am having a little trouble with this function of mine. We are supposed to use bit wise operators only (that means no logical operators and no loops or if statements) and we aren't allowed to use a constant bigger than 0xFF.
I got my function to work, but it uses a huge constant. When I try to implement it with smaller numbers and shifting, I can't get it to work and I'm not sure why.
The function is supposed to check all of the even bits in a given integer, and return 1 if they are all set to 1.
Working code
int allEvenBits(int x) {
/* implements a check for all even-numbered bits in the word set to 1 */
/* if yes, the program outputs 1 WORKING */
int all_even_bits = 0x55555555;
return (!((x & all_even_bits) ^ all_even_bits));
}
Trying to implement with a smaller constant and shifts
int allEvenBits(int x) {
/* implements a check for all even-numbered bits in the word set to 1 */
/* if yes, the program outputs 1 WORKING */
int a, b, c, d, e = 0;
int mask = 0x55;
/* first 8 bits */
a = (x & mask)&1;
/* second eight bits */
b = ((x>>8) & mask)&1;
/* third eight bits */
c = ((x>>16) & mask)&1;
/* fourth eight bits */
d = ((x>>24) & mask)&1;
e = a & b & c & d;
return e;
}
What am I doing wrong here?
When you do, for example, this:
d = ((x>>24) & mask)&1;
..you're actually checking whether the lowest bit (with value 1) is set, not whether any of the the mask bits are set... since the &1 at the end bitwise ANDs the result of the rest with 1. If you change the &1 to == mask, you'll instead get 1 when all of the bits set in mask are set in (x>>24), as intended. And of course, the same problem exists for the other similar lines as well.
If you can't use comparisons like == or != either, then you'll need to shift all the interesting bits into the same position, then AND them together and with a mask to eliminate the other bit positions. In two steps, this could be:
/* get bits that are set in every byte of x */
x = (x >> 24) & (x >> 16) & (x >> 8) & x;
/* 1 if all of bits 0, 2, 4 and 6 are set */
return (x >> 6) & (x >> 4) & (x >> 2) & x & 1;
I don't know why you are ANDing your values with 1. What is the purpose of that?
This code is untested, but I would do something along the lines of the following.
int allEvenBits(int x) {
return (x & 0x55 == 0x55) &&
((x >> 8) & 0x55 == 0x55) &&
((x >> 16) & 0x55 == 0x55) &&
((x >> 24) & 0x55 == 0x55);
}
Say you are checking the first 4 least significant digits, the even ones would make 1010. Now you should AND this with the first 4 bits of the number you're checking against. All 1's should remain there. So the test would be ((number & mask) == mask) (mask is 1010) for the 4 least significant bits, you do this in blocks of 4bits (or you can use 8 since you are allowed).
If you aren't allowed to use constants larger than 0xff and your existing program works, how about replacing:
int all_even_bits = 0x55555555;
by:
int all_even_bits = 0x55;
all_even_bits |= all_even_bits << 8; /* it's now 0x5555 */
all_even_bits |= all_even_bits << 16; /* it's now 0x55555555 */
Some of the other answers here right shift signed integers (i.e. int) which is undefined behaviour.
An alternative route is:
int allevenbitsone(unsigned int a)
{
a &= a>>16; /* superimpose top 16 bits on bottom */
a &= a>>8; /* superimpose top 8 bits on bottom */
a &= a>>4; /* superimpose top 4 bits on bottom */
a &= a>>2; /* and down to last 2 bits */
return a&1; /* return & of even bits */
}
What this is doing is and-ing together the even 16 bits into bit 0, and the odd 16 bits into bit 1, then returning bit 0.
the main problem in your code that you're doing &1, so you take first 8 bits from number, mask them with 0x55 and them use only 1st bit, which is wrong
consider straightforward approach:
int evenBitsIn8BitNumber(int a) {
return (a & (a>>2) & (a>>4) & (a>>6)) & 1;
}
int allEvenBits(int a) {
return evenBitsIn8BitNumber(a) &
evenBitsIn8BitNumber(a>>8) &
evenBitsIn8BitNumber(a>>16) &
evenBitsIn8BitNumber(a>>24);
}
Related
Full disclosure, this is a homework problem and I do not need exact code. I am tasked with reproducing the following code while only using ~ & + <<.
int result = 0;
int i;
for(i = lowbit; i <= highbit; i++)
result |= 1 << i;
return result;
Where lowbit and highbit are parameters between 0 and 31 inclusive. If lowbit is a larger number than highbit, return 0.
