Related
I have a function in this form (From Fastest Implementation of Exponential Function Using SSE):
__m128 FastExpSse(__m128 x)
{
static __m128 const a = _mm_set1_ps(12102203.2f); // (1 << 23) / ln(2)
static __m128i const b = _mm_set1_epi32(127 * (1 << 23) - 486411);
static __m128 const m87 = _mm_set1_ps(-87);
// fast exponential function, x should be in [-87, 87]
__m128 mask = _mm_cmpge_ps(x, m87);
__m128i tmp = _mm_add_epi32(_mm_cvtps_epi32(_mm_mul_ps(a, x)), b);
return _mm_and_ps(_mm_castsi128_ps(tmp), mask);
}
I want to make it C compatible.
Yet the compiler doesn't accept the form static __m128i const b = _mm_set1_epi32(127 * (1 << 23) - 486411); when I use C compiler.
Yet I don't want the first 3 values to be recalculated in each function call.
One solution is to inline it (But sometimes the compilers reject that).
Is there a C style to achieve it in case the function isn't inlined?
Thank You.
Remove static and const.
Also remove them from the C++ version. const is OK, but static is horrible, introducing guard variables that are checked every time, and a very expensive initialization the first time.
__m128 a = _mm_set1_ps(12102203.2f); is not a function call, it's just a way to express a vector constant. No time can be saved by "doing it only once" - it normally happens zero times, with the constant vector being prepared in the data segment of the program and simply being loaded at runtime, without the junk around it that static introduces.
Check the asm to be sure, without static this is what happens: (from godbolt)
FastExpSse(float __vector(4)):
movaps xmm1, XMMWORD PTR .LC0[rip]
cmpleps xmm1, xmm0
mulps xmm0, XMMWORD PTR .LC1[rip]
cvtps2dq xmm0, xmm0
paddd xmm0, XMMWORD PTR .LC2[rip]
andps xmm0, xmm1
ret
.LC0:
.long 3266183168
.long 3266183168
.long 3266183168
.long 3266183168
.LC1:
.long 1262004795
.long 1262004795
.long 1262004795
.long 1262004795
.LC2:
.long 1064866805
.long 1064866805
.long 1064866805
.long 1064866805
_mm_set1_ps(-87); or any other _mm_set intrinsic is not a valid static initializer with current compilers, because it's not treated as a constant expression.
In C++, it compiles to runtime initialization of the static storage location (copying from a vector literal somewhere else). And if it's a static __m128 inside a function, there's a guard variable to protect it.
In C, it simply refuses to compile, because C doesn't support non-constant initializers / constructors. _mm_set is not like a braced initializer for the underlying GNU C native vector, like #benjarobin's answer shows.
This is really dumb, and seems to be a missed-optimization in all 4 mainstream x86 C++ compilers (gcc/clang/ICC/MSVC). Even if it somehow matters that each static const __m128 var have a distinct address, the compiler could achieve that by using initialized read-only storage instead of copying at runtime.
So it seems like constant propagation fails to go all the way to turning _mm_set into a constant initializer even when optimization is enabled.
Never use static const __m128 var = _mm_set... even in C++; it's inefficient.
Inside a function is even worse, but global scope is still bad.
Instead, avoid static. You can still use const to stop yourself from accidentally assigning something else, and to tell human readers that it's a constant. Without static, it has no effect on where/how your variable is stored. const on automatic storage just does compile-time checking that you don't modify the object.
const __m128 var = _mm_set1_ps(-87); // not static
Compilers are good at this, and will optimize the case where multiple functions use the same vector constant, the same way they de-duplicate string literals and put them in read-only memory.
Defining constants this way inside small helper functions is fine: compilers will hoist the constant-setup out of a loop after inlining the function.
It also lets compilers optimize away the full 16 bytes of storage, and load it with vbroadcastss xmm0, dword [mem], or stuff like that.
