Related
int n_b ( char *addr , int i ) {
char char_in_chain = addr [ i / 8 ] ;
return char_in_chain >> i%8 & 0x1;
}
Like what is that : " i%8 & Ox1" ?
Edit: Note that 0x1 is the hexadecimal notation for 1. Also note that :
0x1 = 0x01 = 0x000001 = 0x0...01
i%8 means i modulo 8, ie the rest in the Euclidean division of i by 8.
& 0x1 is a bitwise AND, it converts the number before to binary form then computes the bitwise operation. (it's already in binary but it's just so you understand)
Example : 0x1101 & 0x1001 = 0x1001
Note that any number & 0x1 is either 0 or one.
Example: 0x11111111 & 0x00000001 is 0x1 and 0x11111110 & 0x00000001 is 0x0
Essentially, it is testing the last bit on the number, which the bit determining parity.
Final edit:
I got the precedence wrong, thanks to the comments for pointing it out. Here is the real precedence.
First, we compute i%8.
The result could be 0, 1, 2, 3, 4, 5, 6, 7.
Then, we shift the char by the result, which is maximum 7. That means the i % 8 th bit is now the least significant bit.
Then, we check if the original i % 8 bit is set (equals one) or not. If it is, return 1. Else, return 0.
This function returns the value of a specific bit in a char array as the integer 0 or 1.
addr is the pointer to the first char.
i is the index to the bit. 8 bits are commonly stored in a char.
First, the char at the correct offset is fetched:
char char_in_chain = addr [ i / 8 ] ;
i / 8 divides i by 8, ignoring the remainder. For example, any value in the range from 24 to 31 gives 3 as the result.
This result is used as the index to the char in the array.
Next and finally, the bit is obtained and returned:
return char_in_chain >> i%8 & 0x1;
Let's just look at the expression char_in_chain >> i%8 & 0x1.
It is confusing, because it does not show which operation is done in what sequence. Therefore, I duplicate it with appropriate parentheses: (char_in_chain >> (i % 8)) & 0x1. The rules (operation precedence) are given by the C standard.
First, the remainder of the division of i by 8 is calculated. This is used to right-shift the obtained char_in_chain. Now the interesting bit is in the least significant bit. Finally, this bit is "masked" with the binary AND operator and the second operand 0x1. BTW, there is no need to mark this constant as hex.
Example:
The array contains the bytes 0x5A, 0x23, and 0x42. The index of the bit to retrieve is 13.
i as given as argument is 13.
i / 8 gives 13 / 8 = 1, remainder ignored.
addr[1] returns 0x23, which is stored in char_in_chain.
i % 8 gives 5 (13 / 8 = 1, remainder 5).
0x23 is binary 0b00100011, and right-shifted by 5 gives 0b00000001.
0b00000001 ANDed with 0b00000001 gives 0b00000001.
The value returned is 1.
Note: If more is not clear, feel free to comment.
What the various operators do is explained by any C book, so I won't address that here. To instead analyse the code step by step...
The function and types used:
int as return type is an indication of the programmer being inexperienced at writing hardware-related code. We should always avoid signed types for such purposes. An experienced programmer would have used an unsigned type, like for example uint8_t. (Or in this specific case maybe even bool, depending on what the data is supposed to represent.)
n_b is a rubbish name, we should obviously never give an identifier such a nondescript name. get_bit or similar would have been a better name.
char* is, again, an indication of the programmer being inexperienced. char is particularly problematic when dealing with raw data, since we can't even know if it is signed or unsigned, it depends on which compiler that is used. Had the raw data contained a value of 0x80 or larger and char was negative, we would have gotten a negative type. And then right shifting a negative value is also problematic, since that behavior too is compiler-specific.
char* is proof of the programmer lacking the fundamental knowledge of const correctness. The function does not modify this parameter so it should have been const qualified. Good code would use const uint8_t* addr.
int i is not really incorrect, the signedness doesn't really matter. But good programming practice would have used an unsigned type or even size_t.
With types unsloppified and corrected, the function might look like this:
#include <stdint.h>
uint8_t get_bit (const uint8_t* addr, size_t i ) {
uint8_t char_in_chain = addr [ i / 8 ] ;
return char_in_chain >> i%8 & 0x1;
}
This is still somewhat problematic, because the average C programmer might not remember the precedence of >> vs % vs & on top of their head. It happens to be % over >> over &, but lets write the code a bit more readable still by making precedence explicit: (char_in_chain >> (i%8)) & 0x1.
