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Date: Tue, 05 Nov 1996 20:50:14 GMTServer: NCSA/1.5Content-type: text/htmlLast-modified: Wed, 30 Aug 1995 21:21:34 GMTContent-length: 9458<html><head><title> Lecture notes - Chapter 5 - Integer Arithmetic</title></head><BODY><h1> Chapter 5 -- integer arithmetic</h1><pre>all about integer arithmetic.-------------------------------------operations we'll get to know (and love): addition subtraction multiplication division logical operations (not, and, or, nand, nor, xor, xnor) shiftingthe rules for doing the arithmetic operations vary dependingon what representation is implied.A LITTLE BIT ON ADDING---------------------- an overview. carry in a b | sum carry out ---------------+---------------- 0 0 0 | 0 0 0 0 1 | 1 0 0 1 0 | 1 0 0 1 1 | 0 1 1 0 0 | 1 0 1 0 1 | 0 1 1 1 0 | 0 1 1 1 1 | 1 1 | a 0011 +b +0001 -- ----- sum 0100 its really just like we do for decimal! 0 + 0 = 0 1 + 0 = 1 1 + 1 = 2 which is 10 in binary, sum is 0 and carry the 1. 1 + 1 + 1 = 3 sum is 1, and carry a 1. ADDITION--------unsigned: just like the simple addition given. examples: 100001 00001010 (10) +011101 +00001110 (14) ------- --------- 111110 00011000 (24) ignore (throw away) carry out of the msb.sign magnitude: rules: - add magnitudes only (do not carry into the sign bit) - throw away any carry out of the msb of the magnitude - add only integers of like sign ( + to + or - to -) - sign of result is same as sign of the addends examples: 0 0101 (5) 1 1010 (-10) + 0 0011 (3) + 1 0011 (-3) --------- --------- 0 1000 (8) 1 1101 (-13) 0 01011 (11) + 1 01110 (-14) ---------------- don't add! must do subtraction! one's complement: by example 00111 (7) 111110 (-1) 11110 (-1) + 00101 (5) + 000010 (2) + 11100 (-3) ----------- ------------ ------------ 01100 (12) 1 000000 (0) wrong! 1 11010 (-5) wrong! + 1 + 1 ---------- ---------- 000001 (1) right! 11011 (-4) right! so it seems that if there is a carry out of the msb, then the result will be off by 1, so add 1 again to get the correct result. (Implementation in HW called an "end around carrry.")two's complement: rules: - just add all the bits - throw away any carry out of the msb - (same as for unsigned!) examples 00011 (3) 101000 111111 (-1) + 11100 (-4) + 010000 + 001000 (8) ------------ -------- -------- 11111 (-1) 111000 1 000111 (7)after seeing examples for all these representations, you may seewhy 2's complement addition requires simpler hardware thansign mag. or one's complement addition. SUBTRACTION----------- general rules: 1 - 1 = 0 0 - 0 = 0 1 - 0 = 1 10 - 1 = 1 0 - 1 = borrow!unsigned: - it only makes sense to subtract a smaller number from a larger one examples 1011 (11) must borrow - 0111 (7) ------------ 0100 (4)sign magnitude: - if the signs are the same, then do subtraction - if signs are different, then change the problem to addition - compare magnitudes, then subtract smaller from larger - if the order is switched, then switch the sign too. - when the integers are of the opposite sign, then do a - b becomes a + (-b) a + b becomes a - (-b) examples 0 00111 (7) 1 11000 (-24) - 0 11000 (24) - 1 00010 (-2) -------------- -------------- 1 10110 (-22)do 0 11000 (24) - 0 00111 (7) -------------- 1 10001 (-17) (switch sign since the order of the subtraction was reversed) one's complement: figure it out on your owntwo's complement: - change the problem to addition! a - b becomes a + (-b) - so, get the additive inverse of b, and do addition. examples 10110 (-10) - 00011 (3) --> 00011 ------------ | \|/ 11100 + 1 ------- 11101 (-3) so do 10110 (-10) + 11101 (-3) ------------ 1 10011 (-13) (throw away carry out) OVERFLOW DETECTION IN ADDITION unsigned -- when there is a carry out of the msb 1000 (8) +1001 (9) ----- 1 0001 (1) sign magnitude -- when there is a carry out of the msb of the magnitude 1 1000 (-8) + 