This solution is clearly not portable, it's working with GCC 8 (only tested with this compiler):
#include <stdio.h>
#include <stdint.h>
#include <emmintrin.h>
#include <string.h>
#define INIT_M128(vFloat) {(vFloat), (vFloat), (vFloat), (vFloat)}
#define INIT_M128I(vU32) {((uint64_t)(vU32) | (uint64_t)(vU32) << 32u), ((uint64_t)(vU32) | (uint64_t)(vU32) << 32u)}
static void print128(const void *p)
{
unsigned char buf[16];
memcpy(buf, p, 16);
for (int i = 0; i < 16; ++i)
{
printf("%02X ", buf[i]);
}
printf("\n");
}
int main(void)
{
static __m128 const glob_a = INIT_M128(12102203.2f);
static __m128i const glob_b = INIT_M128I(127 * (1 << 23) - 486411);
static __m128 const glob_m87 = INIT_M128(-87.0f);
__m128 a = _mm_set1_ps(12102203.2f);
__m128i b = _mm_set1_epi32(127 * (1 << 23) - 486411);
__m128 m87 = _mm_set1_ps(-87);
print128(&a);
print128(&glob_a);
print128(&b);
print128(&glob_b);
print128(&m87);
print128(&glob_m87);
return 0;
}
As explained in the answer of #harold (in C only), the following code (build with or without WITHSTATIC) produces exactly the same code.
#include <stdio.h>
#include <stdint.h>
#include <emmintrin.h>
#include <string.h>
#define INIT_M128(vFloat) {(vFloat), (vFloat), (vFloat), (vFloat)}
#define INIT_M128I(vU32) {((uint64_t)(vU32) | (uint64_t)(vU32) << 32u), ((uint64_t)(vU32) | (uint64_t)(vU32) << 32u)}
__m128 FastExpSse2(__m128 x)
{
#ifdef WITHSTATIC
static __m128 const a = INIT_M128(12102203.2f);
static __m128i const b = INIT_M128I(127 * (1 << 23) - 486411);
static __m128 const m87 = INIT_M128(-87.0f);
#else
__m128 a = _mm_set1_ps(12102203.2f);
__m128i b = _mm_set1_epi32(127 * (1 << 23) - 486411);
__m128 m87 = _mm_set1_ps(-87);
#endif
__m128 mask = _mm_cmpge_ps(x, m87);
__m128i tmp = _mm_add_epi32(_mm_cvtps_epi32(_mm_mul_ps(a, x)), b);
return _mm_and_ps(_mm_castsi128_ps(tmp), mask);
}
So in summary it's better to remove static and const keywords (better and simpler code in C++, and in C the code is portable since with my proposed hack the code is not really portable)
I am trying to play around with a manual encryption in CBC mode but still use Crypto++, just to know can I do it manually.
The CBC algorithm is (AFAIK):
Presume we have n block K[1]....k[n]
0. cipher = empty;
1. xor(IV, K1) -> t1
2. encrypt(t1) -> r1
3. cipher += r1
4. xor (r1, K2) -> t2
5. encrypt(t2) -> r2
6. cipher += r2
7. xor(r2, K3)->t3
8. ...
So I tried to implement it with Crypto++. I have a text file with alphanumeric characters only. Test 1 is read file chunk by chunk (16 byte) and encrypt them using CBC mode manually, then sum up the cipher. Test 2 is use Crypto++ built-in CBC mode.
Test 1
char* key;
char* iv;
//Iterate in K[n] array of n blocks
BSIZE = 16;
std::string vectorToString(vector<char> v){
string s ="";
for (int i = 0; i < v.size(); i++){
s[i] = v[i];
}
return s;
}
vector<char> xor( vector<char> s1, vector<char> s2, int len){
vector<char> r;
for (int i = 0; i < len; i++){
int u = s1[i] ^ s2[i];
r.push_back(u);
}
return r;
}
vector<char> byteToVector(byte *b, int len){
vector<char> v;
for (int i = 0; i < len; i++){
v.push_back( b[i]);
}
return v;
}
string cbc_manual(byte [n]){
int i = 0;
//Open a file and read from it, buffer size = 16
// , equal to DEFAULT_BLOCK_SIZE
std::ifstream fin(fileName, std::ios::binary | std::ios::in);
const int BSIZE = 16;
vector<char> encryptBefore;
//This function will return cpc
string cpc ="";
while (!fin.eof()){
char buffer[BSIZE];
//Read a chunk of file
fin.read(buffer, BSIZE);
int sb = sizeof(buffer);
if (i == 0){
encryptBefore = byteToVector( iv, BSIZE);
}
//If i == 0, xor IV with current buffer
//else, xor encryptBefore with current buffer
vector<char> t1 = xor(encryptBefore, byteToVector((byte*) buffer, BSIZE), BSIZE);
//After xored, encrypt the xor result, it will be current step cipher
string r1= encrypt(t1, BSIZE).c_str();
cpc += r1;
const char* end = r1.c_str() ;
encryptBefore = stringToVector( r1);
i++;
}
return cpc;
}
This is my encrypt() function, because we have only one block so I use ECB (?) mode
string encrypt(string s, int size){
ECB_Mode< AES >::Encryption e;
e.SetKey(key, size);
string cipher;
StringSource ss1(s, true,
new StreamTransformationFilter(e,
new StringSink(cipher)
) // StreamTransformationFilter
); // StringSource
return cipher;
}
And this is 100% Crypto++ made solution:
Test 2
encryptCBC(char * plain){
CBC_Mode < AES >::Encryption encryption(key, sizeof(key), iv);
StreamTransformationFilter encryptor(encryption, NULL);
for (size_t j = 0; j < plain.size(); j++)
encryptor.Put((byte)plain[j]);
encryptor.MessageEnd();
size_t ready = encryptor.MaxRetrievable();
string cipher(ready, 0x00);
encryptor.Get((byte*)&cipher[0], cipher.size());
}
Result of Test 1 and Test 2 are different. In the fact, ciphered text from Test 1 is contain the result of Test 2. Example:
Test 1's result aaa[....]bbb[....]ccc[...]...
