# Bit manipulation

### 3/11/2018

Bit manipulation is something I find interesting, to strengthen my understanding I thought it'd be best to put some of the basics in text.

First, an understanding of binary is required:

128 |
64 | 32 | 16 | 8 | 4 | 2 | 1 |
---|---|---|---|---|---|---|---|

0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 |

Got it? I knew you would. If not see my post on binary.

Next you need an understanding of bitwise operations:

AND (&) | copies bit if it is set in both operands 11 AND 5 (1011 AND 0101) = 1 1011 0101 ---- 0001 |
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OR (|) | inclusive OR, copies bit if it is set in either operand 13 OR 7 (1101 OR 0111) = 15 1101 0111 ---- 1111 |

XOR (^) | exclusive OR, copies bit if it is set in either operand but not if it is set in both 13 XOR 7 (1101 XOR 0111) = 10 1101 0111 ---- 1010 |

NOT (~) | flips each bit obtaining the numbers one's complement NOT 5 (101) = 2 101 --- 010 |

Left Shift (<<) | Shifts the bits of the left operand by the number in the right operand to the left 5 << 2 (101 << 010) = 20 000101 <<< (the 101 is moved to the left by 2 bits) ------ 010100 |

Right Shift (>>) | Shifts the bits of the left operand by the number in the right operand to the right, underflow is discarded 22 >> 3 (10110 >> 00011) = 2 10110 >>>>>(the 10110 is move to the right by 3 bits) ----- 00010 |

First note, counting bits starts from the least significant bit (right-most), which is the 0th bit.

Setting a bit | To set a bit to 1 OR the number with 1 left shifted with the desired bit to set number |= (1 << bit) number = 5, bit = 3 5 | (1 << 3) = 13 1 << 3 = 8 0001 <<< ----- 1000 5 | 8 = 13 0101 | 1000 ------ 1101 ^ our 3rd bit (0,1,2,3) is now set |
---|---|

Clearing a bit | To clear a bit AND the number with the one's complement of 1 left shifted with the desired bit to clear number &= ~(1 << bit) number = 9, bit = 3 9 & ~(1 << 3) = 1 1 << 3 = 8 0001 <<< ---- 1000 ~8 = 7 1000 ~ ------ 0111 9 & 7 = 1 1001 & 0111 ------ 0001 ^ our 3rd bit (0,1,2,3) is now 0 |

Getting a bit | To get a bit right shift the number by the desired bit to retrieve, then AND it by 1 result = (number >> bit) & 1 number = 14, bit = 3 result = (14 >> 3) & 1 result = 1 14 >> 3 = 1 1110 >>> ----- 0001 1 & 1 = 1 01 01 -- 01 |

Toggling a bit | To toggle a bit XOR the number by 1 left shifted by the desired bit number ^= 1 << bit number = 12, bit = 4 12 ^= 1 << 4 = 28 1 << 4 = 16 00001 <<<< ----- 10000 12 ^ 16 = 28 01100 ^ 10000 ------- 11100 ^ our 4th bit (0,1,2,3,4) is now set |

#include <stdio.h> int dectobin(int dec) { /* return the binary representation of dec */ if (dec == 0) return 0; if (dec == 1) return 1; return (dec % 2) + 10 * dectobin(dec / 2); } int getbit(int num, int bit) { return (num >> bit) & 1; } int togglebit(int number, int bit) { return number ^= 1 << bit; } int setbit(int number, int bit) { return number |= 1 << bit; } int clearbit(int number, int bit) { return number &= ~(1 << bit); } int main(void) { unsigned char flags = 0; /* char because it is 1 byte, unsigned because we don't need negatives */ printf("flag has value %d, in binary: %08d\n", flags, dectobin(flags)); puts("let's make the number 145"); puts("we need to toggle bits: 0, 4, 7"); flags = setbit(flags, 0); puts("set bit 0"); printf("flag has value %d, in binary: %08d\n", flags, dectobin(flags)); flags = setbit(flags, 4); puts("set bit 4"); printf("flag has value %d, in binary: %08d\n", flags, dectobin(flags)); flags = setbit(flags, 7); puts("set bit 7"); printf("flag has value %d, in binary: %08d\n", flags, dectobin(flags)); puts("let's check the value of bit 7"); printf("bit 7 is %d\n", getbit(flags, 7)); puts("we don't want bit 7 set anymore"); flags = clearbit(flags, 7); puts("cleared bit 7"); printf("flag has value %d, in binary: %08d\n", flags, dectobin(flags)); return 0; }output:

flag has value 0, in binary: 00000000 let's make the number 145 we need to toggle bits: 0, 4, 7 set bit 0 flag has value 1, in binary: 00000001 set bit 4 flag has value 17, in binary: 00010001 set bit 7 flag has value 145, in binary: 10010001 let's check the value of bit 7 bit 7 is 1 we don't want bit 7 set anymore cleared bit 7 flag has value 17, in binary: 00010001