## Bitwise operator for simply flipping all bits in an integer?

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I have to flip all bits in a binary representation of an integer. Given:

10101

The output should be

01010

What is the bitwise operator to accomplish this when used with an integer? For example, if I were writing a method like `int flipBits(int n);`

, what would go in the body? I need to flip only what's already present in the number, not all 32 bits in the integer.

The `~`

unary operator is bitwise negation. If you need fewer bits than what fits in an `int`

then you'll need to mask it with `&`

after the fact.

**Bit Manipulation,** We all know that 1 byte comprises of 8 bits and any integer or character can be binary representation of (x-1) can be obtained by simply flipping all the bits to Simply use the bitwise not operator ~. int flipBits(int n) { return ~n; } To use the k least significant bits, convert it to the right mask. (I assume you want at least 1 bit of course, that's why mask starts at 1) int flipBits(int n, int k) { int mask = 1; for (int i = 1; i < k; ++i) mask |= mask << 1; return ~n & mask; }

Simply use the bitwise not operator `~`

.

int flipBits(int n) { return ~n; }

To use the k least significant bits, convert it to the right mask. (I assume you want at least 1 bit of course, that's why mask starts at 1)

int flipBits(int n, int k) { int mask = 1; for (int i = 1; i < k; ++i) mask |= mask << 1; return ~n & mask; }

As suggested by Lưu Vĩnh Phúc, one can create the mask as `(1 << k) - 1`

instead of using a loop.

int flipBits2(int n, int k) { int mask = (1 << k) - 1; return ~n & mask; }

**Bitwise operators,** This means they look directly at the binary digits or bits of an integer. This all sounds scary, but in truth bitwise operators are quite easy to use. Hence for uint simply convert the flipped number to decimal to get the compliment (220 in this A bit can be turned on by applying bitwise OR with a power of two to the operand. It is turned off by applying bitwise AND with the complement of a power of two to the operand Which power to use and how to distinguish 'on' from 'off' flips is left as an exercise for the reader.

There is a number of ways to flip all the bit using operations

x = ~x; // has been mentioned and the most obvious solution. x = -x - 1; or x = -1 * (x + 1); x ^= -1; or x = x ^ ~0;

**Bitwise operation,** In digital computer programming, a bitwise operation operates on one or more bit patterns or For example, the binary value 0001 (decimal 1) has zeroes at every position For example, for 8-bit unsigned integers, NOT x = 255 - x , which can be may be used to invert selected bits in a register (also called toggle or flip). Flip all bit values in a byte [duplicate] Bitwise operator for simply flipping all bits in an integer? The number of equivalence relation

Well since so far there's only one solution that gives the "correct" result and that's.. really not a nice solution (using a string to count leading zeros? that'll haunt me in my dreams ;) )

So here we go with a nice clean solution that should work - haven't tested it thorough though, but you get the gist. Really, java not having an unsigned type is extremely annoying for this kind of problems, but it should be quite efficient nonetheless (and if I may say so MUCH more elegant than creating a string out of the number)

private static int invert(int x) { if (x == 0) return 0; // edge case; otherwise returns -1 here int nlz = nlz(x); return ~x & (0xFFFFFFFF >>> nlz); } private static int nlz(int x) { // Replace with whatever number leading zero algorithm you want - I can think // of a whole list and this one here isn't that great (large immediates) if (x < 0) return 0; if (x == 0) return 32; int n = 0; if ((x & 0xFFFF0000) == 0) { n += 16; x <<= 16; } if ((x & 0xFF000000) == 0) { n += 8; x <<= 8; } if ((x & 0xF0000000) == 0) { n += 4; x <<= 4; } if ((x & 0xC0000000) == 0) { n += 2; x <<= 2; } if ((x & 0x80000000) == 0) { n++; } return n; }

**Understanding Bitwise Operators - Tuts+ Code,** This means they look directly at the binary digits or bits of an integer. of a binary number, we simply flip all the bits (i.e. do what the ~ operator To flip all bits of a binary number you can run loop from 0 to size of the integer and flip individual bit at a time. However, C language has given bitwise complement ~ operator for the purpose. Bitwise complement ~ evaluates complement of the operand bit. It evaluate to 1 if corresponding bit of the operand is 0 otherwise evaluate to 0.

faster and simpler solution :

/* inverts all bits of n, with a binary length of the return equal to the length of n k is the number of bits in n, eg k=(int)Math.floor(Math.log(n)/Math.log(2))+1 if n is a BigInteger : k= n.bitLength(); */ int flipBits2(int n, int k) { int mask = (1 << k) - 1; return n ^ mask; }

**C program to flip all bits of a binary number,** Write a C program to input a number from user and flip all bits of the given number. How to flip all bits of a number using bitwise operator in C programming. run loop from 0 to size of the integer and flip individual bit at a time. AND operator&, is used to mask out bits. OR operator|, is used to reassemble bit fields. XOR operator^, is used to controllably invert bits. NOT operator~, is used to invert all the bits in a number. Left shift operator<<, makes numbers bigger by shifting their bits to higher places.