Then I would question if the local variable really adds anything to readability. Not really, we might as well write:
uint8_t get_bit (const uint8_t* addr, size_t i ) {
return ((addr[i/8]) >> (i%8)) & 0x1;
}
As for what this code actually does: this happens to be a common design pattern for how to access a specific bit in a raw bit-field.
Any bit-field in C may be accessed as an array of bytes.
Bit number n in that bit-field, will be found at byte n/8.
Inside that byte, the bit will be located at n%8.
Bit masking in C is most readably done as data & (1u << bit). Which can be obfuscated as somewhat equivalent but less readable (data >> bit) & 1u, where the masked bit ends up in the LSB.
For example lets assume we have 64 bits of raw data. Bits are always enumerated from 0 to 63 and bytes (just like any C array) from index 0. We want to access bit 33. Then 33/8 integer division = 4.
So byte[4]. Bit 33 will be found at 33%8 = 1. So we can obtain the value of bit 33 from ordinary bit masking byte[33/8] & (1u << (bit%8)). Or similarly, (byte[33/8] >> (bit%8)) & 1u
An alternative, more readable version of it all:
bool is_bit_set (const uint8_t* data, size_t bit)
{
uint8_t byte = data [bit / 8u];
size_t mask = 1u << (bit % 8u);
return (byte & mask) != 0u;
}
(Strictly speaking we could as well do return byte & mask; since a boolean type is used, but it doesn't hurt to be explicit.)
I have a problem understanding this code. What I know is that we have passed a code into a assembler that has converted code into "byte code". Now I have a Virtual machine that is supposed to read this code. This function is supposed to read the first byte code instruction. I don't understand what is happening in this code. I guess we are trying to read this byte code but don't understand how it is done.
static int32_t bytecode_to_int32(const uint8_t *bytecode, size_t size)
{
int32_t result;
t_bool sign;
int i;
result = 0;
sign = (t_bool)(bytecode[0] & 0x80);
i = 0;
while (size)
{
if (sign)
result += ((bytecode[size - 1] ^ 0xFF) << (i++ * 8));
else
result += bytecode[size - 1] << (i++ * 8);
size--;
}
if (sign)
result = ~(result);
return (result);
}
This code is somewhat badly written, lots of operations on a single line and therefore containing various potential bugs. It looks brittle.
bytecode[0] & 0x80 Simply reads the MSB sign bit, assuming it's 2's complement or similar, then converts it to a boolean.
The loop iterates backwards from most significant byte to least significant.
If the sign was negative, the code will perform an XOR of the data byte with 0xFF. Basically inverting all bits in the data. The result of the XOR is an int.
The data byte (or the result of the above XOR) is then bit shifted i * 8 bits to the left. The data is always implicitly promoted to int, so in case i * 8 happens to give a result larger than INT_MAX, there's a fat undefined behavior bug here. It would be much safer practice to cast to uint32_t before the shift, carry out the shift, then convert to a signed type afterwards.
The resulting int is converted to int32_t - these could be the same type or different types depending on system.
i is incremented by 1, size is decremented by 1.
If sign was negative, the int32_t is inverted to some 2's complement negative number that's sign extended and all the data bits are inverted once more. Except all zeros that got shifted in with the left shift are also replaced by ones. If this is intentional or not, I cannot tell. So for example if you started with something like 0x0081 you now have something like 0xFFFF01FF. How that format makes sense, I have no idea.
My take is that the bytecode[size - 1] ^ 0xFF (which is equivalent to ~) was made to toggle the data bits, so that they would later toggle back to their original values when ~ is called later. A programmer has to document such tricks with comments, if they are anything close to competent.
Anyway, don't use this code. If the intention was merely to swap the byte order (endianess) of a 4 byte integer, then this code must be rewritten from scratch.
That's properly done as:
static int32_t big32_to_little32 (const uint8_t* bytes)
{
uint32_t result = (uint32_t)bytes[0] << 24 |
(uint32_t)bytes[1] << 16 |
(uint32_t)bytes[2] << 8 |
(uint32_t)bytes[3] << 0 ;
return (int32_t)result;
}
Anything more complicated than the above is highly questionable code. We need not worry about signs being a special case, the above code preserves the original signedness format.