What I have tried so for is the following code
int result = 0;
int negone = ~0x0;
int first = 1 << (lowbit + negone); //the first 1 bit is at the lowbit th location
int last = 1 << (highbit + negone); //the last 1 bit is at the highbit th location
int tick = ~(first + last); //attempting to get all bits in the range of low and highbit.
result = ~(~first & ~tick); //bitwise | without using |
result = ~(~last & ~result);
return result + 1; //the first bit should always be on.
So is there something fundamental I am missing here? In addition to what I have not working this also goes over my limit of 12 operators that I am allowed to use, but I'd like to try and get it working before I even begin to limit the operators.
When I run the test script on this I get errors on most of the tests it is put against including lowbit and highbit being equal to each other. Cases where highbit is the max size and lowbit is the least size seem to work though.
Any help would be much appreciated.
negone should be initialized this way:
uint32_t negone = ~0UL;
You are adding the bit number with a bit pattern in:
int first = 1 << (lowbit + negone); //the first 1 bit is at the lowbit th location
int last = 1 << (highbit + negone);
You should instead compute the 32 bit masks
uint32_t first = negone << lowbit; // all bits below lowbit are 0, others are 1
uint32_t last = negone << highbit << 1; // all bits above highbit are 1, other are 0
The result is obtained by masking the complement of first with last:
uint32_t result = ~first & last;
Combining the above steps gives is a direct solution with 7 operators (12 including the parentheses and the assignment), no addition, and no subtraction:
uint32_t result = ~(~0UL << highbit << 1) & (~0UL << lowbit);
I use 0UL because type unsigned long is guaranteed to have at least 32 bits, whereas type unsigned int might have just 16 bits.
1) Create a mask with the bits low to high set:
uint32_t mask = ~(~0ul << highbit << 1) & (~0ul << lowbit)
Example: lowbit = 4, highbit = 12 (9 bits)
mask = ~(0xffffffff << 12 << 1) & (0xffffffff << 4)
= ~(0xffff7000) & 0xfffffff0
= 0x00001fff & 0xfffffff0
= 0x00001ff0
2) Apply the mask to the value to be modified, this most simply an | operation, but that is not a valid operator in this exercise, so must be transformed using De Morgan's forum:
A|B -> ~(~A & ~B) :
result = ~(~result & ~mask) ;
It is of course possible to combining the two steps, but perhaps clarity would not then be served.
The original code generates a block of 1 from lowbit on until highbit (inclusive).
This can be achieved without a loop as follows:
int nrOfBits = highbit + ~lowbit + 2; // highbit - lowbit + 1
int shift = (nrOfBits & 0x1f + 1);
int result = ~(~(1 << shift)+1) << lowbit;
The idea is that, for example a range of 8 bits filled up with 1 means a number of 255, whereas 2^8 is 256. So - as operator - is not allowed, we use 2-complement to get -256, add 1 to get -255, and turn it back to +255 using 2-complement operator ~. Then, we just have to shift the block lowbits left.
The problem could be that tick = ~(first+last) does not flip the bit from the lowbit to the highbit.
Maybe we can do something like this:
/* supposed that lowbit = 1, highbit = 2 */
uint32_t negone = ~(0u); /* negone = all 1s */
uint32_t first = negone << lowbit; /* first = ...111110 */
uint32_t last = (1 << (highbit + 1)) + negone; /* last = ...0000111 */
uint32_t tick = last & first; /* tick = ...000110 */
result = ~(~result&~tick); /* Bitwise Or without | as you mentioned. */
It takes 11 bit operations to do this.
p.s. I am wondering why the first bit should be always on.
Edit: In order to avoid undefined operation, we should use unsigned type, like uint32_t.
I have a long sequence of bits stored in an array of unsigned long integers, like this
struct bit_array
{
int size; /* nr of bits */
unsigned long *array; /* the container that stores bits */
}
I am trying to design an algorithm to reverse the order of bits in *array. Problems:
size can be anything, i.e. not necessarily a multiple of 8 or 32 etc, so the first bit in the input array can end up at any position within the unsigned long in the output array;
the algorithm should be platform-independent, i.e. work for any sizeof(unsigned long).
