make the moving bits more efficient - c

in our company's latest project, we want to move a char arry to left half a byte, e.g.
char buf[] = {0x12, 0x34, 0x56, 0x78, 0x21}
we want to make the buf like
0x23, 0x45, 0x67, 0x82, 0x10
how do I make the process more efficient, can you make the time complexity less than O(N) if there are N bytes to be processed?
SOS...

Without more context, I would go even as far as questioning the need for an actual array. If you have 4 bytes, that can easily be represented using a uint32_t, and then you can perform an O(1) shift operation:
uint32_t x = 0x12345678;
uint32_t offByHalf = x << 4;
This way, you would replace array access with bit masking, like this:
array[i]
would be equivalent with
(x >> 8 * (3 - i)) & 0xff
And who knows, arithmetic may even be faster than memory access. But don't take my word for it, benchmark it.

No, if you want to actually shift the array, you'll need to hit every element at least once so it'll be O(n). There's no getting around that. You can do it with something like the following:
#include <stdio.h>
void shiftNybbleLeft (unsigned char *arr, size_t sz) {
for (int i = 1; i < sz; i++)
arr[i-1] = ((arr[i-1] & 0x0f) << 4) | (arr[i] >> 4);
arr[sz-1] = (arr[sz-1] & 0x0f) << 4;
}
int main (int argc, char *argv[]) {
unsigned char buf[] = {0x12, 0x34, 0x56, 0x78};
shiftNybbleLeft (buf, sizeof (buf));
for (int i = 0; i < sizeof (buf); i++)
printf ("0x%02x ", buf[i]);
putchar ('\n');
return 0;
}
which gives you:
0x23 0x45 0x67 0x80
That's not to say you can't make it more efficient (a). If you instead modify your extraction code so that it behaves differently, you can avoid the shifting operation.
In other words, don't shift the array, simply set an offset variable and use that to modify the extraction process. Examine the following code:
#include <stdio.h>
unsigned char getByte (unsigned char *arr, size_t index, size_t shiftSz) {
if ((shiftSz % 2) == 0)
return arr[index + shiftSz / 2];
return ((arr[index + shiftSz / 2] & 0x0f) << 4)
| (arr[index + shiftSz / 2 + 1] >> 4);
}
int main (int argc, char *argv[]) {
unsigned char buf[] = {0x12, 0x34, 0x56, 0x78};
//shiftNybbleLeft (buf, sizeof (buf));
for (int i = 0; i < 4; i++)
printf ("buf[1] with left shift %d nybbles -> 0x%02x\n",
i, getByte (buf, 1, i));
return 0;
}
With shiftSz set to 0, it's as if the array isn't shifted. By setting shiftSz to non-zero, an O(1) operation, getByte() will actually return the element as if you had shifted it by that amount. The output is as you would expect:
Index 1 with left shift 0 nybbles -> 0x34
Index 1 with left shift 1 nybbles -> 0x45
Index 1 with left shift 2 nybbles -> 0x56
Index 1 with left shift 3 nybbles -> 0x67
Now that may seem a contrived example (because it is) but there's ample precedent in using tricks like that to avoid potentially costly operations. You'd probably also want to add some bounds checking to catch problems with referencing outside the array.
Keep in mind that there's a trade-off. What you gain by not having to shift the array may be offset to some degree by the calculations done during extraction. Whether it's actually worth it depends on how you use the data. If the arrays is large but you don't extract that many values from it, this trick may be worth it.
As another example of using "tricks" to prevent costly operations, I've seen text editors that don't bother shifting the contents of lines either (when deleting a character for example). Instead they simply set the character to a 0 code point and take care of it when displaying the line (ignoring the 0 code points).
They'll generally clean up eventually but often in the background where it won't interfere with your editing speed.
(a) Though you may want to actually make sure this is necessary.
One of your comments stated that your arrays are about 500 entries in length and I can tell you that my not-supremely-grunty development box can shift that array one nybble to the left at the rate of about half a million times every single second.
So, even if your profiler states that a large proportion of time is being spent in there, that doesn't necessarily mean it's a large amount of time.
You should only look into optimising code if there's a specific, identified bottleneck.