So the A^0xFF toggles the bits set in A, so if you have 10101100 xored with 11111111.. it will become 01010011. I am not sure why they didn't use ~ here. The ^ is a xor operator, so you are xoring with 0xFF.
The << is a bitshift "up" or left. In other words, A<<1 is equivalent to multiplying A by 2.
the >> moves down so is equivalent to bitshifting right, or dividing by 2.
The ~ inverts the bits in a byte.
Note it's better to initialise variables at declaration it costs no additional processing whatsoever to do it that way.
sign = (t_bool)(bytecode[0] & 0x80); the sign in the number is stored in the 8th bit (or position 7 counting from 0), which is where the 0x80 is coming from. So it's literally checking if the signed bit is set in the first byte of bytecode, and if so then it stores it in the sign variable.
Essentially if it's unsigned then it's copying the bytes from from bytecode into result one byte at a time.
If the data is signed then it flips the bits then copies the bytes, then when it's done copying, it flips the bits back.
Personally with this kind of thing i prefer to get the data, stick in htons() format (network byte order) and then memcpy it to an allocated array, store it in a endian agnostic way, then when i retrieve the data i use ntohs() to convert it back to the format used by the computer. htons() and ntohs() are standard C functions and are used in networking and platform agnostic data formatting / storage / communication all the time.
This function is a very naive version of the function which converts form the big endian to little endian.
The parameter size is not needed as it works only with the 4 bytes data.
It can be much easier archived by the union punning (and it allows compilers to optimize it - in this case to the simple instruction):
#define SWAP(a,b,t) do{t c = (a); (a) = (b); (b) = c;}while(0)
int32_t my_bytecode_to_int32(const uint8_t *bytecode)
{
union
{
int32_t i32;
uint8_t b8[4];
}i32;
uint8_t b;
i32.b8[3] = *bytecode++;
i32.b8[2] = *bytecode++;
i32.b8[1] = *bytecode++;
i32.b8[0] = *bytecode++;
return i32.i32;
}
int main()
{
union {
int32_t i32;
uint8_t b8[4];
}i32;
uint8_t b;
i32.i32 = -4567;
SWAP(i32.b8[0], i32.b8[3], uint8_t);
SWAP(i32.b8[1], i32.b8[2], uint8_t);
printf("%d\n", bytecode_to_int32(i32.b8, 4));
i32.i32 = -34;
SWAP(i32.b8[0], i32.b8[3], uint8_t);
SWAP(i32.b8[1], i32.b8[2], uint8_t);
printf("%d\n", my_bytecode_to_int32(i32.b8));
}
https://godbolt.org/z/rb6Na5
If the purpose of the code is to sign-extend a 1-, 2-, 3-, or 4-byte sequence in network/big-endian byte order to a signed 32-bit int value, it's doing things the hard way and reimplementing the wheel along the way.
This can be broken down into a three-step process: convert the proper number of bytes to a 32-bit integer value, sign-extend bytes out to 32 bits, then convert that 32-bit value from big-endian to the host's byte order.
The "wheel" being reimplemented in this case is the the POSIX-standard ntohl() function that converts a 32-bit unsigned integer value in big-endian/network byte order to the local host's native byte order.
The first step I'd do is to convert 1, 2, 3, or 4 bytes into a uint32_t:
#include <stdint.h>
#include <limits.h>
#include <arpa/inet.h>
#include <errno.h>
// convert the `size` number of bytes starting at the `bytecode` address
// to a uint32_t value
static uint32_t bytecode_to_uint32( const uint8_t *bytecode, size_t size )
{
uint32_t result = 0;
switch ( size )
{
case 4:
result = bytecode[ 0 ] << 24;
case 3:
result += bytecode[ 1 ] << 16;
case 2:
result += bytecode[ 2 ] << 8;
case 1:
result += bytecode[ 3 ];
break;
default:
// error handling here
break;
}
return( result );
}
Then, sign-extend it (borrowing from this answer):
static uint32_t sign_extend_uint32( uint32_t in, size_t size );
{
if ( size == 4 )
{
return( in );
}
// being pedantic here - the existence of `[u]int32_t` pretty
// much ensures 8 bits/byte
size_t bits = size * CHAR_BIT;
uint32_t m = 1U << ( bits - 1 );
uint32_t result = ( in ^ m ) - m;
return ( result );
}
Put it all together:
static int32_t bytecode_to_int32( const uint8_t *bytecode, size_t size )
{
uint32_t result = bytecode_to_uint32( bytecode, size );
result = sign_extend_uint32( result, size );
// set endianness from network/big-endian to
// whatever this host's endianness is
result = ntohl( result );
// converting uint32_t here to signed int32_t
// can be subject to implementation-defined
// behavior
return( result );
}
Note that the conversion from uint32_t to int32_t implicitly performed by the return statement in the above code can result in implemenation-defined behavior as there can be uint32_t values that can not be mapped to int32_t values. See this answer.
