2-2二進制運算.ppt

1、Cha2. Addition and Subtraction of Binary I,原則：,逢二進一（加） 借一當二（減）,(i-1)s Carry/Borrow,Augend/Minuend,Addend/Subtrahend,Sum,carry,difference,borrow,無符號加減,電子科技大學光電信息學院 陳德軍,Cha2. Addition and Subtraction of Binary II,EX9：,1 0 0 1 1 0 0 1,0 1 1 1 1 0 0 1,+,0,0,1,1,0,0,0,0,1,1,1,0,1,0,1,0,1,1,逢2進1,153,+,121

2、,274,1 0 0 1 1 0 0 1,0 1 1 1 1 0 0 1,-,0,0,0,0,0,0,0,0,0,0,0,1,1,0,1,0,借1當2,153,-,121,32,2,2,電子科技大學光電信息學院 陳德軍,Cha2: Representation of Negative Numbers III,Signed-magnitude system：,A symbol (+ or -) + magnitude原碼,S-M碼,EX：,Decimal: +98 53 100 +56,S,M,The S M of binary numbers：,S：定義在最高有效位（MSB）,0=正+,1=負

3、-,M：數(shù)值位 magnitude,電子科技大學光電信息學院 陳德軍,Cha2: Representation of Negative Numbers IV,EX：,+410=,01002,+ Signed,-410=,11002,- Signed,（4-bit）,+410=,000001002,-410=,100001002,（8-bit）,+12710=011111112,-12710=111111112,010=000000002=100000002,+0 Negative: “1”+radix-complement Twos complement can +,Ones Compleme

4、nt Representation(反碼)：,positive：=S-M system,符號位0 Magnitude, change it into its complement; 2. Implement the addition by addition rules,The result exceeds the range of the number system. The range : -2n-1,2n-1-1, n is the number of digit;,Overflow：,電子科技大學光電信息學院 陳德軍,數(shù)制與編碼.二進制補碼的加減法（2）,EX：,+53 +49,+,Pe

5、rform the addition operation with 8-bit twos complement,+102,-53 -49,+,-102,+53 -49,+,+4,+49 -53,+,-4,00110101 00110001,+,01100110,11001011 11001111,+,10011010,00110101 11001111,+,00000100,00110001 11001011,+,11111100,電子科技大學光電信息學院 陳德軍,數(shù)制與編碼.二進制補碼的加減法（3）,減法規(guī)則：,首先將減號變負號，再對數(shù)值其求補數(shù)，在依據(jù)加法規(guī)則進行運算,前提：,不超過記數(shù)系

6、統(tǒng)的范圍,EX：,Perform the operation of subtration with 4-bit twos complement,+2 +5,-,-3,+2 -5,-,+7,-2 -5,-,+3,-5 +2,-,-7,0010,+,0101,1011,1101,0010,+,1011,0101,0111,1110,+,1011,0101,0011,1011,+,0010,1110,1001,數(shù)制與編碼.二進制補碼加減法（4）,+3,+4,+,+7,0011,0100,+,0111,兩個正數(shù)相加,-2,-6,+,-8,1110,1010,+,1000,兩個正數(shù)相加,1,+3,-4,

7、+,-1,0011,1100,+,1111,一正一負相加,+3,+7,-4,0011,0111,+,1100,兩個數(shù)相減,-,1001,求補,-3,-5,+2,1101,1011,+,0010,兩個數(shù)相減,-,0101,求補,1,數(shù)制與編碼.二進制補碼的加減法溢出（overflow）判定（1）,The reason of Overflow,The addition operation produces a result that exceeds the range of the number system,電子科技大學光電信息學院 陳德軍,How to detect the overflow

8、in addition,1.If the augends sign is different from the addends, no overflow；otherwise, perhaps overflow,2.If augends sign = addends is 0 or 1, but the sum s sign is 1 or 0, two signs are different, overflow!,3.If the carry bits cin into and cout out of the sign position(MSB) are different, overflow

9、!,數(shù)制與編碼.二進制補碼的加減法溢出（overflow）判定（2）,EX10：,Indicate whether or not overflow occurs when adding the following with 8-bits twos complement numbers,1、53+（-79）,無溢出，相加二數(shù)符號相反,2、（-53）+（-79）,寫成二進制補碼相加形式：,11001011,10110001,+,01111100,1,該數(shù)的出現(xiàn)不是判定溢出的依據(jù),因為兩個加數(shù)符號相同，且和為正，加數(shù)為負，所以存在溢出,Cin=0,Cout=1,因為和的符號位 ，所以存在溢出,電子科技大學光電