Test 2 (Crypto++ built-in CBC)'s result: aaabbbccc...
I know the xor() function may cause a problem relate to "sameChar ^ sameChar = 0", but is there any problem relate to algorithm in my code?
This is my Test 2.1 after the 1st solution of jww.
static string auto_cbc2(string plain, long size){
CBC_Mode< AES >::Encryption e;
e.SetKeyWithIV(key, sizeof(key), iv, sizeof(iv));
string cipherText;
CryptoPP::StringSource ss(plain, true,
new CryptoPP::StreamTransformationFilter(e,
new CryptoPP::StringSink(cipherText)
, BlockPaddingSchemeDef::NO_PADDING
) // StreamTransformationFilter
); // StringSource
return cipherText;
}
It throw an error:
Unhandled exception at 0x7407A6F2 in AES-CRPP.exe: Microsoft C++
exception: CryptoPP::InvalidDataFormat at memory location 0x00EFEA74
I only got this error when use BlockPaddingSchemeDef::NO_PADDING, tried to remove BlockPaddingSchemeDef or using BlockPaddingSchemeDef::DEFAULT_PADDING, I got no error . :?
StringSource ss1(s, true,
new StreamTransformationFilter(e,
new StringSink(cipher)));
This uses PKCS padding by default. It takes a 16-byte input and produces a 32-byte output due to padding. You should do one of two things.
First, you can use BlockPaddingScheme::NO_PADDING. Something like:
StringSource ss1(s, true,
new StreamTransformationFilter(e,
new StringSink(cipher)
BlockPaddingScheme::NO_PADDING));
Second, you can process blocks manually, 16 bytes at a time. Something like:
AES::Encryption encryptor(key, keySize);
byte ibuff[<some size>] = ...;
byte obuff[<some size>];
ASSERT(<some size> % AES::BLOCKSIZE == 0);
unsigned int BLOCKS = <some size>/AES::BLOCKSIZE;
for (unsigned int i=0; i<BLOCKS; i==)
{
encryptor.ProcessBlock(&ibuff[i*16], &obuff[i*16]);
// Do the CBC XOR thing...
}
You may be able to call ProcessAndXorBlock from the BlockCipher base class and do it in one shot.
Hi,
I am coding in OpenCL.
I am converting a "C function" having 2D array starting from i=1 and j=1 .PFB .
cv::Mat input; //Input :having some data in it ..
//Image input size is :input.rows=288 ,input.cols =640
cv::Mat output(input.rows-2,input.cols-2,CV_32F); //Output buffer
//Image output size is :output.rows=286 ,output.cols =638
This is a code Which I want to modify in OpenCL:
for(int i=1;i<output.rows-1;i++)
{
for(int j=1;j<output.cols-1;j++)
{
float xVal = input.at<uchar>(i-1,j-1)-input.at<uchar>(i-1,j+1)+ 2*(input.at<uchar>(i,j-1)-input.at<uchar>(i,j+1))+input.at<uchar>(i+1,j-1) - input.at<uchar>(i+1,j+1);
float yVal = input.at<uchar>(i-1,j-1) - input.at<uchar>(i+1,j-1)+ 2*(input.at<uchar>(i-1,j) - input.at<uchar>(i+1,j))+input.at<uchar>(i-1,j+1)-input.at<uchar>(i+1,j+1);
output.at<float>(i-1,j-1) = xVal*xVal+yVal*yVal;
}
}
...