Any decent compiler should optimize that well into inline functions.
I personally think this also needs much better error handling/input validation.
I would to implement a function like this:
int read_single_bit(unsigned char* buffer, unsigned int index)
where index is the offset of the bit that I would want to read.
How do I use bit shifting or masking to achieve this?
You might want to split this into three separate tasks:
Determining which char contains the bit that you're looking for.
Determining the bit offset into that char that you need to read.
Actually selecting that bit out of that char.
I'll leave parts (1) and (2) as exercises, since they're not too bad. For part (3), one trick you might find useful would be to do a bitwise AND between the byte in question and a byte with a single 1 bit at the index that you want. For example, suppose you want to get the fourth bit out of a byte. You could then do something like this:
Byte: 11011100
Mask: 00001000
----------------
AND: 00001000
So think about the following: how would you generate the mask that you need given that you know the bit index? And how would you convert the AND result back to a single bit?
Good luck!
buffer[index/8] & (1u<<(index%8))
should do it (that is, view buffer as a bit array and test the bit at index).
Similarly:
buffer[index/8] |= (1u<<(index%8))
should set the index-th bit.
Or you could store a table of the eight shift states of 1 and & against that
unsigned char bits[] = { 1u<<0, 1u<<1, 1u<<2, 1u<<3, 1u<<4, 1u<<5, 1u<<6, 1u<<7 };
If your compiler doesn't optimize those / and % to bit ops (more efficient), then:
unsigned_int / 8 == unsigned_int >> 3
unsigned_int % 8 == unsigned_int & 0x07 //0x07 == 0000 0111
so
buffer[index>>3] & (1u<<(index&0x07u)) //test
buffer[index>>3] |= (1u<<(index&0x07u)) //set
One possible implementation of your function might look like this:
int read_single_bit(unsigned char* buffer, unsigned int index)
{
unsigned char c = buffer[index / 8]; //getting the byte which contains the bit
unsigned int bit_position = index % 8; //getting the position of that bit within the byte
return ((c >> (7 - bit_position)) & 1);
//shifting that byte to the right with (7 - bit_position) will move the bit whose value you want to know at "the end" of the byte.
//then, by doing bitwise AND with the new byte and 1 (whose binary representation is 00000001) will yield 1 or 0, depending on the value of the bit you need.
}
I want to create a very large array on which I write '0's and '1's. I'm trying to simulate a physical process called random sequential adsorption, where units of length 2, dimers, are deposited onto an n-dimensional lattice at a random location, without overlapping each other. The process stops when there is no more room left on the lattice for depositing more dimers (lattice is jammed).
Initially I start with a lattice of zeroes, and the dimers are represented by a pair of '1's. As each dimer is deposited, the site on the left of the dimer is blocked, due to the fact that the dimers cannot overlap. So I simulate this process by depositing a triple of '1's on the lattice. I need to repeat the entire simulation a large number of times and then work out the average coverage %.
I've already done this using an array of chars for 1D and 2D lattices. At the moment I'm trying to make the code as efficient as possible, before working on the 3D problem and more complicated generalisations.