Code, pseudocode, algo description etc. -- anything better than bruteforce ("bit by bit") approach is welcome.
My favorite solution is to fill a lookup-table that does bit-reversal on a single byte (hence 256 byte entries).
You apply the table to 1 to 4 bytes of the input operand, with a swap. If the size isn't a multiple of 8, you will need to adjust by a final right shift.
This scales well to larger integers.
Example:
11 10010011 00001010 -> 01010000 11001001 11000000 -> 01 01000011 00100111
To split the number into bytes portably, you need to use bitwise masking/shifts; mapping of a struct or array of bytes onto the integer can make it more efficient.
For brute performance, you can think of mapping up to 16 bits at a time, but this doesn't look quite reasonable.
I like the idea of lookup table. Still it's also a typical task for log(n) group bit tricks that may be very fast. Like:
unsigned long reverseOne(unsigned long x) {
x = ((x & 0xFFFFFFFF00000000) >> 32) | ((x & 0x00000000FFFFFFFF) << 32);
x = ((x & 0xFFFF0000FFFF0000) >> 16) | ((x & 0x0000FFFF0000FFFF) << 16);
x = ((x & 0xFF00FF00FF00FF00) >> 8) | ((x & 0x00FF00FF00FF00FF) << 8);
x = ((x & 0xF0F0F0F0F0F0F0F0) >> 4) | ((x & 0x0F0F0F0F0F0F0F0F) << 4);
x = ((x & 0xCCCCCCCCCCCCCCCC) >> 2) | ((x & 0x3333333333333333) << 2);
x = ((x & 0xAAAAAAAAAAAAAAAA) >> 1) | ((x & 0x5555555555555555) << 1);
return x;
}
The underlying idea is that when we aim to reverse the order of some sequence we may swap the head and tail halves of this sequence and then separately reverse each of halves (which is done here by applying the same procedure recursively to each half).
Here is a more portable version supporting unsigned long widths of 4,8,16 or 32 bytes.
#include <limits.h>
#define ones32 0xFFFFFFFFUL
#if (ULONG_MAX >> 128)
#define fill32(x) (x|(x<<32)|(x<<64)|(x<<96)|(x<<128)|(x<<160)|(x<<192)|(x<<224))
#define patt128 (ones32|(ones32<<32)|(ones32<<64) |(ones32<<96))
#define patt64 (ones32|(ones32<<32)|(ones32<<128)|(ones32<<160))
#define patt32 (ones32|(ones32<<64)|(ones32<<128)|(ones32<<192))
#else
#if (ULONG_MAX >> 64)
#define fill32(x) (x|(x<<32)|(x<<64)|(x<<96))
#define patt64 (ones32|(ones32<<32))
#define patt32 (ones32|(ones32<<64))
#else
#if (ULONG_MAX >> 32)
#define fill32(x) (x|(x<<32))
#define patt32 (ones32)
#else
#define fill32(x) (x)
#endif
#endif
#endif
unsigned long reverseOne(unsigned long x) {
#if (ULONG_MAX >> 32)
#if (ULONG_MAX >> 64)
#if (ULONG_MAX >> 128)
x = ((x & ~patt128) >> 128) | ((x & patt128) << 128);
#endif
x = ((x & ~patt64) >> 64) | ((x & patt64) << 64);
#endif
x = ((x & ~patt32) >> 32) | ((x & patt32) << 32);
#endif
x = ((x & fill32(0xffff0000UL)) >> 16) | ((x & fill32(0x0000ffffUL)) << 16);
x = ((x & fill32(0xff00ff00UL)) >> 8) | ((x & fill32(0x00ff00ffUL)) << 8);
x = ((x & fill32(0xf0f0f0f0UL)) >> 4) | ((x & fill32(0x0f0f0f0fUL)) << 4);
x = ((x & fill32(0xccccccccUL)) >> 2) | ((x & fill32(0x33333333UL)) << 2);
x = ((x & fill32(0xaaaaaaaaUL)) >> 1) | ((x & fill32(0x55555555UL)) << 1);
return x;
}
In a collection of related topics which can be found here, the bits of an individual array entry could be reversed as follows.
unsigned int v; // input bits to be reversed
unsigned int r = v; // r will be reversed bits of v; first get LSB of v
int s = sizeof(v) * CHAR_BIT - 1; // extra shift needed at end
for (v >>= 1; v; v >>= 1)
{
r <<= 1;
r |= v & 1;
s--;
}
r <<= s; // shift when v's highest bits are zero
The reversal of the entire array could be done afterwards by rearranging the individual positions.
