I have some code below that is supposed to be converting a C (Arduino) 8-bit byte array to a 16-bit int array, but it only seems to partially work. I'm not sure what I'm doing wrong.
The byte array is in little endian byte order. How do I convert it to an int (two bytes per enty) array?
In layman's terms, I want to merge every two bytes.
Currently it is outputting for an input BYTE ARRAY of: {0x10, 0x00, 0x00, 0x00, 0x30, 0x00}. The output INT ARRAY is: {1,0,0}. The output should be an INT ARRAY is: {1,0,3}.
The code below is what I currently have:
I wrote this function based on a solution in Stack Overflow question Convert bytes in a C array as longs.
I also have this solution based off the same code which works fine for byte array to long (32-bits) array http://pastebin.com/TQzyTU2j.
/**
* Convert the retrieved bytes into a set of 16 bit ints
**/
int * byteA2IntA(byte * byte_slice, int sizeOfB, int * ret_array){
//Variable that stores the addressed int to be stored in SRAM
int currentInt;
int sizeOfI = sizeOfB / 2;
if(sizeOfB % 2 != 0) ++sizeOfI;
for(int i = 0; i < sizeOfB; i+=2){
currentInt = 0;
if(byte_slice[i]=='\0') {
break;
}
if(i + 1 < sizeOfB)
currentInt = (currentInt << 8) + byte_slice[i+1];
currentInt = (currentInt << 8) + byte_slice[i+0];
*ret_array = currentInt;
ret_array++;
}
//Pointer to the return array in the parent scope.
return ret_array;
}
What is the meaning of this line of code?
if(i + 1 < sizeOfB) currentInt = (currentInt << 8) + byte_slice[i+1];
Here currentInt is always 0 and 0 << 8 = 0.
Also what you do is, for each couple of bytes (let me call them uint8_t from now on), you pack an int (let me call it uint16_t from now on) by doing the following:
You take the rightmost uint8_t
You shift it 8 positions to the left
You add the leftmost uint8_t
Is this really what you want?
Supposing you have byte_slice[] = {1, 2}, you pack a 16 bit integer with the value 513 (2<<8 + 1)!
Also, you don't need to return the pointer to the array of uint16_t as the caller has already provided it to the function.
If you use the return of your function, as Joachim said, you get a pointer starting from a position of the uint16_t array which is not position [0].
Vincenzo has a point (or two), you need to be clear what you're trying to do;
Combine two bytes to one 16-bit int, one byte being the MSB and one byte being the LSB
int16 result = (byteMSB << 8) | byteLSB;
Convert an array of bytes into 16-bit
for(i = 0; i < num_of_bytes; i++)
{
myint16array[i] = mybytearray[i];
}
Copy an array of data into another one
memcpy(dest, src, num_bytes);
That will (probably, platform/compiler dependent) have the same effect as my 1st example.
Also, beware of using ints as that suggests signed values, use uints, safer and probably faster.
The problem is most likely that you increase ret_array and then return it. When you return it, it will point to one place beyond the destination array.
Save the pointer at the start of the function, and use that pointer instead.
Consider using a struct. This is kind of a hack, though.
Off the top of my head it would look like this.
struct customINT16 {
byte ByteHigh;
byte ByteLow;
}
So in your case you would write:
struct customINT16 myINT16;
myINT16.ByteHigh = BYTEARRAY[0];
myINT16.ByteLow = BYTEARRAY[1];
You'll have to go through a pointer to cast it, though:
intpointer = (int*)(&myINT16);
INTARRAY[0] = *intpointer;
Related
I am trying to speed up a program with bitshifts.
For doing that I am allocating some space for the bits to store.
Now I have to set the bit in the middle to 1.
So for example I want:
00000010000000
to be in my memory. (The length is defined by the user on run-time).
Here is my Code:
int *state = malloc(b); // Where 'b' is a user defnied value
*state |= (1 << b);
But when I print out the bits I get something like this (for b = 10):
000000000010000000000000000000000000000000100000000000000000000000000000001000000000000000000000000
So it keeps placing many 1's instead of one 1 in the middle.
What am I doing wrong here?
I belived it has to do with me using int, maybe the workaround would be not using int. But I don't know how to allocate a bitarray in another way.
This is how I print the bits (requestet in comments):
64 for(i =0; i < b; i++){
65 printf("%d", ((*state & (1 << i)) >> i));
66 }
You are mixing up bits and bytes.
To allocate b bits, you need
int *state = malloc(sizeof(int)*((b+sizeof(int)-1)/sizeof(int))); // Where 'b' is a user defnied value
To set a bit in this array, decompose into int position and bit position:
state[b/sizeof(int)] |= (1 << (b % sizeof(int)));
If you are going to work with single bits, it might be easier to use unsigned char instead of int as the base array.
Edit: just changed the allocation from (b+sizeof(int)-1)/sizeof(int) -- this calculates the required byte size. Of course you need to allocate room for int sized members, so the allocated memory size is the number of bytes times sizeof(int).