Host code :
//Input Image size is :input.rows=288 ,input.cols =640
//Output Image size is :output.rows=286 ,output.cols =638
OclStr->global_work_size[0] =(input.cols);
OclStr->global_work_size[1] =(input.rows);
size_t outBufSize = (output.rows) * (output.cols) * 4;//4 as I am copying all 4 uchar values into one float variable space
cl_mem cl_input_buffer = clCreateBuffer(
OclStr->context, CL_MEM_READ_ONLY | CL_MEM_USE_HOST_PTR ,
(input.rows) * (input.cols),
static_cast<void *>(input.data), &OclStr->returnstatus);
cl_mem cl_output_buffer = clCreateBuffer(
OclStr->context, CL_MEM_WRITE_ONLY| CL_MEM_USE_HOST_PTR ,
(output.rows) * (output.cols) * sizeof(float),
static_cast<void *>(output.data), &OclStr->returnstatus);
OclStr->returnstatus = clSetKernelArg(OclStr->objkernel, 0, sizeof(cl_mem), (void *)&cl_input_buffer);
OclStr->returnstatus = clSetKernelArg(OclStr->objkernel, 1, sizeof(cl_mem), (void *)&cl_output_buffer);
OclStr->returnstatus = clEnqueueNDRangeKernel(
OclStr->command_queue,
OclStr->objkernel,
2,
NULL,
OclStr->global_work_size,
NULL,
0,
NULL,
NULL
);
clEnqueueMapBuffer(OclStr->command_queue, cl_output_buffer, true, CL_MAP_READ, 0, outBufSize, 0, NULL, NULL, &OclStr->returnstatus);
kernel Code :
__kernel void Sobel_uchar (__global uchar *pSrc, __global float *pDstImage)
{
const uint cols = get_global_id(0)+1;
const uint rows = get_global_id(1)+1;
const uint width= get_global_size(0);
uchar Opsoble[8];
Opsoble[0] = pSrc[(cols-1)+((rows-1)*width)];
Opsoble[1] = pSrc[(cols+1)+((rows-1)*width)];
Opsoble[2] = pSrc[(cols-1)+((rows+0)*width)];
Opsoble[3] = pSrc[(cols+1)+((rows+0)*width)];
Opsoble[4] = pSrc[(cols-1)+((rows+1)*width)];
Opsoble[5] = pSrc[(cols+1)+((rows+1)*width)];
Opsoble[6] = pSrc[(cols+0)+((rows-1)*width)];
Opsoble[7] = pSrc[(cols+0)+((rows+1)*width)];
float gx = Opsoble[0]-Opsoble[1]+2*(Opsoble[2]-Opsoble[3])+Opsoble[4]-Opsoble[5];
float gy = Opsoble[0]-Opsoble[4]+2*(Opsoble[6]-Opsoble[7])+Opsoble[1]-Opsoble[5];
pDstImage[(cols-1)+(rows-1)*width] = gx*gx + gy*gy;
}
Here I am not able to get the output as expected.
I am having some questions that
My for loop is starting from i=1 instead of zero, then How can I get proper index by using the global_id() in x and y direction
What is going wrong in my above kernel code :(
I am suspecting there is a problem in buffer stride but not able to further break my head as already broke it throughout a day :(
I have observed that with below logic output is skipping one or two frames after some 7/8 frames sequence.
I have added the screen shot of my output which is compared with the reference output.
My above logic is doing partial sobelling on my input .I changed the width as -
const uint width = get_global_size(0)+1;
PFB
Your suggestions are most welcome !!!
It looks like you may be fetching values in (y,x) format in your opencl version. Also, you need to add 1 to the global id to replicate your for loops starting from 1 rather than 0.
I don't know why there is an unused iOffset variable. Maybe your bug is related to this? I removed it in my version.
Does this kernel work better for you?