This is basically what the code looks like in 1D, simplified:
int main()
{
/* Define lattice */
array = (char*)malloc(N * sizeof(char));
total_c = 0;
/* Carry out RSA multiple times */
for (i = 0; i < 1000; i++)
rand_seq_ads();
/* Calculate average coverage efficiency at jamming */
printf("coverage efficiency = %lf", total_c/1000);
return 0;
}
void rand_seq_ads()
{
/* Initialise array, initial conditions */
memset(a, 0, N * sizeof(char));
available_sites = N;
count = 0;
/* While the lattice still has enough room... */
while(available_sites != 0)
{
/* Generate random site location */
x = rand();
/* Deposit dimer (if site is available) */
if(array[x] == 0)
{
array[x] = 1;
array[x+1] = 1;
count += 1;
available_sites += -2;
}
/* Mark site left of dimer as unavailable (if its empty) */
if(array[x-1] == 0)
{
array[x-1] = 1;
available_sites += -1;
}
}
/* Calculate coverage %, and add to total */
c = count/N
total_c += c;
}
For the actual project I'm doing, it involves not just dimers but trimers, quadrimers, and all sorts of shapes and sizes (for 2D and 3D).
I was hoping that I would be able to work with individual bits instead of bytes, but I've been reading around and as far as I can tell you can only change 1 byte at a time, so either I need to do some complicated indexing or there is a simpler way to do it?
Thanks for your answers
If I am not too late, this page gives awesome explanation with examples.
An array of int can be used to deal with array of bits. Assuming size of int to be 4 bytes, when we talk about an int, we are dealing with 32 bits. Say we have int A[10], means we are working on 10*4*8 = 320 bits and following figure shows it: (each element of array has 4 big blocks, each of which represent a byte and each of the smaller blocks represent a bit)
So, to set the kth bit in array A:
// NOTE: if using "uint8_t A[]" instead of "int A[]" then divide by 8, not 32
void SetBit( int A[], int k )
{
int i = k/32; //gives the corresponding index in the array A
int pos = k%32; //gives the corresponding bit position in A[i]
unsigned int flag = 1; // flag = 0000.....00001
flag = flag << pos; // flag = 0000...010...000 (shifted k positions)
A[i] = A[i] | flag; // Set the bit at the k-th position in A[i]
}
or in the shortened version
void SetBit( int A[], int k )
{
A[k/32] |= 1 << (k%32); // Set the bit at the k-th position in A[i]
}
similarly to clear kth bit:
void ClearBit( int A[], int k )
{
A[k/32] &= ~(1 << (k%32));
}
and to test if the kth bit:
int TestBit( int A[], int k )
{
return ( (A[k/32] & (1 << (k%32) )) != 0 ) ;
}
As said above, these manipulations can be written as macros too:
// Due order of operation wrap 'k' in parentheses in case it
// is passed as an equation, e.g. i + 1, otherwise the first
// part evaluates to "A[i + (1/32)]" not "A[(i + 1)/32]"
#define SetBit(A,k) ( A[(k)/32] |= (1 << ((k)%32)) )
#define ClearBit(A,k) ( A[(k)/32] &= ~(1 << ((k)%32)) )
#define TestBit(A,k) ( A[(k)/32] & (1 << ((k)%32)) )
typedef unsigned long bfield_t[ size_needed/sizeof(long) ];
// long because that's probably what your cpu is best at
// The size_needed should be evenly divisable by sizeof(long) or
// you could (sizeof(long)-1+size_needed)/sizeof(long) to force it to round up
Now, each long in a bfield_t can hold sizeof(long)*8 bits.
You can calculate the index of a needed big by:
bindex = index / (8 * sizeof(long) );
and your bit number by
b = index % (8 * sizeof(long) );
You can then look up the long you need and then mask out the bit you need from it.
result = my_field[bindex] & (1<<b);
or
result = 1 & (my_field[bindex]>>b); // if you prefer them to be in bit0
The first one may be faster on some cpus or may save you shifting back up of you need
to perform operations between the same bit in multiple bit arrays. It also mirrors
the setting and clearing of a bit in the field more closely than the second implemention.
set:
my_field[bindex] |= 1<<b;
clear:
my_field[bindex] &= ~(1<<b);
You should remember that you can use bitwise operations on the longs that hold the fields
and that's the same as the operations on the individual bits.
You'll probably also want to look into the ffs, fls, ffc, and flc functions if available. ffs should always be avaiable in strings.h. It's there just for this purpose -- a string of bits.
