[Study Notes] [C Language] bit operation, learning note operation

[Study Notes] [C Language] bit operation, learning note operation
1. & bitwise AND

1> Functions

The result bit is 1 only when the two binary numbers are 1. Otherwise, the result bit is 0.

2> for example, 9 & 5 is actually 1001 & 101 = 1, so 9 & 5 = 1

3> rule

In binary, the Phase 1 is in the original position, and the phase 0 is 0.

2. | by bit or

1> Functions

If one of the two binary numbers is 1, The result bit is 1. Otherwise, the result bit is 0.

2> example: 9 | 5 is actually 1001 | 101 = 1101, So 9 | 5 = 13

3. ^ bitwise OR

1> Functions

When the binary numbers are different (different), the result is 1. Otherwise, the result is 0.

2> for example, 9 ^ 5 is actually 1001 ^ 101 = 1100, So 9 ^ 5 = 12

3> rule

The result of the same integer ^ is 0. For example, 5 ^ 5 = 0

The result of multiple integers ^ is irrelevant to the order. For example, 5 ^ 6 ^ 7 = 5 ^ 7 ^ 6

Therefore, it is concluded that a ^ B ^ a = B

4 .~ Invert

Returns the decimal digits of the binary digits of integer a, and the sign bit is also reversed (0 to 1, 1 to 0)

 

5. <move left

Remove all the binary numbers of integer a from the left n bits, discard the high bits, and add 0 to the low bits. Shifting n places to the left is actually multiplied by the n power of 2

Since the Left shift is to discard the highest bit, and the 0 complement the second bit, the symbol bit is also discarded, and the result value removed from the Left shift may change the positive and negative values.

6.> right shift

Shifts all the binary numbers of integer a to the right n bits to keep the symbol bit unchanged. Shifting n places to the right is actually dividing by the n power of 2

When the number is positive, the sign bit is 0, and the maximum bit is 0.

When it is a negative number, the sign bit is 1, and the maximum bit is 0 or 1, depending on the requirements of the compilation system

 

7. Learn the code

1 # include <stdio. h> 2 3 4 int main () 5 {6/* by bit and & 7 8 10101010000 9 0000010000010 --------------- 11 0000000000012 13 1011101114 1010110115 --------- 16 1010100117 18 100119 010120 ----- 21 000122 */23 24/* 25 by bit or | 26 100127 010128 ----- 29 110130 */31 32 33/* 34 by bit or ^ 35 1. returns 0 if the same value is exclusive or, for example, 9 ^ 936 2. switch 9 ^ 5 ^ 6 = 9 ^ 6 ^ 537 3. returns the original value if any value is different from 0. 9 ^ 0 = 938 4. a ^ B ^ a = a ^ B = 0 ^ B = b39 40 100141 010142 ----- 43 110044 45 100146 100147 ----- 48 0000049 50 010151 000052 ---- 53 010154 55 9 ^ 5 ^ 9 = 9 ^ 9 ^ 5 = 0 ^ 5 = 556 57 a ^ B ^ = b58 */59 // printf ("% d \ n ", 9 ^ 9); 60 61 // printf ("% d \ n", 9 ^ 5); 62 63/* 64 bitwise inversion ~ 65 ~ 0000 0000 0000 0000 0000 0000 0000 100166 1111 1111 1111 1111 1111 1111 1111 */68 // printf ("% d \ n ",~ 9 ); 69 70/* 71 shift left <72 73 0000 0000 0000 0000 0000 0000 0000 000074 00 0000 0000 0000 0000 0000 0000 10010075 76 9 <1-> 9*2 to the power of 1 = = 1877 9 <2-> 9*2 to the power of 2 = 3678 9 <n-> 9*2 to the power of Npower 79 */80 81 // printf ("% d \ n ", 9 <1 ); 82 83/* 84 right shift> 85 0000 0000 0000 0000 0000 0000 0000 000086 000000 0000 0000 0000 0000 0000 0000 1087 111111 1111 1111 1111 1111 10 88 89 8> 1-> 8/2 = 490 8> 2-> 8/2 to the power of 2 = 291 8> n-> 8/2 to the power of Npower 92 */93 94 printf ("% d \ n", 8> 3); 95 96 return 0; 97}

1 # include <stdio. h> 2 3/* 4 Use bitwise OR operator to exchange the values of two variables 5 */6 7 int main () 8 {9 int a = 10; 10 int B = 11; 11 12/* use the third-party variable 13 int temp = a; 14 a = B; 15 B = temp; 16 */17 18/* 19 a = B-; 20 B = B-a; 21 a = B +; 22 */23 24 // a ^ B ^ a = b25 26 // a --> 10 ^ 1127/B --> 1028 a = a ^ B; 29 B = a ^ B; 30 a = a ^ B; 31 32 printf ("a = % d, B = % d \ n", a, B ); 33 34 return 0; 35}

1 # include <stdio. h> 2/* 3 Use bitwise AND & operator to determine the parity of the variable 4 */5 int main () 6 {7/* 8 15: 1111 9 9: 100110 11 14: 111012 10: 101013 */14 int a = 15; 15 16 a & 1 = 1 // odd 17 a & 1 = 0 // even 18 19/* 20 if (a % 2) {21 printf ("odd \ n"); 22} else {23 printf ("even \ n"); 24} */25 26 // a % 2 = 0? Printf ("even \ n"): printf ("odd \ n"); 27 28 // a % 2? Printf ("odd \ n"): printf ("even \ n"); 29 30 31 32 return 0; 33}

1/* 2 Write a function to output the binary form of integers in the memory. 3 */4 5 # include <stdio. h> 6 void printBinary (int number); 7 8 int main () 9 {10/* 11 0000 0000 0000 0000 0000 0000 0000 000012 0000 0000 0000 0000 0000 0000 0000 111113 0000 14 9: 0000 0000 0000 0000 0000 0000 100115-10: 1111 1111 1111 1111 1111 1111 1111 */17 18 19 // printf ("% d \ n ",~ 9); 20 21 22 printBinary (-10); 23 return 0; 24} 25 26 void printBinary (int number) 27 {28 29 // The record is now moved to the nth 30 // (sizeof (number) * 8)-1 = 3131 int temp = (sizeof (number) <3) -1; 32 33 while (temp> = 0) 34 {35 // first move the bit, and then & 1, get the value of the corresponding bit 36 int value = (number> temp) & 1; 37 printf ("% d", value); 38 39 // 40 temp --; 41 42 // 4 bits per output, output a space 43 if (temp + 1) % 4 = 0) 44 {45 printf (""); 46} 47} 48 49 printf ("\ n"); 50}