__kernel void simple(__global uchar *pSrc, __global float *pDstImage)
{
const uint i = get_global_id(0) +1;
const uint j = get_global_id(1) +1;
const uint width = get_global_size(0) +2;
uchar Opsoble[8];
Opsoble[0] = pSrc[(i-1) + (j - 1)*width];
Opsoble[1] = pSrc[(i-1) + (j + 1)*width];
Opsoble[2] = pSrc[i + (j-1)*width];
Opsoble[3] = pSrc[i + (j+1)*width];
Opsoble[4] = pSrc[(i+1) + (j - 1)*width];
Opsoble[5] = pSrc[(i+1) + (j + 1)*width];
Opsoble[6] = pSrc[(i-1) + (j)*width];
Opsoble[7] = pSrc[(i+1) + (j)*width];
float gx = Opsoble[0]-Opsoble[1]+2*(Opsoble[2]-Opsoble[3])+Opsoble[4]-Opsoble[5];
float gy = Opsoble[0]-Opsoble[4]+2*(Opsoble[6]-Opsoble[7])+Opsoble[1]-Opsoble[5];
pDstImage[(i-1) + (j-1)*width] = gx*gx + gy*gy ;
}
I am a bit apprehensive about posting an answer suggesting optimizations to your kernel, seeing as the original output has not been reproduced exactly as of yet. There is a major improvement available to be made for problems related to image processing/filtering.
Using local memory will help you out by reducing the number of global reads by a factor of eight, as well as grouping the global writes together for potential gains with the single write-per-pixel output.
The kernel below reads a block of up to 34x34 from pSrc, and outputs a 32x32(max) area of the pDstImage. I hope the comments in the code are enough to guide you in using the kernel. I have not been able to give this a complete test, so there could be changes required. Any comments are appreciated as well.
__kernel void sobel_uchar_wlocal (__global uchar *pSrc, __global float *pDstImage, __global uint2 dimDstImage)
{
//call this kernel 1-dimensional work group size: 32x1
//calculates 32x32 region of output with 32 work items
const uint wid = get_local_id(0);
const uint wid_1 = wid+1; // corrected for the calculation step
const uint2 gid = (uint2)(get_group_id(0),get_group_id(1));
const uint localDim = get_local_size(0);
const uint2 globalTopLeft = (uint2)(localDim * gid.x, localDim * gid.y); //position in pSrc to copy from/to
//dimLocalBuff is used for the right and bottom edges of the image, where the work group may run over the border
const uint2 dimLocalBuff = (uint2)(localDim,localDim);
if(dimDstImage.x - globalTopLeft.x < dimLocalBuff.x){
dimLocalBuff.x = dimDstImage.x - globalTopLeft.x;
}
if(dimDstImage.y - globalTopLeft.y < dimLocalBuff.y){
dimLocalBuff.y = dimDstImage.y - globalTopLeft.y;
}
int i,j;
//save region of data into local memory
__local uchar srcBuff[34][34]; //34^2 uchar = 1156 bytes
for(j=-1;j<dimLocalBuff.y+1;j++){
for(i=x-1;i<dimLocalBuff.x+1;i+=localDim){
srcBuff[i+1][j+1] = pSrc[globalTopLeft.x+i][globalTopLeft.y+j];
}
}
mem_fence(CLK_LOCAL_MEM_FENCE);
//compute output and store locally
__local float dstBuff[32][32]; //32^2 float = 4096 bytes
if(wid_1 < dimLocalBuff.x){
for(i=0;i<dimLocalBuff.y;i++){
float gx = srcBuff[(wid_1-1)+ (i - 1)]-srcBuff[(wid_1-1)+ (i + 1)]+2*(srcBuff[wid_1+ (i-1)]-srcBuff[wid_1+ (i+1)])+srcBuff[(wid_1+1)+ (i - 1)]-srcBuff[(wid_1+1)+ (i + 1)];
float gy = srcBuff[(wid_1-1)+ (i - 1)]-srcBuff[(wid_1+1)+ (i - 1)]+2*(srcBuff[(wid_1-1)+ (i)]-srcBuff[(wid_1+1)+ (i)])+srcBuff[(wid_1-1)+ (i + 1)]-srcBuff[(wid_1+1)+ (i + 1)];
dstBuff[wid][i] = gx*gx + gy*gy;
}
}
mem_fence(CLK_LOCAL_MEM_FENCE);
//copy results to output
for(j=0;j<dimLocalBuff.y;j++){
for(i=0;i<dimLocalBuff.x;i+=localDim){
srcBuff[i][j] = pSrc[globalTopLeft.x+i][globalTopLeft.y+j];
}
}
}
Background:
I have a section of memory, 1024 bytes. The last 1020 bytes will always be the same. The first 4 bytes will change (serial number of a product). I need to calculate the CRC-16 CCITT (0xFFFF starting, 0x1021 mask) for the entire section of memory, CRC_WHOLE.
