0) = Mux(shamt<2>, stage1<7:0>, 0000 stage1<3:0>) For Barrel shift used wid muxes CS/ECE 552 Lecture Notes: Chapt 4 27 CS/ECE 552 Lecture Notes: Chapt 4 28

8 Shift All Togeth Mux d 0 0 d 7 d 6 d 1 d 0 shift based on 0 th bit by 0 or 1 stage0 s0 7 s0 0 s0 7 s0 6 s0 0 s0 2 s0 1 0 s0 0 0 Mux shift based on 1 st bit by 0 or 2 stage1 s1 7 s1 0 s1 7 s1 3 s1 4 s1 0 s1 3 0 s1 0 0 Mux dout shift based on 2 nd bit by 0 or 4 shamt0 shamt1 shamt2 a b invt Mux carry in opation Mux result dout 7 dout 7 CS/ECE 552 Lecture Notes: Chapt 4 29 CS/ECE 552 Lecture Notes: Chapt 4 30 Ovflow Ovflow with n-bits only 2 n combinations Unsigned [0, 2 n -1], 2 s Complement [-2 n-1, 2 n-1-1] Unsigned > More involved for 2 s Complement = = -2 is correct => can t just use carry-out 1011 f(3:0) = a(2:0) + b(2:0) => ovflow = f(3) ;; carryout CS/ECE 552 Lecture Notes: Chapt 4 31 CS/ECE 552 Lecture Notes: Chapt 4 32

9 ition Ovflow When is ovflow NOT possible? p1, p2 > 0 and n1, n2 < 0 p1 + p2 p1 + n1 not possible n1 + p2 not possible n1 + n2 ovflow = X * a(2) * b(2) + Y * a(2) * b(2) What are X and Y? ition Ovflow = 5 > = -3 < 0! In genal, X = f(2) which is 011 > 0 In genal Y = f(2) Ovflow = f(2) * a(2) * b(2) + f(2) * a(2) * b(2) CS/ECE 552 Lecture Notes: Chapt 4 33 CS/ECE 552 Lecture Notes: Chapt 4 34 Subtraction Ovflow What to do on ovflow No ovflow on a-b if signs are same neg - pos ==> neg ;; ovflow othwise pos - neg ==> pos ;; ovflow othwise ovflow = f(2) * (a2) * b(2) + f(2) * a(2) * (b2) Ignore! Flag - condition code that may be tested by software sticky flag - e.g., for floating point trap - possibly with mask CS/ECE 552 Lecture Notes: Chapt 4 35 CS/ECE 552 Lecture Notes: Chapt 4 36

10 zo = f(2) + f(1) + f(0) can t also look at f(3) because So, negative = f(2) Zo and Negative Zo and Negative May also want correct answ even on ovflow negative = (a < b) = (a-b < 0) even if ovflow E.g., is -4 < 2? => ovflow If you work it out, negative = f(2) XOR ovflow CS/ECE 552 Lecture Notes: Chapt 4 37 CS/ECE 552 Lecture Notes: Chapt 4 38 MMX MMX, cont. MMX [Peleg & Weis, IEEE Micro, Aug. 96] Goal 2x pformance in audio, video, etc. Key technique: SIMD - single instruction multiple data 1999 Streaming SIMD Extensions in same spirit Data types 1 x 64 bit quad word 2 x 32 bit double-word 4 x 16 bit word 8 x 8 bit byte E.g., ADDB (for byte) more more == 300 mod 256 or 255 == maximum value E.g., 16 element dot product from matrix multiply [a1... a16] x [b1... b16] = a1*b a16*b16 IA-32: 32 loads, 16 *, 15 +, 12 loop control = 76 instr. MMX: 16 instr. Cycles 200 for int, 76 for FP, & 12 for MMX (6x ov FP!) CS/ECE 552 Lecture Notes: Chapt 4 39 CS/ECE 552 Lecture Notes: Chapt 4 40

11 MMX, cont. Oths: MOV, (UN)PACK, & MASK (e.g., next) FF FF FF FF Why? Weathpson at 00 s & weathmap at FF s Comments Backward compatible & no OS changes (ovload FP regs) Oths have similar: Sun, HP, and now Intel SSE ISVs (i.e., for games) have not (yet) embraced CS/ECE 552 Lecture Notes: Chapt 4 41

Lecture Topics. Announcements. Today: Integer Arithmetic (P&H ) Next: continued. Consulting hours. Introduction to Sim. Milestone #1 (due 1/26)

Lecture Topics. Announcements. Today: Integer Arithmetic (P&H ) Next: continued. Consulting hours. Introduction to Sim. Milestone #1 (due 1/26) Lecture Topics Today: Integer Arithmetic (P&H 3.1-3.4) Next: continued 1 Announcements Consulting hours Introduction to Sim Milestone #1 (due 1/26) 2 1 Overview: Integer Operations Internal representation