I'll tackle the only objectively answerable part of the question, which is:
can you make the time complexity less than O(N) if there are N bytes to be processed?
If you need the entire output array then no, you cannot do better than O(N).
If you only need certain elements of the output array, then you can compute just those.

It may not compile well due to alignment, but you can try using a bitfield offset in a struct.
struct __attribute__((packed)) shifted{
char offset:4; // dump data
char data[N]; // rest of data
};
or on some systems
struct __attribute__((packed)) shifted{
char offset:4; // dump data
char data[N]; // rest of data
char last:4; // to make an even byte
};
struct shifted *shifted_buf=&buf;
//now operate on shifted_buf->data
Or you can try making it a union
union __attribute__((packed)) {
char old[N];
struct{
char offset:4;
char buf[N];
char last:4; // to make an even byte
}shifted;
}data;
The alternative would be to cast to an array of int and <<4 for each int, reducing it to N/4, but this is dependent on endianness.

Related

How do I find unsigned BB[] such that (char *) (BB + 1) = "Red Ross!"

I have this short code fragment:
unsigned BB[] = ???;
printf("%s\n", (char *) (BB + 1));
I want the output of that printf to be "Red Ross!". I do not know how to approach this kind of problem, I think it has to do with ASCII table.
Here are some assumptions that can be used:
32-bit little-endian platform
sizeof(char) == 1
sizeof(unsigned) == 4
Start by taking the target string and printing out the value of each byte:
char *str = "Red Ross!";
int i;
for (i=0; i<strlen(str); i++) {
printf("%02x ", str[i]);
}
This will tell you what values to use. Then you can use those values to populate your unsigned array.
Because the unsigned values are implemented at 32-bit little endian, each element can store 4 bytes, and those bytes needs to be stored in the reverse order they should be displayed in due to little-endian byte ordering.
For example, if you wanted to store 0x01, 0x02, 0x03, and 0x04 in that order in a single unsigned, you would do so as follows:
unsigned value = 0x04030201;
So using those two pieces of information, you can construct your unsigned array. Also, because you need to start with BB + 1, the value first element of the array doesn't matter.
Lastly, make sure the last element of the array is 0 so that your string is null terminated.
Well, given that you've sacrificed a lot of portability already (around sizes of built-in types, complementing schemes, and endianness), why not go the whole hog and use multicharacter constants ?
int main(void) {
unsigned BB[] = {[1] = ' deR', 'ssoR', '!'}; // little endian
printf("%s\n", (char *) (BB + 1));
}
will do it, where I've made some fairly commonplace assumptions about your compiler's implementation of multicharacter constants. Note the use of the designated initialiser [1].
Saves all that fishing around in your character tables.
Working example at https://ideone.com/9DiFgy
Taking a look at which codes each letter corresponds to in the ASCII table might be helpful, but not necessary. You can assign character literals to an integer just fine, by using bitwise OR and bit shift:
uint32_t x = ('A' << 24) | ('B' << 16) | ('C' << 8) | 'D';
This puts 'A' in the most significant byte. Where that is depends on endianess. On little endian, the above would result in "DCBA".
This should be enough to solve the assignment. Do remember that strings are null terminated, so you need to end the "string" with a zero.

Converting a byte array to an int array in C

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;

How to convert from integer to unsigned char in C, given integers larger than 256?

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.