Anyway, it is find first set and essentially:
int ffs(int x) {
int c = 0;
while (!(x&1) ) {
c++;
x>>=1;
}
return c; // except that it handles x = 0 differently
}
This is a common operation for processors to have an instruction for and your compiler will probably generate that instruction rather than calling a function like the one I wrote. x86 has an instruction for this, by the way. Oh, and ffsl and ffsll are the same function except take long and long long, respectively.
You can use & (bitwise and) and << (left shift).
For example, (1 << 3) results in "00001000" in binary. So your code could look like:
char eightBits = 0;
//Set the 5th and 6th bits from the right to 1
eightBits &= (1 << 4);
eightBits &= (1 << 5);
//eightBits now looks like "00110000".
Then just scale it up with an array of chars and figure out the appropriate byte to modify first.
For more efficiency, you could define a list of bitfields in advance and put them in an array:
#define BIT8 0x01
#define BIT7 0x02
#define BIT6 0x04
#define BIT5 0x08
#define BIT4 0x10
#define BIT3 0x20
#define BIT2 0x40
#define BIT1 0x80
char bits[8] = {BIT1, BIT2, BIT3, BIT4, BIT5, BIT6, BIT7, BIT8};
Then you avoid the overhead of the bit shifting and you can index your bits, turning the previous code into:
eightBits &= (bits[3] & bits[4]);
Alternatively, if you can use C++, you could just use an std::vector<bool> which is internally defined as a vector of bits, complete with direct indexing.
bitarray.h:
#include <inttypes.h> // defines uint32_t
//typedef unsigned int bitarray_t; // if you know that int is 32 bits
typedef uint32_t bitarray_t;
#define RESERVE_BITS(n) (((n)+0x1f)>>5)
#define DW_INDEX(x) ((x)>>5)
#define BIT_INDEX(x) ((x)&0x1f)
#define getbit(array,index) (((array)[DW_INDEX(index)]>>BIT_INDEX(index))&1)
#define putbit(array, index, bit) \
((bit)&1 ? ((array)[DW_INDEX(index)] |= 1<<BIT_INDEX(index)) \
: ((array)[DW_INDEX(index)] &= ~(1<<BIT_INDEX(index))) \
, 0 \
)
Use:
bitarray_t arr[RESERVE_BITS(130)] = {0, 0x12345678,0xabcdef0,0xffff0000,0};
int i = getbit(arr,5);
putbit(arr,6,1);
int x=2; // the least significant bit is 0
putbit(arr,6,x); // sets bit 6 to 0 because 2&1 is 0
putbit(arr,6,!!x); // sets bit 6 to 1 because !!2 is 1
EDIT the docs:
"dword" = "double word" = 32-bit value (unsigned, but that's not really important)
RESERVE_BITS: number_of_bits --> number_of_dwords
RESERVE_BITS(n) is the number of 32-bit integers enough to store n bits
DW_INDEX: bit_index_in_array --> dword_index_in_array
DW_INDEX(i) is the index of dword where the i-th bit is stored.
Both bit and dword indexes start from 0.
BIT_INDEX: bit_index_in_array --> bit_index_in_dword
If i is the number of some bit in the array, BIT_INDEX(i) is the number
of that bit in the dword where the bit is stored.
And the dword is known via DW_INDEX().
getbit: bit_array, bit_index_in_array --> bit_value
putbit: bit_array, bit_index_in_array, bit_value --> 0
getbit(array,i) fetches the dword containing the bit i and shifts the dword right, so that the bit i becomes the least significant bit. Then, a bitwise and with 1 clears all other bits.
putbit(array, i, v) first of all checks the least significant bit of v; if it is 0, we have to clear the bit, and if it is 1, we have to set it.
To set the bit, we do a bitwise or of the dword that contains the bit and the value of 1 shifted left by bit_index_in_dword: that bit is set, and other bits do not change.
To clear the bit, we do a bitwise and of the dword that contains the bit and the bitwise complement of 1 shifted left by bit_index_in_dword: that value has all bits set to one except the only zero bit in the position that we want to clear.
The macro ends with , 0 because otherwise it would return the value of dword where the bit i is stored, and that value is not meaningful. One could also use ((void)0).
It's a trade-off:
(1) use 1 byte for each 2 bit value - simple, fast, but uses 4x memory
(2) pack bits into bytes - more complex, some performance overhead, uses minimum memory
If you have enough memory available then go for (1), otherwise consider (2).