Question:
Is it possible to calculate the CRC for only the first 4 bytes, CRC_A, then apply a function such as the one below to calculate the full CRC? We can assume that the checksum for the last 1020 bytes, CRC_B, is already known.
CRC_WHOLE = XOR(CRC_A, CRC_B)
I know that this formula does not work (tried it), but I am hoping that something similar exists.
Yes. You can see how in zlib's crc32_combine(). If you have two sequences A and B, then the pure CRC of AB is the exclusive-or of the CRC of A0 and the CRC of 0B, where the 0's represent a series of zero bytes with the length of the corresponding sequence, i.e. B and A respectively.
For your application, you can pre-compute a single operator that applies 1020 zeros to the CRC of your first four bytes very rapidly. Then you can exclusive-or that with the pre-computed CRC of the 1020 bytes.
Update:
Here is a post of mine from 2008 with a detailed explanation that #ArtemB discovered (that I had forgotten about):
crc32_combine() in zlib is based on two key tricks. For what follows,
we set aside the fact that the standard 32-bit CRC is pre and post-
conditioned. We can deal with that later. Assume for now a CRC that
has no such conditioning, and so starts with the register filled with
zeros.
Trick #1: CRCs are linear. So if you have stream X and stream Y of
the same length and exclusive-or the two streams bit-by-bit to get Z,
i.e. Z = X ^ Y (using the C notation for exclusive-or), then CRC(Z) =
CRC(X) ^ CRC(Y). For the problem at hand we have two streams A and B
of differing length that we want to concatenate into stream Z. What
we have available are CRC(A) and CRC(B). What we want is a quick way
to compute CRC(Z). The trick is to construct X = A concatenated with
length(B) zero bits, and Y = length(A) zero bits concatenated with B.
So if we represent concatenation simply by juxtaposition of the
symbols, X = A0, Y = 0B, then X^Y = Z = AB. Then we have CRC(Z) =
CRC(A0) ^ CRC(0B).
Now we need to know CRC(A0) and CRC(0B). CRC(0B) is easy. If we feed
a bunch of zeros to the CRC machine starting with zero, the register
is still filled with zeros. So it's as if we did nothing at all.
Therefore CRC(0B) = CRC(B).
CRC(A0) requires more work however. Taking a non-zero CRC and feeding
zeros to the CRC machine doesn't leave it alone. Every zero changes
the register contents. So to get CRC(A0), we need to set the register
to CRC(A), and then run length(B) zeros through it. Then we can
exclusive-or the result of that with CRC(B) = CRC(0B), and we get what
we want, which is CRC(Z) = CRC(AB). Voila!
Well, actually the voila is premature. I wasn't at all satisfied with
that answer. I didn't want a calculation that took a time
proportional to the length of B. That wouldn't save any time compared
to simply setting the register to CRC(A) and running the B stream
through. I figured there must be a faster way to compute the effect
of feeding n zeros into the CRC machine (where n = length(B)). So
that leads us to:
Trick #2: The CRC machine is a linear state machine. If we know the
linear transformation that occurs when we feed a zero to the machine,
then we can do operations on that transformation to more efficiently
find the transformation that results from feeding n zeros into the
machine.
The transformation of feeding a single zero bit into the CRC machine
is completely represented by a 32x32 binary matrix. To apply the
transformation we multiply the matrix by the register, taking the
register as a 32 bit column vector. For the matrix multiplication in
binary (i.e. over the Galois Field of 2), the role of multiplication
is played by and'ing, and the role of addition is played by exclusive-
or'ing.
There are a few different ways to construct the magic matrix that
represents the transformation caused by feeding the CRC machine a
single zero bit. One way is to observe that each column of the matrix
is what you get when your register starts off with a single one in
it. So the first column is what you get when the register is 100...
and then feed a zero, the second column comes from starting with
0100..., etc. (Those are referred to as basis vectors.) You can see
this simply by doing the matrix multiplication with those vectors.
The matrix multiplication selects the column of the matrix
corresponding to the location of the single one.
Now for the trick. Once we have the magic matrix, we can set aside
the initial register contents for a while, and instead use the
transformation for one zero to compute the transformation for n
zeros. We could just multiply n copies of the matrix together to get
the matrix for n zeros. But that's even worse than just running the n
zeros through the machine. However there's an easy way to avoid most
of those matrix multiplications to get the same answer. Suppose we
want to know the transformation for running eight zero bits, or one
byte through. Let's call the magic matrix that represents running one
zero through: M. We could do seven matrix multiplications to get R =
MxMxMxMxMxMxMxM. Instead, let's start with MxM and call that P. Then
PxP is MxMxMxM. Let's call that Q. Then QxQ is R. So now we've
reduced the seven multiplications to three. P = MxM, Q = PxP, and R =
QxQ.
