Let's say i have this number
int x = 65535;
Which is the decimal representation of:
ÿÿ
I know how i can do it from single char
#include <stdio.h>
int main() {
int f = 65535;
printf("%c", f);
}
But this will only give me "ÿ"
I would like to do this without using any external library, and preferably using C type strings.
#include <stdio.h>
int main() {
unsigned f = 65535; // initial value
// this will do the printf and ff >>= 8 until f <= 0 ( =0 actually)
do {
printf("%c", f & 0xff); // print once char. The &0xff keeps only the bits for one byte (8 bits)
f >>= 8; // shifts f right side for 8 bits
} while (f > 0);
}
Consider the value 65535, or 0xffff in hexadecimal, meaning its positive value takes 2 bytes that are 0xff and 0xff
print of f & 0xff keeps only the 8 LSb, (0xffff & 0xff = 0xff)
f >> = 8 shifts the value 8 bits to the right, 0xffff becomes 0x00ff (the 'ff' right side are gone
f > 0 is true since f == 0xff now
Next loop is the same, but f >>= 8 shifts 0x00ff to the right => 0x0000, and f is null.
Thus the f > 0 condition is wrong and the loop ends.
What you're looking for is bit masking: (x >> 8) & 0xFF for
the high order byte, and (x & 0xFF) for the lower. (Actually,
if int has 32 bits, it's (x >> 24) & 0xFF for the high order
byte. But given the values, and what you say your expecting,
you probably want the second byte, and not the high order byte.)
What you have is a 16 bit (two bytes) unsigned number. A char is 8 bits (one byte). This means you have to extract the two bytes in the number to get them as separate character.
This is done with the bitwise operators. You can use bitwise and & and bitwise shift >> to accomplish that.
Something like
char buffer[9];
long value = 65535;
char *cur = buffer;
while (value > 0)
{
*cur++ = value % 0x100;
value /= 0x100;
}
*cur = 0;
Related
I know you can get the first byte by using
int x = number & ((1<<8)-1);
or
int x = number & 0xFF;
But I don't know how to get the nth byte of an integer.
For example, 1234 is 00000000 00000000 00000100 11010010 as 32bit integer
How can I get all of those bytes? first one would be 210, second would be 4 and the last two would be 0.
int x = (number >> (8*n)) & 0xff;
where n is 0 for the first byte, 1 for the second byte, etc.
For the (n+1)th byte in whatever order they appear in memory (which is also least- to most- significant on little-endian machines like x86):
int x = ((unsigned char *)(&number))[n];
For the (n+1)th byte from least to most significant on big-endian machines:
int x = ((unsigned char *)(&number))[sizeof(int) - 1 - n];
For the (n+1)th byte from least to most significant (any endian):
int x = ((unsigned int)number >> (n << 3)) & 0xff;
Of course, these all assume that n < sizeof(int), and that number is an int.
int nth = (number >> (n * 8)) & 0xFF;
Carry it into the lowest byte and take it in the "familiar" manner.
If you are wanting a byte, wouldn't the better solution be:
byte x = (byte)(number >> (8 * n));
This way, you are returning and dealing with a byte instead of an int, so we are using less memory, and we don't have to do the binary and operation & 0xff just to mask the result down to a byte. I also saw that the person asking the question used an int in their example, but that doesn't make it right.
I know this question was asked a long time ago, but I just ran into this problem, and I think that this is a better solution regardless.
//was trying to do inplace, would have been better if I had swapped higher and lower bytes somehow
uint32_t reverseBytes(uint32_t value) {
uint32_t temp;
size_t size=sizeof(uint32_t);
for(int i=0; i<size/2; i++){
//get byte i
temp = (value >> (8*i)) & 0xff;
//put higher in lower byte
value = ((value & (~(0xff << (8*i)))) | (value & ((0xff << (8*(size-i-1)))))>>(8*(size-2*i-1))) ;
//move lower byte which was stored in temp to higher byte
value=((value & (~(0xff << (8*(size-i-1)))))|(temp << (8*(size-i-1))));
}
return value;
}
I know you can get the first byte by using
int x = number & ((1<<8)-1);
or
int x = number & 0xFF;
But I don't know how to get the nth byte of an integer.
For example, 1234 is 00000000 00000000 00000100 11010010 as 32bit integer
How can I get all of those bytes? first one would be 210, second would be 4 and the last two would be 0.
int x = (number >> (8*n)) & 0xff;
where n is 0 for the first byte, 1 for the second byte, etc.
For the (n+1)th byte in whatever order they appear in memory (which is also least- to most- significant on little-endian machines like x86):
int x = ((unsigned char *)(&number))[n];
For the (n+1)th byte from least to most significant on big-endian machines:
int x = ((unsigned char *)(&number))[sizeof(int) - 1 - n];
For the (n+1)th byte from least to most significant (any endian):
int x = ((unsigned int)number >> (n << 3)) & 0xff;
Of course, these all assume that n < sizeof(int), and that number is an int.
int nth = (number >> (n * 8)) & 0xFF;
Carry it into the lowest byte and take it in the "familiar" manner.