I want to shift the contents of an array of bytes by 12-bit to the left.
For example, starting with this array of type uint8_t shift[10]:
{0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x0A, 0xBC}
I'd like to shift it to the left by 12-bits resulting in:
{0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0xAB, 0xC0, 0x00}
Hurray for pointers!
This code works by looking ahead 12 bits for each byte and copying the proper bits forward. 12 bits is the bottom half (nybble) of the next byte and the top half of 2 bytes away.
unsigned char length = 10;
unsigned char data[10] = {0x0,0x0,0x0,0x0,0x0,0x0,0x0,0x0,0x0A,0xBC};
unsigned char *shift = data;
while (shift < data+(length-2)) {
*shift = (*(shift+1)&0x0F)<<4 | (*(shift+2)&0xF0)>>4;
shift++;
}
*(data+length-2) = (*(data+length-1)&0x0F)<<4;
*(data+length-1) = 0x00;
Justin wrote:
#Mike, your solution works, but does not carry.
Well, I'd say a normal shift operation does just that (called overflow), and just lets the extra bits fall off the right or left. It's simple enough to carry if you wanted to - just save the 12 bits before you start to shift. Maybe you want a circular shift, to put the overflowed bits back at the bottom? Maybe you want to realloc the array and make it larger? Return the overflow to the caller? Return a boolean if non-zero data was overflowed? You'd have to define what carry means to you.
unsigned char overflow[2];
*overflow = (*data&0xF0)>>4;
*(overflow+1) = (*data&0x0F)<<4 | (*(data+1)&0xF0)>>4;
while (shift < data+(length-2)) {
/* normal shifting */
}
/* now would be the time to copy it back if you want to carry it somewhere */
*(data+length-2) = (*(data+length-1)&0x0F)<<4 | (*(overflow)&0x0F);
*(data+length-1) = *(overflow+1);
/* You could return a 16-bit carry int,
* but endian-ness makes that look weird
* if you care about the physical layout */
unsigned short carry = *(overflow+1)<<8 | *overflow;
Here's my solution, but even more importantly my approach to solving the problem.
I approached the problem by
drawing the memory cells and drawing arrows from the destination to the source.
made a table showing the above drawing.
labeling each row in the table with the relative byte address.
This showed me the pattern:
let iL be the low nybble (half byte) of a[i]
let iH be the high nybble of a[i]
iH = (i+1)L
iL = (i+2)H
This pattern holds for all bytes.
Translating into C, this means:
a[i] = (iH << 4) OR iL
a[i] = ((a[i+1] & 0x0f) << 4) | ((a[i+2] & 0xf0) >> 4)
We now make three more observations:
since we carry out the assignments left to right, we don't need to store any values in temporary variables.
we will have a special case for the tail: all 12 bits at the end will be zero.
we must avoid reading undefined memory past the array. since we never read more than a[i+2], this only affects the last two bytes
So, we
handle the general case by looping for N-2 bytes and performing the general calculation above
handle the next to last byte by it by setting iH = (i+1)L
handle the last byte by setting it to 0
given a with length N, we get:
for (i = 0; i < N - 2; ++i) {
a[i] = ((a[i+1] & 0x0f) << 4) | ((a[i+2] & 0xf0) >> 4);
}
a[N-2] = (a[N-1) & 0x0f) << 4;
a[N-1] = 0;
And there you have it... the array is shifted left by 12 bits. It could easily be generalized to shifting N bits, noting that there will be M assignment statements where M = number of bits modulo 8, I believe.
The loop could be made more efficient on some machines by translating to pointers
for (p = a, p2=a+N-2; p != p2; ++p) {
*p = ((*(p+1) & 0x0f) << 4) | (((*(p+2) & 0xf0) >> 4);
}
and by using the largest integer data type supported by the CPU.
(I've just typed this in, so now would be a good time for somebody to review the code, especially since bit twiddling is notoriously easy to get wrong.)
Lets make it the best way to shift N bits in the array of 8 bit integers.
N - Total number of bits to shift
F = (N / 8) - Full 8 bit integers shifted
R = (N % 8) - Remaining bits that need to be shifted
I guess from here you would have to find the most optimal way to make use of this data to move around ints in an array. Generic algorithms would be to apply the full integer shifts by starting from the right of the array and moving each integer F indexes. Zero fill the newly empty spaces. Then finally perform an R bit shift on all of the indexes, again starting from the right.