I have a char* string coming in. I need to store it accordingly.
The string can be any of those values { UK, GD, BD, ER, WR, FL}
If I want to keep them as enumerated type, which data type is the best to use. Like for 6 values three bits is enough, but how to store three bits in C?
What you want is a Bit Field:
typedef struct {
unsigned char val : 2; //use 2 bits
unsigned char : 6; // remaining 6 bits
} valContainer;
...
valContainer x;
x.val = GD;
Do note that there isn't really a way to store less than one byte, as the definition of a byte is the smallest amount of memory the computer can address. This is just a method of having names associated with different bits in a byte.
Also, of course, 2 bits is not enough for 6 values (2 bits hold 4 distinct values). So you really want at least 3 bits (8 distinct values).
Just store them as an unsigned short. Unless you're storing other things in your struct to fill out a whole word, you're WAY prematurely optimizing. The compiler will have to pad out your data anyway.
As the answer by Eric Finn suggests, you can use bit fields to store a data element of 3 bits. However, this is only good if you have something else to store in the same byte.
struct {
unsigned char value: 3;
unsigned char another: 4;
unsigned char yet_another: 5;
// 12 bits declared so far; 4 more "padding" bits are unusable
} whatever;
If you want to store an array of many such small elements, you have to do it in a different way, for example, clumping 10 elements in each 32-bit word.
int n = ...; // number of elements to store
uint32_t *data = calloc(n / 10, sizeof(*data));
for (int i = 0; i < n; i++)
{
int value = read_string_and_convert_to_int();
data[i / 10] &= ~(7 << (i % 10 * 3));
data[i / 10] |= value << (i % 10 * 3);
}
If you want to have only one element (or a few), just use enum or int.
What is the best way to implement a bitwise memmove? The method should take an additional destination and source bit-offset and the count should be in bits too.
I saw that ARM provides a non-standard _membitmove, which does exactly what I need, but I couldn't find its source.
Bind's bitset includes isc_bitstring_copy, but it's not efficient
I'm aware that the C standard library doesn't provide such a method, but I also couldn't find any third-party code providing a similar method.
Assuming "best" means "easiest", you can copy bits one by one. Conceptually, an address of a bit is an object (struct) that has a pointer to a byte in memory and an index of a bit in the byte.
struct pointer_to_bit
{
uint8_t* p;
int b;
};
void membitmovebl(
void *dest,
const void *src,
int dest_offset,
int src_offset,
size_t nbits)
{
// Create pointers to bits
struct pointer_to_bit d = {dest, dest_offset};
struct pointer_to_bit s = {src, src_offset};
// Bring the bit offsets to range (0...7)
d.p += d.b / 8; // replace division by right-shift if bit offset can be negative
d.b %= 8; // replace "%=8" by "&=7" if bit offset can be negative
s.p += s.b / 8;
s.b %= 8;
// Determine whether it's OK to loop forward
if (d.p < s.p || d.p == s.p && d.b <= s.b)
{
// Copy bits one by one
for (size_t i = 0; i < nbits; i++)
{
// Read 1 bit
int bit = (*s.p >> s.b) & 1;
// Write 1 bit
*d.p &= ~(1 << d.b);
*d.p |= bit << d.b;
// Advance pointers
if (++s.b == 8)
{
s.b = 0;
++s.p;
}
if (++d.b == 8)
{
d.b = 0;
++d.p;
}
}
}
else
{
// Copy stuff backwards - essentially the same code but ++ replaced by --
}
}
If you want to write a version optimized for speed, you will have to do copying by bytes (or, better, words), unroll loops, and handle a number of special cases (memmove does that; you will have to do more because your function is more complicated).
P.S. Oh, seeing that you call isc_bitstring_copy inefficient, you probably want the speed optimization. You can use the following idea:
Start copying bits individually until the destination is byte-aligned (d.b == 0). Then, it is easy to copy 8 bits at once, doing some bit twiddling. Do this until there are less than 8 bits left to copy; then continue copying bits one by one.
// Copy 8 bits from s to d and advance pointers
*d.p = *s.p++ >> s.b;
*d.p++ |= *s.p << (8 - s.b);
P.P.S Oh, and seeing your comment on what you are going to use the code for, you don't really need to implement all the versions (byte/halfword/word, big/little-endian); you only want the easiest one - the one working with words (uint32_t).
Here is a partial implementation (not tested). There are obvious efficiency and usability improvements.
Copy n bytes from src to dest (not overlapping src), and shift bits at dest rightwards by bit bits, 0 <= bit <= 7. This assumes that the least significant bits are at the right of the bytes
void memcpy_with_bitshift(unsigned char *dest, unsigned char *src, size_t n, int bit)
{
int i;
memcpy(dest, src, n);
for (i = 0; i < n; i++) {
dest[i] >> bit;
}
for (i = 0; i < n; i++) {
dest[i+1] |= (src[i] << (8 - bit));
}
}
Some improvements to be made:
Don't overwrite first bit bits at beginning of dest.