Turn a large chunk of memory backwards, fast

I need to rewrite about 4KB of data in reverse order, at bit level (last bit of last byte becoming first bit of first byte), as fast as possible. Are there any clever sniplets to do it?
Rationale: The data is display contents of LCD screen in an embedded device that is usually positioned in a way that the screen is on your shoulders level. The screen has "6 o'clock" orientation, that is to be viewed from below - like lying flat or hanging above your eyes level. This is fixable by rotating the screen 180 degrees, but then I need to reverse the screen data (generated by library), which is 1 bit = 1 pixel, starting with upper left of the screen. The CPU isn't very powerful, and the device has enough work already, plus several frames a second would be desirable so performance is an issue; RAM not so much.
edit:
Single core, ARM 9 series. 64MB, (to be scaled down to 32MB later), Linux. The data is pushed from system memory to the LCD driver over 8-bit IO port.
The CPU is 32bit and performs much better at this word size than at byte level.
There's a classic way to do this. Let's say unsigned int is your 32-bit word. I'm using C99 because the restrict keyword lets the compiler perform extra optimizations in this speed-critical code that would otherwise be unavailable. These keywords inform the compiler that "src" and "dest" do not overlap. This also assumes you are copying an integral number of words, if you're not, then this is just a start.
I also don't know which bit shifting / rotation primitives are fast on the ARM and which are slow. This is something to consider. If you need more speed, consider disassembling the output from the C compiler and going from there. If using GCC, try O2, O3, and Os to see which one is fastest. You might reduce stalls in the pipeline by doing two words at the same time.
This uses 23 operations per word, not counting load and store. However, these 23 operations are all very fast and none of them access memory. I don't know if a lookup table would be faster or not.
void
copy_rev(unsigned int *restrict dest,
unsigned int const *restrict src,
unsigned int n)
{
unsigned int i, x;
for (i = 0; i < n; ++i) {
x = src[i];
x = (x >> 16) | (x << 16);
x = ((x >> 8) & 0x00ff00ffU) | ((x & 0x00ff00ffU) << 8);
x = ((x >> 4) & 0x0f0f0f0fU) | ((x & 0x0f0f0f0fU) << 4);
x = ((x >> 2) & 0x33333333U) | ((x & 0x33333333U) << 2);
x = ((x >> 1) & 0x55555555U) | ((x & 0x555555555) << 1);
dest[n-1-i] = x;
}
}
This page is a great reference: http://graphics.stanford.edu/~seander/bithacks.html#BitReverseObvious
Final note: Looking at the ARM assembly reference, there is a "REV" opcode which reverses the byte order in a word. This would shave 7 operations per loop off the above code.
Fastest way would probably to store the reverse of all possible byte values in a look-up table. The table would take only 256 bytes.
Build a 256 element lookup table of byte values that are bit-reversed from their index.
{0x00, 0x80, 0x40, 0xc0, etc}
Then iterate through your array copying using each byte as an index into your lookup table.
If you are writing assembly language, the x86 instruction set has an XLAT instruction that does just this sort of lookup. Although it may not actually be faster than C code on modern processors.
You can do this in place if you iterate from both ends towards the middle. Because of cache effects, you may find it's faster to swap in 16 byte chunks (assuming a 16 byte cache line).
Here's the basic code (not including the cache line optimization)
// bit reversing lookup table
typedef unsigned char BYTE;
extern const BYTE g_RevBits[256];
void ReverseBitsInPlace(BYTE * pb, int cb)
{
int iter = cb/2;
for (int ii = 0, jj = cb-1; ii < iter; ++ii, --jj)
{
BYTE b1 = g_RevBits[pb[ii]];
pb[ii] = g_RevBits[pb[jj]];
pb[jj] = b1;
}
if (cb & 1) // if the number of bytes was odd, swap the middle one in place
{
pb[cb/2] = g_RevBits[pb[cb/2]];
}
}
// initialize the bit reversing lookup table using macros to make it less typing.
#define BITLINE(n) \
0x0##n, 0x8##n, 0x4##n, 0xC##n, 0x2##n, 0xA##n, 0x6##n, 0xE##n,\
0x1##n, 0x9##n, 0x5##n, 0xD##n, 0x3##n, 0xB##n, 0x7##n, 0xF##n,
const BYTE g_RevBits[256] = {
BITLINE(0), BITLINE(8), BITLINE(4), BITLINE(C),
BITLINE(2), BITLINE(A), BITLINE(6), BITLINE(E),
BITLINE(1), BITLINE(9), BITLINE(5), BITLINE(D),
BITLINE(3), BITLINE(B), BITLINE(7), BITLINE(F),