1 1001 (-9) ----- 1 0001 (-1) (carry out of msb of magnitude) 2's complement -- when the signs of the addends are the same, and the sign of the result is different 0011 (3) + 0110 (6) ---------- 1001 (-7) (note that a correct answer would be 9, but 9 cannot be represented in 4 bit 2's complement) a detail -- you will never get overflow when adding 2 numbers of opposite signsOVERFLOW DETECTION IN SUBTRACTION unsigned -- never sign magnitude -- never happen when doing subtraction 2's complement -- we never do subtraction, so use the addition rule on the addition operation done. MULTIPLICATION of integers 0 x 0 = 0 0 x 1 = 0 1 x 0 = 0 1 x 1 = 1 -- longhand, it looks just like decimal -- the result can require 2x as many bits as the larger multiplicand -- in 2's complement, to always get the right answer without thinking about the problem, sign extend both multiplicands to 2x as many bits (as the larger). Then take the correct number of result bits from the least significant portion of the result. 2's complement example: 1111 1111 -1 x 1111 1001 x -7 ---------------- ------ 11111111 7 00000000 00000000 11111111 11111111 11111111 11111111 + 11111111 ---------------- 1 00000000111 -------- (correct answer underlined) 0011 (3) 0000 0011 (3) x 1011 (-5) x 1111 1011 (-5) ------ ----------- 0011 00000011 0011 00000011 0000 00000000 + 0011 00000011 --------- 00000011 0100001 00000011 not -15 in any 00000011 representation! + 00000011 ------------------ 1011110001 take the least significant 8 bits 11110001 = -15 DIVISION of integers unsigned only! example of 15 / 3 1111 / 011LOGICAL OPERATIONS done bitwise X = 0011 Y = 1010 X AND Y is 0010 X OR Y is 1011 X NOR Y is 0100 X XOR Y is 1001 etc.SHIFT OPERATIONS a way of moving bits around within a word 3 types: logical, arithmetic and rotate (each type can go either left or right) logical left - move bits to the left, same order - throw away the bit that pops off the msb - introduce a 0 into the lsb 00110101 01101010 (logically left shifted by 1 bit) logical right - move bits to the right, same order - throw away the bit that pops off the lsb - introduce a 0 into the msb 00110101 00011010 (logically right shifted by 1 bit) arithmetic left - move bits to the left, same order - throw away the bit that pops off the msb - introduce a 0 into the lsb - SAME AS LOGICAL LEFT SHIFT! arithmetic right - move bits to the right, same order - throw away the bit that pops off the lsb - reproduce the original msb into the new msb - another way of thinking about it: shift the bits, and then do sign extension 00110101 -> 00011010 (arithmetically right shifted by 1 bit) 1100 -> 1110 (arithmetically right shifted by 1 bit) rotate left - move bits to the left, same order - put the bit that pops off the msb into the lsb, so no bits are thrown away or lost. 00110101 -> 01101010 (rotated left by 1 place) 1100 -> 1001 (rotated left by 1 place) rotate right - move bits to the right, same order - put the bit that pops off the lsb into the msb, so no bits are thrown away or lost. 00110101 -> 10011010 (rotated right by 1 place) 1100 -> 0110 (rotated right by 1 place)SAL INSTRUCTIONS FOR LOGICAL AND SHIFT OPERATIONS SAL has instructions that do bitwise logical operations and shifting operations (ON .word OPERANDS!). not x, y x <- NOT (y) and x, y, z x <- (y) AND (z) or x, y, z x <- (y) OR (z) nor x, y, z x <- (y) NOR (z) nand x, y, z x <- (y) NAND (z) xor x, y, z x <- (y) XOR (z) xnor x, y, z x <- (y) XNOR (z) sll x, y, value x <- (y), logically left shifted by value bits srl x, y, value x <- (y), logically right shifted by value bits sra x, y, value x <- (y), arithmetically right shifted by value bits rol x, y, value x <- (y), rotated left by value bits ror x, y, value x <- (y), rotated right by value bits</pre>⌨️ 快捷键说明
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