Merge loops
Have a way to copy a number of bits not divisible by 8
Fix for >8 bits in a char
As part of my CS course I've been given some functions to use. One of these functions takes a pointer to unsigned chars to write some data to a file (I have to use this function, so I can't just make my own purpose built function that works differently BTW). I need to write an array of integers whose values can be up to 4095 using this function (that only takes unsigned chars).
However am I right in thinking that an unsigned char can only have a max value of 256 because it is 1 byte long? I therefore need to use 4 unsigned chars for every integer? But casting doesn't seem to work with larger values for the integer. Does anyone have any idea how best to convert an array of integers to unsigned chars?
Usually an unsigned char holds 8 bits, with a max value of 255. If you want to know this for your particular compiler, print out CHAR_BIT and UCHAR_MAX from <limits.h> You could extract the individual bytes of a 32 bit int,
#include <stdint.h>
void
pack32(uint32_t val,uint8_t *dest)
{
dest[0] = (val & 0xff000000) >> 24;
dest[1] = (val & 0x00ff0000) >> 16;
dest[2] = (val & 0x0000ff00) >> 8;
dest[3] = (val & 0x000000ff) ;
}
uint32_t
unpack32(uint8_t *src)
{
uint32_t val;
val = src[0] << 24;
val |= src[1] << 16;
val |= src[2] << 8;
val |= src[3] ;
return val;
}
Unsigned char generally has a value of 1 byte, therefore you can decompose any other type to an array of unsigned chars (eg. for a 4 byte int you can use an array of 4 unsigned chars). Your exercise is probably about generics. You should write the file as a binary file using the fwrite() function, and just write byte after byte in the file.
The following example should write a number (of any data type) to the file. I am not sure if it works since you are forcing the cast to unsigned char * instead of void *.
int homework(unsigned char *foo, size_t size)
{
int i;
// open file for binary writing
FILE *f = fopen("work.txt", "wb");
if(f == NULL)
return 1;
// should write byte by byte the data to the file
fwrite(foo+i, sizeof(char), size, f);
fclose(f);
return 0;
}
I hope the given example at least gives you a starting point.
Yes, you're right; a char/byte only allows up to 8 distinct bits, so that is 2^8 distinct numbers, which is zero to 2^8 - 1, or zero to 255. Do something like this to get the bytes:
int x = 0;
char* p = (char*)&x;
for (int i = 0; i < sizeof(x); i++)
{
//Do something with p[i]
}
(This isn't officially C because of the order of declaration but whatever... it's more readable. :) )
Do note that this code may not be portable, since it depends on the processor's internal storage of an int.
If you have to write an array of integers then just convert the array into a pointer to char then run through the array.
int main()
{
int data[] = { 1, 2, 3, 4 ,5 };
size_t size = sizeof(data)/sizeof(data[0]); // Number of integers.
unsigned char* out = (unsigned char*)data;
for(size_t loop =0; loop < (size * sizeof(int)); ++loop)
{
MyProfSuperWrite(out + loop); // Write 1 unsigned char
}
}
Now people have mentioned that 4096 will fit in less bits than a normal integer. Probably true. Thus you can save space and not write out the top bits of each integer. Personally I think this is not worth the effort. The extra code to write the value and processes the incoming data is not worth the savings you would get (Maybe if the data was the size of the library of congress). Rule one do as little work as possible (its easier to maintain). Rule two optimize if asked (but ask why first). You may save space but it will cost in processing time and maintenance costs.
The part of the assignment of: integers whose values can be up to 4095 using this function (that only takes unsigned chars should be giving you a huge hint. 4095 unsigned is 12 bits.
You can store the 12 bits in a 16 bit short, but that is somewhat wasteful of space -- you are only using 12 of 16 bits of the short. Since you are dealing with more than 1 byte in the conversion of characters, you may need to deal with endianess of the result. Easiest.
You could also do a bit field or some packed binary structure if you are concerned about space. More work.
It sounds like what you really want to do is call sprintf to get a string representation of your integers. This is a standard way to convert from a numeric type to its string representation. Something like the following might get you started:
char num[5]; // Room for 4095
// Array is the array of integers, and arrayLen is its length
for (i = 0; i < arrayLen; i++)
{
sprintf (num, "%d", array[i]);
// Call your function that expects a pointer to chars
printfunc (num);
}
Without information on the function you are directed to use regarding its arguments, return value and semantics (i.e. the definition of its behaviour) it is hard to answer. One possibility is:
Given:
void theFunction(unsigned char* data, int size);
then
int array[SIZE_OF_ARRAY];
theFunction((insigned char*)array, sizeof(array));
or
theFunction((insigned char*)array, SIZE_OF_ARRAY * sizeof(*array));
or
theFunction((insigned char*)array, SIZE_OF_ARRAY * sizeof(int));
All of which will pass all of the data to theFunction(), but whether than makes any sense will depend on what theFunction() does.
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.