brief introduction ： The numbers in the program exist in binary form in the computer memory , Bit operation is directly to the integer in memory corresponding to the binary operation , Generally, the operation is carried out after digitizing into binary numbers .
Application scenarios ： Use bit operations in general operations and bit operations , Can improve performance . But it will make the code difficult to understand , It is suggested to make rational use of .
1、 Count odd numbers
2、 Statistical even number
3、 Statistics of different numbers, etc
4、 Find the opposite number
Bit operation has 6 Kind of ：
1、 Bitwise AND ： Both of them are 1 when , The result is 1（ Count odd numbers ） I.e. complete 1 by 1.
2、 Press bit or ： Both of them are 0 when , The result is 0（ Statistical even number ） I.e. complete 0 by 0.
3、 Bitwise XOR ： The two bits are the same 0, Dissimilarity is 1( Common statistics are different ） The difference is 1.
4、 According to the not ：0 change 1,1 change 0, amount to -x-1
5、 Left shift operation ： Each binary moves several bits to the left , High level discard , Low complement 0.
6、 Right shift operation ： Each binary moves several bits to the right , To an unsigned number , High compensation 0, Sign digit complement of signed numbers （ Arithmetic shift right ）, Or make up 0（ Logical shift right ）.
Source code of case ：
# -*- coding: utf-8 -*-
# time: 2022/5/22 17:56
# file: bitwise.py
# official account : Play with test development
# &： Both of them are 1 when , The result is 1（ Count odd numbers ） I.e. complete 1 by 1.
a1 = 10
b1 = 9
""" 10 = 0b1010 9 = 0b1001 8 = 0b1000 """
print(bin(a1))
print(bin(b1))
print(a1 & b1) # 8
print(int("0b1000", 2))
# |： Both of them are 0 when , The result is 0（ Statistical even number ） I.e. complete 0 by 0.
a2 = 10
b2 = 9
""" 10 = 0b1010 9 = 0b1001 11 = 0b1011 """
print(bin(a2))
print(bin(b2))
print(a2 | b2) # 11
print(int("0b1011", 2))
# ^： The two bits are the same 0, Dissimilarity is 1( Common statistics are different ） The difference is 1.
a3 = 10
b3 = 9
""" 10 = 0b1010 9 = 0b1001 3 = 0b0011 """
print(bin(a3))
print(bin(b3))
print(a3 ^ b3) # 11
print(int("0b0011", 2))
# ~：0 change 1,1 change 0, amount to -x-1
a4 = 10
""" 10 = 0b1010 -x-1 = -11 """
print(bin(a4))
print(~a4) # -11
print(int("-0b1011", 2))
# Find the opposite number
print(~a4 + 1) # -10
# <<： Each binary moves several bits to the left , High level discard , Low complement 0, namely ：x << n = x * (2 ** n)
a5 = 10
""" 10 = 0b1010 x = 10 * 2 ** 3 = 10 * 2 * 2 * 2 """
b5 = a5 << 3
print(bin(b5))
print(b5) # 80
print(int("0b1010000", 2))
# >>： Each binary moves several bits to the right , To an unsigned number , High compensation 0, Sign digit complement of signed numbers （ Arithmetic shift right ）, Or make up 0（ Logical shift right ）.
# namely ：x << n = x / (2 ** n)
a6 = 64
""" 64 = 0b1000000 x = 64 / (2 ** 3) = 64 / (2 * 2 * 2) """
b6 = a6 >> 3
print(bin(b6))
print(b6) # 8
# Classic case ： Use ^ Find out the number of times
a7 = 1 ^ 1 ^ 2
a8 = 1 ^ 2 ^ 1
a9 = 2 ^ 1 ^ 1
print(a7)
print(a8)
print(a9)
# Count the time spent by the original method and the bit operation method
import time
loop = 30000000
start1 = time.time()
odd_list1 = []
for i in range(loop):
if i & 1 == 1:
odd_list1.append(i)
end1 = time.time()
print(f"time1:{
end1 - start1}")
start2 = time.time()
odd_list2 = []
for i in range(loop):
if i % 2 == 1:
odd_list1.append(i)
end2 = time.time()
print(f"time2:{
end2 - start2}")
# time1:5.262001037597656
# time2:4.736037492752075
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