You must define what is the order of bits in an unsigned long. You might assume that bit n is corresponds to array[x] & (1 << n) but this needs to be specified. If so, you need to handle the byte ordering (little or big endian) if you are going to use access the array as bytes instead of unsigned long.
I would definitely implement brute force first and measure whether the speed is an issue. No need to waste time trying to optimize this if it is not used a lot on large arrays. An optimized version can be tricky to implement correctly. If you end up trying anyway, the brute force version can be used to verify correctness on test values and benchmark the speed of the optimized version.
The fact that the size is not multiple of sizeof(long) is the hardest part of the problem. This can result in a lot of bit shifting.
But, you don't have to do that if you can introduce new struct member:
struct bit_array
{
int size; /* nr of bits */
int offset; /* First bit position */
unsigned long *array; /* the container that stores bits */
}
Offset would tell you how many bits to ignore at the beginning of the array.
Then you only only have to do following steps:
Reverse array elements.
Swap bits of each element. There are many hacks for in the other answers, but your compiler might also provide intrisic functions to do it in fewer instructions (like RBIT instruction on some ARM cores).
Calculate new starting offset. This is equal to unused bits the last element had.
I would split the problem into two parts.
First, I would ignore the fact that the number of used bits is not a multiple of 32. I would use one of the given methods to swap around the whole array like that.
pseudocode:
for half the longs in the array:
take the first longword;
take the last longword;
swap the bits in the first longword
swap the bits in the last longword;
store the swapped first longword into the last location;
store the swapped last longword into the first location;
and then fix up the fact that the first few bits (call than number n) are actually garbage bits from the end of the longs:
for all of the longs in the array:
split the value in the leftmost n bits and the rest;
store the leftmost n bits into the righthand part of the previous word;
shift the rest bits to the left over n positions (making the rightmost n bits zero);
store them back;
You could try to fold that into one pass over the whole array of course. Something like this:
for half the longs in the array:
take the first longword;
take the last longword;
swap the bits in the first longword
swap the bits in the last longword;
split both value in the leftmost n bits and the rest;
for the new first longword:
store the leftmost n bits into the righthand side of the previous word;
store the remaining bits into the first longword, shifted left;
for the new last longword:
remember the leftmost n bits for the next iteration;
store the remembered leftmost n bits, combined with the remaining bits, into the last longword;
store the swapped first longword into the last location;
store the swapped last longword into the first location;
I'm abstracting from the edge cases here (first and last longword), and you may need to reverse the shifting direction depending on how the bits are ordered inside each longword.
I want to create a function that returns the next multiple of 2^p.
Here's an attempt, but not sure if it works...:
#define P 20
#define ALIGN_FORWARD(x, alignment) ((((int)x) + ((alignment)-1)) & (~((alignment)-1)))
int round_up(int x)
{
return ALIGN_FORWARD(x, P);
}
This snippet first fills up all bits below the highest set bit. After v |= v >> 1 the first two bits can be copied and so on. Finally the value is incremented by one.
uint32_t v = ...;
v |= v >> 1;
v |= v >> 2;
v |= v >> 4;
v |= v >> 8;
v |= v >> 16;
v += 1;
The unsigned part of uint32_t is important, because otherwise the result would be always 0, because the sign bit gets extended by the shift operation. If the value is a uint64_t, then you have to add a further shift, if it is a uint16_t, then you have to remove a shift.
For an input of 8 the result would be 16. You need to test if the input is a power of two if you don't like that. Knowing that the binary representation of a power of two is one character shorter when you decrement it (8=0b1000, 7=0b111), you can use this guard:
if ((v & (v - 1)) > 0) {
…
}
(Reproduced from memory. Original from https://graphics.stanford.edu/~seander/bithacks.html#RoundUpPowerOf2, which contains many more interesting tricks.)
Using only:
! ~ & ^ | + << >>
I need to find out if a signed 32 bit integer can be represented as a 16 bit, two's complement integer.