**Bits manipulation (Important tactics),** Convert binary code directly into an integer in C++. Simple approach to flip the bits of a number: It can be done by a simple way, just simply subtract the number from the value obtained when all the bits are equal to 1 . For example: Number : Given Number Value : A number with all bits set in given number. Flipped Invert actual bits of a number. Given a non-negative integer n. The problem is to invert the bits of n and print the number obtained after inverting the bits. Note that the actual binary representation of the number is being considered for inverting the bits, no leading 0’s are being considered.

**O.2,** right shift, >>, x >> y, all bits in x shifted right y bits. bitwise NOT The bitwise left shift (<<) operator shifts bits to the left. The left int main(). {. unsigned int x { 0b0100 }; It simply flips each bit from a 0 to a 1, or vice versa. Bitwise operator for simply flipping all bits in an integer? (10) As we are only required to flip the minimum bits required for the integer (say 50 is 110010 and when inverted, it becomes 001101 which is 13), we can invert individual bits one at a time from the LSB to MSB, and keep shifting the bits to the right and accordingly apply the power of 2.

**Bitwise Operators in C and C++,** A comprehensive tutorial on bit manipulations and bitwise operators in C and C++ with The leftshift operator is the equivalent of moving all the bits of a number a Note that a bitwise right-shift will be the equivalent of integer division by 2. In essence, a binary AND simply takes the logical AND of the bits in each position Count set bits in an integer; Count number of bits to be flipped to convert A to B; Count of subarrays of an Array having all unique digits; Find XOR of all elements in an Array; Set, Clear and Toggle a given bit of a number in C; Bitwise AND of all the elements of array; Check if a Number is Odd or Even using Bitwise Operators

##### Comments

- What does OP mean by "I need to flip only what's already present in the number, not all 32 bits in the integer."? If the number is "000101", does he expect "111010", or "000" as it is followed by "010" because the 1st one starts from 3rd LSB? Either way, it is inconsistent to the earlier statement "I have to flip all bits".
- Unfortunately, that doesn't give me the expected value. The bitwise reversal of 26 should be 11, but I'm getting some crazy values when using
`~`

. Is there a way to make it only use the number of bits actually used in an integer? - In java,
`int`

s are always 32 bits (2's compliment) regardless of the size of the number represented - This was established as a given in the problem I need to solve.
- Btw the bitwise reversal of 26 is not 11, but 5. 26: 11010, ~26: 00101 = 5.
- to get a mask with k low bits set, use
`(1 << k) - 1`

instead of looping and set each bit. - @Vuntic Well like I said, it probably isn't the best way to do it, but it would get the job done. It's really just a question of how you want to represent the data. It would also work to left-shift the number until the leading zeros are gone, flip the bits, and then right-shift it back, but that wouldn't end up being much simpler. When you have to do something in a high-level language like Java that's better suited for a low-level language like C, the solution isn't always going to be as elegant.
- @Ben Apart from the fact that Java doesn't have an unsigned type (not really problematic here but a bit annoying) the solution is the same in C or any other language you can think of if it allows bit twiddling - cheap excuse ;) But sure taking a few ns more to execute such a function won't matter and the solution is easy and simple to understand.. not bad in itself - it just misses this certain elegance of a good mathematical solution imho
- @Voo, In Java using Strings like this would take tens of micro-seconds.
`~`

takes hundreds of pico-seconds. Strings make more of difference in Java than they do in C. - @Peter I assumed I wouldn't be taken literally with "won't take too long". Also considering that the frequency of a modern CPU is still only about <5*10^9Hz picoseconds are a bit "optimistic" ;)
- @Voo, most modern CPUs are better 2-3.3 GHz some as high as 5.1 GHz. A bit wise invert is typically a single clock cycle instruction.
- your way is extremely inefficient. To get 2^c simply use
`1 << c`

, which is hundreds of times faster than log, ceil and pow. Moreover it's completely exact while dealing with floating-point maths may get you into trouble - While code-only answers are valid for simple contexts, if they are correct; please have in mind that they are unrecommended. Always try to provide explanation of at least
*what*the code is doing, and*how*or*why*it's doing it, in at least the most critical parts of the code. In this specific case, links to the official documentation for`BigInteger(String)`

and`Scanner.nextInt()`

are also recommended. - This is a legitimate suggestion, but something equivalent has already been posted