If you are wanting a byte, wouldn't the better solution be:
byte x = (byte)(number >> (8 * n));
This way, you are returning and dealing with a byte instead of an int, so we are using less memory, and we don't have to do the binary and operation & 0xff just to mask the result down to a byte. I also saw that the person asking the question used an int in their example, but that doesn't make it right.
I know this question was asked a long time ago, but I just ran into this problem, and I think that this is a better solution regardless.
//was trying to do inplace, would have been better if I had swapped higher and lower bytes somehow
uint32_t reverseBytes(uint32_t value) {
uint32_t temp;
size_t size=sizeof(uint32_t);
for(int i=0; i<size/2; i++){
//get byte i
temp = (value >> (8*i)) & 0xff;
//put higher in lower byte
value = ((value & (~(0xff << (8*i)))) | (value & ((0xff << (8*(size-i-1)))))>>(8*(size-2*i-1))) ;
//move lower byte which was stored in temp to higher byte
value=((value & (~(0xff << (8*(size-i-1)))))|(temp << (8*(size-i-1))));
}
return value;
}
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).
I have a variable with "x" number of bits. How can I extract a specific group of bits and then work on them in C?
You would do this with a series of 2 bitwise logical operations.
[[Terminology MSB (msb) is the most-significant-bit; LSB (lsb) is the least-significant-bit. Assume bits are numbered from lsb==0 to some msb (e.g. 31 on a 32-bit machine). The value of the bit position i represents the coefficient of the 2^i component of the integer.]]
For example if you have int x, and you want to extract some range of bits x[msb..lsb] inclusive, for example a 4-bit field x[7..4] out of the x[31..0] bits, then:
By shifting x right by lsb bits, e.g. x >> lsb, you put the lsb bit of x in the 0th (least significant) bit of the expression, which is where it needs to be.
Now you have to mask off any remaining bits above those designated by msb. The number of such bits is msb-lsb + 1. We can form a bit mask string of '1' bits that long with the expression ~(~0 << (msb-lsb+1)). For example ~(~0 << (7-4+1)) == ~0b11111111111111111111111111110000 == 0b1111.
Putting it all together, you can extract the bit vector you want with into a new integer with this expression:
(x >> lsb) & ~(~0 << (msb-lsb+1))
For example,
int x = 0x89ABCDEF;
int msb = 7;
int lsb = 4;
int result = (x >> lsb) & ~(~0 << (msb-lsb+1));
// == 0x89ABCDE & 0xF
// == 0xE (which is x[7..4])
Make sense?
Happy hacking!
If you're dealing with a primitive then just use bitwise operations:
int bits = 0x0030;
bool third_bit = bits & 0x0004; // bits & 00000100
bool fifth_bit = bits & 0x0010; // bits & 00010000
If x can be larger than a trivial primitive but is known at compile-time then you can use std::bitset<> for the task:
#include<bitset>
#include<string>
// ...
std::bitset<512> b(std::string("001"));
b.set(2, true);
std::cout << b[1] << ' ' << b[2] << '\n';
std::bitset<32> bul(0x0010ul);
If x is not known at compile-time then you can use std::vector<unsigned char> and then use bit-manipulation at runtime. It's more work, the intent reads less obvious than with std::bitset and it's slower, but that's arguably your best option for x varying at runtime.
#include<vector>
// ...
std::vector<unsigned char> v(256);
v[2] = 1;
bool eighteenth_bit = v[2] & 0x02; // second bit of third byte
work on bits with &, |. <<, >> operators.
For example, if you have a value of 7 (integer) and you want to zero out the 2nd bit:
7 is 111
(zero-ing 2nd bit AND it with 101 (5 in decimal))
111 & 101 = 101 (5)
here's the code:
#include <stdio.h>
main ()
{
int x=7;
x= x&5;
printf("x: %d",x);
}
You can do with other operators like the OR, shift left, shift right,etc.
You can use bitfields in a union:
typedef union {
unsigned char value;
struct { unsigned b0:1,b1:1,b2:1,b3:1,b4:1,b5:1,b6:1,b7:1; } b;
struct { unsigned b0:2,b1:2,b2:2,b3:2; } b2;
struct { unsigned b0:4,b1:4; } b4;
} CharBits;
CharBits b={0},a={0};
printf("\n%d",b.value);
b.b.b0=1; printf("\n%d",b.value);
b.b.b1=1; printf("\n%d",b.value);
printf("\n%d",a.value);
a.b4.b1=15; printf("\n%d",a.value); /* <- set the highest 4-bit-group with one statement */
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;
}