Shift operator (detailed)

Shift Operators

Move left （<<）

Positive numbers ： In binary form , Move all the numbers to the left by the corresponding digits , High removal , Low vacancy filling 0

      for example 12<<2=48

1. 12 Corresponding binary 0000 1100
2. Move two bits to the left to become 0011 0000
3. Convert to decimal 2^4+2^5=16+32=48

negative ： for example -12<<2=-48

1. 12 Corresponding binary 0000 1100, be -12 The corresponding original code is 1000 1100
2. Original code ——> Inverse code , The sign bits remain the same , The rest is reversed , become 1111 0011
3. Inverse code ——> Complement code , Counter code plus 1, Has become a 1111 0100
4. therefore -12 The corresponding binary is 1111 0100（ Complement code ）
5. Move two places to the left , Low complement 0, become 1101 0000
6. Complement code ——> Inverse code , Subtract one from the corresponding complement , become 1100 1111
7. Inverse code ——> Original code , The sign bits remain the same , The rest is reversed 1011 0000
8. Convert binary to decimal ,2^4+2^5=16+32=48, The highest position is 1, Then for -48

1. The first ① Situated 0 Not enough to reduce , towards ② Loan processing “1” When “2”,2-1=1
2. Because the first ② Situated 0 Borrowed “1”, said “-1”, but ② The place itself does not , And to ③ Loan processing “1” When “2”,2-1=1
3. Because the first ③ Situated 0 Borrowed “1”, said “-1”, But not enough , And to ④ Loan processing “1” When “2”,2-1=1
4. The first ④ Situated 0 Borrowed “1”, said “-1”, But not enough , And to ⑤ Loan processing “1” When “2”,2-1=1
5. The first ⑤ Situated 1 Borrowed , be 1-1=0

Move right （>>）： Move all the numbers to the right by the corresponding shift digits in binary form , Move out low ( Abandon ), The empty bits in the high order are complemented by positive numbers 0, Negative numbers make up 1

    

   Positive numbers ：20>>2=5

1. take 20 Convert to binary to 0001 0100
2. Move two bits to the right to become 0000 0101
3. Convert binary to decimal 2^0+2^2=5

   negative ：-20>>2=-5

1. 20 Corresponding binary 0001 0100,-20 Then the corresponding original code is 1001 0100
2. Original code ——> Inverse code , The sign bits remain the same , The rest is reversed , become 1110 1011
3. Inverse code ——> Complement code , Counter code plus 1, Has become a 1110 1100
4. therefore -20 The corresponding binary is 1110 1100（ Complement code ）
5. Move two places to the right , High compensation 1, become 1111 1011
6. Complement code ——> Inverse code , Subtract one from the corresponding complement , become 1111 1010

1. Inverse code ——> Original code , The sign bits remain the same , The rest is reversed 1000 0101
2. Convert binary to decimal ,, The highest position is 1, Then for 2^0+2^2=5, The highest bit is 1, Then for -5

Unsigned operator （>>>）

   Move all numbers in binary form to the right by the corresponding number of digits , Move out low ( Abandon ), Fill in the high vacancy . For positive numbers and signed right shift （>>） identical , But it's different for negative numbers .

  Positive numbers ：15>>>2=3

1. Convert to binary 0000 1111
2. Move two places to the right , High compensation 0, become 0000 0011
3. Convert binary to decimal , become 2^0+2^1=3

  negative ：-15>>>2

1. 15 Convert to binary 0000 0000 0000 0000 00000 0000 0000 1111,

-15 Corresponding to the original code 1000 0000 0000 0000 00000 0000 0000 1111

1. Original code ——> Inverse code , The sign bits remain the same , The rest is reversed ,1111 1111 1111 1111 1111 1111 1111 0000
2. Inverse code ——> Complement code , Counter code plus 1, become 1111 1111 1111 1111 1111 1111 1111 0001
3. therefore -15 The binary of is 1111 1111 1111 1111 1111 1111 1111 0001（ Complement code ）
4. Move two places to the right , High compensation 0, become 0011 1111 1111 1111 1111 1111 1111 1100
5. Complement code ——> Inverse code , Complement subtraction 1,0011 1111 1111 1111 1111 1111 1111 1011
6. Inverse code ——> Original code , The sign bits remain the same , The rest is reversed , become

0100 0000 0000 0000 00000 0000 0000 0100

1. Convert binary to decimal ,2^2+2^30=1073741828

Because the sign bits of negative numbers （ highest ） by 1, The unsigned right shift should be filled in the highest position 0, therefore 32 Bit binary must be written completely . Move right with symbol （>>）, The high bit is the sign bit of complement , front 24 All places are 1( The number in the example is small ,8 Bits are enough to show ; If the number is large , want 16 Words that can only be expressed by bits , That's before 16 All places are 1), Omitting does not affect the calculation ; But this is a special case , The complement bit is different from the sign bit .