};
The Bit Twiddling Hacks site is alwas a good starting point for these kind of problems. Take a look here for fast bit reversal. Then its up to you to apply it to each byte/word of your memory block.
EDIT:
Inspired by Dietrich Epps answer and looking at the ARM instruction set, there is a RBIT opcode that reverses the bits contained in a register. So if performance is critical, you might consider using some assembly code.
Loop through the half of the array, convert and exchange bytes.
for( int i = 0; i < arraySize / 2; i++ ) {
char inverted1 = invert( array[i] );
char inverted2 = invert( array[arraySize - i - 1] );
array[i] = inverted2;
array[arraySize - i - 1] = inverted1;
}
For conversion use a precomputed table - an array of 2CHAR_BIT (CHAR_BIT will most likely be 8) elements where at position "I" the result of byte with value "I" inversion is stored. This will be very fast - one pass - and consume only 2CHAR_BIT for the table.
It looks like this code takes about 50 clocks per bit swap on my i7 XPS 8500 machine. 7.6 seconds for a million array flips. Single threaded. It prints some ASCI art based on patterns of 1s and 0s. I rotated the pic left 180 degrees after reversing the bit array, using a graphic editor, and they look identical to me. A double-reversed image comes out the same as the original.
As for pluses, it's a complete solution. It swaps bits from the back of a bit array to the front, vs operating on ints/bytes and then needing to swap ints/bytes in an array.
Also, this is a general purpose bit library, so you might find it handy in the future for solving other, more mundane problems.
Is it as fast as the accepted answer? I think it's close, but without working code to benchmark it's impossible to say. Feel free to cut and paste this working program.
// Reverse BitsInBuff.cpp : Defines the entry point for the console application.
#include "stdafx.h"
#include "time.h"
#include "memory.h"
//
// Manifest constants
#define uchar unsigned char
#define BUFF_BYTES 510 //400 supports a display of 80x40 bits
#define DW 80 // Display Width
// ----------------------------------------------------------------------------
uchar mask_set[] = { 0x01, 0x02, 0x04, 0x08, 0x10, 0x20, 0x40, 0x80 };
uchar mask_clr[] = { 0xfe, 0xfd, 0xfb, 0xf7, 0xef, 0xdf, 0xbf, 0x7f };
//
// Function Prototypes
static void PrintIntBits(long x, int bits);
void BitSet(uchar * BitArray, unsigned long BitNumber);
void BitClr(uchar * BitArray, unsigned long BitNumber);
void BitTog(uchar * BitArray, unsigned long BitNumber);
uchar BitGet(uchar * BitArray, unsigned long BitNumber);
void BitPut(uchar * BitArray, unsigned long BitNumber, uchar value);
//
uchar *ReverseBitsInArray(uchar *Buff, int BitKnt);
static void PrintIntBits(long x, int bits);
// -----------------------------------------------------------------------------
// Reverse the bit ordering in an array
uchar *ReverseBitsInArray(uchar *Buff, int BitKnt) {
unsigned long front=0, back = BitKnt-1;
uchar temp;
while( front<back ) {
temp = BitGet(Buff, front); // copy front bit to temp before overwriting
BitPut(Buff, front, BitGet(Buff, back)); // copy back bit to front bit
BitPut(Buff, back, temp); // copy saved value of front in temp to back of bit arra)
front++;
back--;
}
return Buff;
}
// ---------------------------------------------------------------------------
// ---------------------------------------------------------------------------
int _tmain(int argc, _TCHAR* argv[]) {
int i, j, k, LoopKnt = 1000001;
time_t start;
uchar Buff[BUFF_BYTES];
memset(Buff, 0, sizeof(Buff));
// make an ASCII art picture
for(i=0, k=0; i<(sizeof(Buff)*8)/DW; i++) {
for(j=0; j<DW/2; j++) {
BitSet(Buff, (i*DW)+j+k);
}
k++;
}
// print ASCII art picture
for(i=0; i<sizeof(Buff); i++) {
if(!(i % 10)) printf("\n"); // print bits in blocks of 80
PrintIntBits(Buff[i], 8);
}
i=LoopKnt;
start = clock();
while( i-- ) {
ReverseBitsInArray((uchar *)Buff, BUFF_BYTES * 8);
}
// print ASCII art pic flipped upside-down and rotated left
printf("\nMilliseconds elapsed = %d", clock() - start);
for(i=0; i<sizeof(Buff); i++) {
if(!(i % 10)) printf("\n"); // print bits in blocks of 80
PrintIntBits(Buff[i], 8);
}
printf("\n\nBenchmark time for %d loops\n", LoopKnt);
getchar();
return 0;
}
// -----------------------------------------------------------------------------
// Scaffolding...
static void PrintIntBits(long x, int bits) {