My first thoughts were to separate the MSB 16 bits and the LSB 16 bits and then use a mask to and the last 16 bits so if its not zero, it wont be able to be represented and then use that number to check the MSB bits.
An example of the function I need to write is: fitsInShort(33000) = 0 (cant be represented) and fitsInShort(-32768) = 1 (can be represented)
bool fits16(int x)
{
short y = x;
return y == x;
}
Just kidding :) Here's the real answer, assuming int is 32 bits and short is 16 bits and two's complement represantation:
Edit: Please see the last edit for the correct answer!
bool fits16(int x)
{
/* Mask out the least significant word */
int y = x & 0xffff0000;
if (x & 0x00008000) {
return y == 0xffff0000;
} else {
return y == 0;
}
}
Without if statements i beleive that should do it:
return (
!(!(x & 0xffff0000) || !(x & 0x00008000)) ||
!((x & 0xffff0000) || (x & 0x00008000))
);
Edit: Oli's right. I somehow thought that they were allowed. Here's the last attempt, with explanation:
We need the 17 most significant bits of x to be either all ones or all zeroes. So let's start by masking other bits out:
int a = x & 0xffff8000; // we need a to be either 0xffff8000 or 0x00000000
int b = a + 0x00008000; // if a == 0xffff8000 then b is now 0x00000000
// if a == 0x00000000 then b is now 0x00008000
// in any other case b has a different value
int c = b & 0xffff7fff; // all zeroes if it fits, something else if it doesn't
return c;
Or more concisely:
return ((x & 0xffff8000) + 0x8000) & 0xffff7fff;
If a 32-bit number is in the range [-32768,+32767], then the 17 msbs will all be the same.
Here's a crappy way of telling if a 3-bit number is all ones or all zeros using only your operations (I'm assuming that you're not allowed conditional control structures, because they require implicit logical operations):
int allOnes3(int x)
{
return ((x >> 0) & (x >> 1) & (x >> 2)) & 1;
}
int allTheSame3(int x)
{
return allOnes3(x) | allOnes3(~x);
}
I'll leave you to extend/improve this concept.
Here's a solution without casting, if-statements and using only the operators you asked for:
#define fitsInShort(x) !(((((x) & 0xffff8000) >> 15) + 1) & 0x1fffe)
short fitsInShort(int x)
{
int positiveShortRange = (int) ((short) 0xffff / (short) 2);
int negativeShortRange = (int) ((short) 0xffff / (short) 2) + 1;
if(x > negativeShortRange && x < positiveShortRange)
return (short) x;
else
return (short) 0;
}
if (!(integer_32 & 0x8000000))
{
/* if +ve number */
if (integer_32 & 0xffff8000)
/* cannot fit */
else
/* can fit */
}
else if (integer_32 & 0x80000000)
{
/* if -ve number */
if ( ~((integer_32 & 0xffff8000) | 0x00007fff))
/* cannot fit */
else
/* can fit */
}
First if Checks for +ve number first by checking the signed bit. If +ve , then it checks if the bit 15 to bit 31 are 0, if 0, then it cannot fit into short, else it can.
The negative number is withing range if bit 15 to 31 are all set (2's complement method representation).
Therefore The second if it is a -ve number, then the bit 15 to 31 are masked out and the remaining lower bits (0 to 14) are set. If this is 0xffffffff then only the one's complement will be 0, which indicates the bit 15 to 31 are all set, therefore it can fit (the else part), otherwise it cannot fit (the if condition).
Let's say I have a byte with six unknown values:
???1?0??
and I want to swap bits 2 and 4 (without changing any of the ? values):
???0?1??
But how would I do this in one operation in C?
I'm performing this operation thousands of times per second on a microcontroller so performance is the top priority.
It would be fine to "toggle" these bits. Even though this is not the same as swapping the bits, toggling would work fine for my purposes.
Try:
x ^= 0x14;
That toggles both bits. It's a little bit unclear in question as you first mention swap and then give a toggle example. Anyway, to swap the bits:
x = precomputed_lookup [x];
where precomputed_lookup is a 256 byte array, could be the fastest way, it depends on the memory speed relative to the processor speed. Otherwise, it's:
x = (x & ~0x14) | ((x & 0x10) >> 2) | ((x & 0x04) << 2);
EDIT: Some more information about toggling bits.