Now I'm sure you get the idea for an arbitrary n number of zeros. We
can very rapidly generate transformation matrices Mk, where Mk is the
transformation for running 2k zeros through. (In the
paragraph above M3 is R.) We can make M1 through Mk with only k
matrix multiplications, starting with M0 = M. k only has to be as
large as the number of bits in the binary representation of n. We can
then pick those matrices where there are ones in the binary
representation of n and multiply them together to get the
transformation of running n zeros through the CRC machine. So if n =
13, compute M0 x M2 x M3.
If j is the number of one's in the binary representation of n, then we
just have j - 1 more matrix multiplications. So we have a total of k
j - 1 matrix multiplications, where j <= k = floor(logbase2(n)).
Now we take our rapidly constructed matrix for n zeros, and multiply
that by CRC(A) to get CRC(A0). We can compute CRC(A0) in O(log(n))
time, instead of O(n) time. We exclusive or that with CRC(B) and
Voila! (really this time), we have CRC(Z).
That's what zlib's crc32_combine() does.
I will leave it as an exercise for the reader as to how to deal with
the pre and post conditioning of the CRC register. You just need to
apply the linearity observations above. Hint: You don't need to know
length(A). In fact crc32_combine() only takes three arguments:
CRC(A), CRC(B), and length(B) (in bytes).
Below is example C code for an alternative approach for CRC(A0). Rather than working with a matrix, a CRC can be cycled forward n bits by muliplying (CRC ยท ((2^n)%POLY)%POLY . So the repeated squaring is performed on an integer rather than a matrix. If n is constant, then (2^n)%POLY can be pre-computed.
/* crcpad.c - crc - data has a large number of trailing zeroes */
#include <stdio.h>
#include <stdlib.h>
typedef unsigned char uint8_t;
typedef unsigned int uint32_t;
#define POLY (0x04c11db7u)
static uint32_t crctbl[256];
void GenTbl(void) /* generate crc table */
{
uint32_t crc;
uint32_t c;
uint32_t i;
for(c = 0; c < 0x100; c++){
crc = c<<24;
for(i = 0; i < 8; i++)
/* assumes twos complement */
crc = (crc<<1)^((0-(crc>>31))&POLY);
crctbl[c] = crc;
}
}
uint32_t GenCrc(uint8_t * bfr, size_t size) /* generate crc */
{
uint32_t crc = 0u;
while(size--)
crc = (crc<<8)^crctbl[(crc>>24)^*bfr++];
return(crc);
}
/* carryless multiply modulo crc */
uint32_t MpyModCrc(uint32_t a, uint32_t b) /* (a*b)%crc */
{
uint32_t pd = 0;
uint32_t i;
for(i = 0; i < 32; i++){
/* assumes twos complement */
pd = (pd<<1)^((0-(pd>>31))&POLY);
pd ^= (0-(b>>31))&a;
b <<= 1;
}
return pd;
}
/* exponentiate by repeated squaring modulo crc */
uint32_t PowModCrc(uint32_t p) /* pow(2,p)%crc */
{
uint32_t prd = 0x1u; /* current product */
uint32_t sqr = 0x2u; /* current square */
while(p){
if(p&1)
prd = MpyModCrc(prd, sqr);
sqr = MpyModCrc(sqr, sqr);
p >>= 1;
}
return prd;
}
/* # data bytes */
#define DAT ( 32)
/* # zero bytes */
#define PAD (992)
/* DATA+PAD */
#define CNT (1024)
int main()
{
uint32_t pmc;
uint32_t crc;
uint32_t crf;
uint32_t i;
uint8_t *msg = malloc(CNT);
for(i = 0; i < DAT; i++) /* generate msg */
msg[i] = (uint8_t)rand();
for( ; i < CNT; i++)
msg[i] = 0;
GenTbl(); /* generate crc table */
crc = GenCrc(msg, CNT); /* generate crc normally */