unsigned long long z=1;
int i=0;
z = z << (bits-1);
for (; z > 0; z >>= 1) {
printf("%s", ((x & z) == z) ? "#" : ".");
}
}
// These routines do bit manipulations on a bit array of unsigned chars
// ---------------------------------------------------------------------------
void BitSet(uchar *buff, unsigned long BitNumber) {
buff[BitNumber >> 3] |= mask_set[BitNumber & 7];
}
// ----------------------------------------------------------------------------
void BitClr(uchar *buff, unsigned long BitNumber) {
buff[BitNumber >> 3] &= mask_clr[BitNumber & 7];
}
// ----------------------------------------------------------------------------
void BitTog(uchar *buff, unsigned long BitNumber) {
buff[BitNumber >> 3] ^= mask_set[BitNumber & 7];
}
// ----------------------------------------------------------------------------
uchar BitGet(uchar *buff, unsigned long BitNumber) {
return (uchar) ((buff[BitNumber >> 3] >> (BitNumber & 7)) & 1);
}
// ----------------------------------------------------------------------------
void BitPut(uchar *buff, unsigned long BitNumber, uchar value) {
if(value) { // if the bit at buff[BitNumber] is true.
BitSet(buff, BitNumber);
} else {
BitClr(buff, BitNumber);
}
}
Below is the code listing for an optimization using a new buffer, instead of swapping bytes in place. Given that only 2030:4080 BitSet()s are needed because of the if() test, and about half the GetBit()s and PutBits() are eliminated by eliminating TEMP, I suspect memory access time is a large, fixed cost to these kinds of operations, providing a hard limit to optimization.
Using a look-up approach, and CONDITIONALLY swapping bytes, rather than bits, reduces by a factor of 8 the number of memory accesses, and testing for a 0 byte gets amortized across 8 bits, rather than 1.
Using these two approaches together, testing to see if the entire 8-bit char is 0 before doing ANYTHING, including the table lookup, and the write, is likely going to be the fastest possible approach, but would require an extra 512 bytes for the new, destination bit array, and 256 bytes for the lookup table. The performance payoff might be quite dramatic though.
// -----------------------------------------------------------------------------
// Reverse the bit ordering in new array
uchar *ReverseBitsInNewArray(uchar *Dst, const uchar *Src, const int BitKnt) {
int front=0, back = BitKnt-1;
memset(Dst, 0, BitKnt/BitsInByte);
while( front < back ) {
if(BitGet(Src, back--)) { // memset() has already set all bits in Dst to 0,
BitSet(Dst, front); // so only reset if Src bit is 1
}
front++;
}
return Dst;
To reverse a single byte x you can handle the bits one at a time:
unsigned char a = 0;
for (i = 0; i < 8; ++i) {
a += (unsigned char)(((x >> i) & 1) << (7 - i));
}
You can create a cache of these results in an array so that you can quickly reverse a byte just by making a single lookup instead of looping.
Then you just have to reverse the byte array, and when you write the data apply the above mapping. Reversing a byte array is a well documented problem, e.g. here.
Single Core?
How much memory?
Is the display buffered in memory and pushed to the device, or is the only copy of the pixels in the screens memory?
The data is pushed from system memory to the LCD driver over 8-bit IO
port.
Since you'll be writing to the LCD one byte at a time, I think the best idea is to perform the bit reversal right when sending the data to the LCD driver rather than as a separate pre-pass. Something along those lines should be faster than any of the other answers:
void send_to_LCD(uint8_t* data, int len, bool rotate) {
if (rotate)
for (int i=len-1; i>=0; i--)
write(reverse(data[i]));
else
for (int i=0; i<len; i++)
write(data[i]);
}
Where write() is the function that sends a byte to the LCD driver and reverse() one of the single-byte bit reversal methods described in the other answers.
This approach avoids the need to store two copies of the video data in ram and also avoids the read-invert-write roundtrip. Also note that this is the simplest implementation: it could be trivially adapted to load, say, 4 bytes at a time from memory if this were to yield better performance. A smart vectorizing compiler may be even able to do it for you.

How to shift an array of bytes by 12-bits

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.

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