When you xor (^) two integer values together, the xor is performed at the bit level, like this:
for each (bit in value 1 and value 2)
result bit = value 1 bit xor value 2 bit
so that bit 0 of the first value is xor'ed with bit 0 of the second value, bit 1 with bit 1 and so on. The xor operation doesn't affect the other bits in the value. In effect, it's a parallel bit xor on many bits.
Looking at the truth table for xor, you will see that xor'ing a bit with the value '1' effectively toggles the bit.
a b a^b
0 0 0
0 1 1
1 0 1
1 1 0
So, to toggle bits 1 and 3, write a binary number with a one where you want the bit to toggle and a zero where you want to leave the value unchanged:
00001010
convert to hex: 0x0a. You can toggle as many bits as you want:
0x39 = 00111001
will toggle bits 0, 3, 4 and 5
You cannot "swap" two bits (i.e. the bits change places, not value) in a single instruction using bit-fiddling.
The optimum approach if you want to really swap them is probably a lookup table. This holds true for many 'awkward' transformations.
BYTE lookup[256] = {/* left this to your imagination */};
for (/*all my data values */)
newValue = lookup[oldValue];
The following method is NOT a single C instruction, it's just another bit fiddling method. The method was simplified from Swapping individual bits with XOR.
As stated in Roddy's answer, a lookup table would be best. I only suggest this in case you didn't want to use one. This will indeed swap bits also, not just toggle (that is, whatever is in bit 2 will be in 4 and vice versa).
b: your original value - ???1?0?? for instance
x: just a temp
r: the result
x = ((b >> 2) ^ (b >> 4)) & 0x01
r = b ^ ((x << 2) | (x << 4))
Quick explanation: get the two bits you want to look at and XOR them, store the value to x. By shifting this value back to bits 2 and 4 (and OR'ing together) you get a mask that when XORed back with b will swap your two original bits. The table below shows all possible cases.
bit2: 0 1 0 1
bit4: 0 0 1 1
x : 0 1 1 0 <-- Low bit of x only in this case
r2 : 0 0 1 1
r4 : 0 1 0 1
I did not fully test this, but for the few cases I tried quickly it seemed to work.
This might not be optimized, but it should work:
unsigned char bit_swap(unsigned char n, unsigned char pos1, unsigned char pos2)
{
unsigned char mask1 = 0x01 << pos1;
unsigned char mask2 = 0x01 << pos2;
if ( !((n & mask1) != (n & mask2)) )
n ^= (mask1 | mask2);
return n;
}
The function below will swap bits 2 and 4. You can use this to precompute a lookup table, if necessary (so that swapping becomes a single operation):
unsigned char swap24(unsigned char bytein) {
unsigned char mask2 = ( bytein & 0x04 ) << 2;
unsigned char mask4 = ( bytein & 0x10 ) >> 2;
unsigned char mask = mask2 | mask4 ;
return ( bytein & 0xeb ) | mask;
}
I wrote each operation on a separate line to make it clearer.
void swap_bits(uint32_t& n, int a, int b) {
bool r = (n & (1 << a)) != 0;
bool s = (n & (1 << b)) != 0;
if(r != s) {
if(r) {
n |= (1 << b);
n &= ~(1 << a);
}
else {
n &= ~(1 << b);
n |= (1 << a);
}
}
}
n is the integer you want to be swapped in, a and b are the positions (indexes) of the bits you want to be swapped, counting from the less significant bit and starting from zero.
Using your example (n = ???1?0??), you'd call the function as follows:
swap_bits(n, 2, 4);
Rationale: you only need to swap the bits if they are different (that's why r != s). In this case, one of them is 1 and the other is 0. After that, just notice you want to do exactly one bit set operation and one bit clear operation.
Say your value is x i.e, x=???1?0??
The two bits can be toggled by this operation:
x = x ^ ((1<<2) | (1<<4));
#include<stdio.h>
void printb(char x) {
int i;
for(i =7;i>=0;i--)
printf("%d",(1 & (x >> i)));
printf("\n");
}
int swapb(char c, int p, int q) {
if( !((c & (1 << p)) >> p) ^ ((c & (1 << q)) >> q) )
printf("bits are not same will not be swaped\n");
else {
c = c ^ (1 << p);
c = c ^ (1 << q);
}
return c;
}
int main()
{
char c = 10;
printb(c);
c = swapb(c, 3, 1);
printb(c);
return 0;
}