In the case of shifting 0xBC by R bits you can calculate the overflow by doing a bitwise AND, and the shift using the bitshift operator:
// 0xAB shifted 4 bits is:
(0xAB & 0x0F) >> 4 // is the overflow (0x0A)
0xAB << 4 // is the shifted value (0xB0)
Keep in mind that the 4 bits is just a simple mask: 0x0F or just 0b00001111. This is easy to calculate, dynamically build, or you can even use a simple static lookup table.
I hope that is generic enough. I'm not good with C/C++ at all so maybe someone can clean up my syntax or be more specific.
Bonus: If you're crafty with your C you might be able to fudge multiple array indexes into a single 16, 32, or even 64 bit integer and perform the shifts. But that is prabably not very portable and I would recommend against this. Just a possible optimization.
Here a working solution, using temporary variables:
void shift_4bits_left(uint8_t* array, uint16_t size)
{
int i;
uint8_t shifted = 0x00;
uint8_t overflow = (0xF0 & array[0]) >> 4;
for (i = (size - 1); i >= 0; i--)
{
shifted = (array[i] << 4) | overflow;
overflow = (0xF0 & array[i]) >> 4;
array[i] = shifted;
}
}
Call this function 3 times for a 12-bit shift.
Mike's solution maybe faster, due to the use of temporary variables.
The 32 bit version... :-) Handles 1 <= count <= num_words
#include <stdio.h>
unsigned int array[] = {0x12345678,0x9abcdef0,0x12345678,0x9abcdef0,0x66666666};
int main(void) {
int count;
unsigned int *from, *to;
from = &array[0];
to = &array[0];
count = 5;
while (count-- > 1) {
*to++ = (*from<<12) | ((*++from>>20)&0xfff);
};
*to = (*from<<12);
printf("%x\n", array[0]);
printf("%x\n", array[1]);
printf("%x\n", array[2]);
printf("%x\n", array[3]);
printf("%x\n", array[4]);
return 0;
}
#Joseph, notice that the variables are 8 bits wide, while the shift is 12 bits wide. Your solution works only for N <= variable size.
If you can assume your array is a multiple of 4 you can cast the array into an array of uint64_t and then work on that. If it isn't a multiple of 4, you can work in 64-bit chunks on as much as you can and work on the remainder one by one.
This may be a bit more coding, but I think it's more elegant in the end.
There are a couple of edge-cases which make this a neat problem:
the input array might be empty
the last and next-to-last bits need to be treated specially, because they have zero bits shifted into them
Here's a simple solution which loops over the array copying the low-order nibble of the next byte into its high-order nibble, and the high-order nibble of the next-next (+2) byte into its low-order nibble. To save dereferencing the look-ahead pointer twice, it maintains a two-element buffer with the "last" and "next" bytes:
void shl12(uint8_t *v, size_t length) {
if (length == 0) {
return; // nothing to do
}
if (length > 1) {
uint8_t last_byte, next_byte;
next_byte = *(v + 1);
for (size_t i = 0; i + 2 < length; i++, v++) {
last_byte = next_byte;
next_byte = *(v + 2);
*v = ((last_byte & 0x0f) << 4) | (((next_byte) & 0xf0) >> 4);
}
// the next-to-last byte is half-empty
*(v++) = (next_byte & 0x0f) << 4;
}
// the last byte is always empty
*v = 0;
}
Consider the boundary cases, which activate successively more parts of the function:
When length is zero, we bail out without touching memory.
When length is one, we set the one and only element to zero.
When length is two, we set the high-order nibble of the first byte to low-order nibble of the second byte (that is, bits 12-16), and the second byte to zero. We don't activate the loop.
When length is greater than two we hit the loop, shuffling the bytes across the two-element buffer.
If efficiency is your goal, the answer probably depends largely on your machine's architecture. Typically you should maintain the two-element buffer, but handle a machine word (32/64 bit unsigned integer) at a time. If you're shifting a lot of data it will be worthwhile treating the first few bytes as a special case so that you can get your machine word pointers word-aligned. Most CPUs access memory more efficiently if the accesses fall on machine word boundaries. Of course, the trailing bytes have to be handled specially too so you don't touch memory past the end of the array.