crf = GenCrc(msg, DAT); /* generate crc for data */
pmc = PowModCrc(PAD*8); /* pmc = pow(2,PAD*8)%crc */
crf = MpyModCrc(crf, pmc); /* crf = (crf*pmc)%crc */
printf("%08x %08x\n", crc, crf);
free(msg);
return 0;
}
Example C code using intrinsic for carryless multiply, pclmulqdq == _mm_clmulepi64_si128:
/* crcpadm.c - crc - data has a large number of trailing zeroes */
/* pclmulqdq intrinsic version */
#include <stdio.h>
#include <stdlib.h>
#include <intrin.h>
typedef unsigned char uint8_t;
typedef unsigned int uint32_t;
typedef unsigned long long uint64_t;
#define POLY (0x104c11db7ull)
#define POLYM ( 0x04c11db7u)
static uint32_t crctbl[256];
static __m128i poly; /* poly */
static __m128i invpoly; /* 2^64 / POLY */
void GenMPoly(void) /* generate __m12i8 poly info */
{
uint64_t N = 0x100000000ull;
uint64_t Q = 0;
for(size_t i = 0; i < 33; i++){
Q <<= 1;
if(N&0x100000000ull){
Q |= 1;
N ^= POLY;
}
N <<= 1;
}
poly.m128i_u64[0] = POLY;
invpoly.m128i_u64[0] = Q;
}
void GenTbl(void) /* generate crc table */
{
uint32_t crc;
uint32_t c;
uint32_t i;
for(c = 0; c < 0x100; c++){
crc = c<<24;
for(i = 0; i < 8; i++)
/* assumes twos complement */
crc = (crc<<1)^((0-(crc>>31))&POLYM);
crctbl[c] = crc;
}
}
uint32_t GenCrc(uint8_t * bfr, size_t size) /* generate crc */
{
uint32_t crc = 0u;
while(size--)
crc = (crc<<8)^crctbl[(crc>>24)^*bfr++];
return(crc);
}
/* carryless multiply modulo crc */
uint32_t MpyModCrc(uint32_t a, uint32_t b) /* (a*b)%crc */
{
__m128i ma, mb, mp, mt;
ma.m128i_u64[0] = a;
mb.m128i_u64[0] = b;
mp = _mm_clmulepi64_si128(ma, mb, 0x00); /* p[0] = a*b */
mt = _mm_clmulepi64_si128(mp, invpoly, 0x00); /* t[1] = (p[0]*((2^64)/POLY))>>64 */
mt = _mm_clmulepi64_si128(mt, poly, 0x01); /* t[0] = t[1]*POLY */
return mp.m128i_u32[0] ^ mt.m128i_u32[0]; /* ret = p[0] ^ t[0] */
}
/* exponentiate by repeated squaring modulo crc */
uint32_t PowModCrc(uint32_t p) /* pow(2,p)%crc */
{
uint32_t prd = 0x1u; /* current product */
uint32_t sqr = 0x2u; /* current square */
while(p){
if(p&1)
prd = MpyModCrc(prd, sqr);
sqr = MpyModCrc(sqr, sqr);
p >>= 1;
}
return prd;
}
/* # data bytes */
#define DAT ( 32)
/* # zero bytes */
#define PAD (992)
/* DATA+PAD */
#define CNT (1024)
int main()
{
uint32_t pmc;
uint32_t crc;
uint32_t crf;
uint32_t i;
uint8_t *msg = malloc(CNT);
GenMPoly(); /* generate __m128 polys */
GenTbl(); /* generate crc table */
for(i = 0; i < DAT; i++) /* generate msg */
msg[i] = (uint8_t)rand();
for( ; i < CNT; i++)
msg[i] = 0;
crc = GenCrc(msg, CNT); /* generate crc normally */
crf = GenCrc(msg, DAT); /* generate crc for data */
pmc = PowModCrc(PAD*8); /* pmc = pow(2,PAD*8)%crc */
crf = MpyModCrc(crf, pmc); /* crf = (crf*pmc)%crc */
printf("%08x %08x\n", crc, crf);
free(msg);
return 0;
}
I'm trying to modify a simple dynamic vector in CUDA using the thrust library of CUDA. But I'm getting "launch_closure_by_value" error on the screen indicatiing that the error is related to some synchronization process.
A simple 1D dynamic array modification is not possible due to this error.
My code segment which is causing the error is as follows.
from a .cpp file I call setIndexedGrid, which is defined in System.cu
float* a= (float*)(malloc(8*sizeof(float)));
a[0]= 0; a[1]= 1; a[2]= 2; a[3]= 3; a[4]= 4; a[5]= 5; a[6]= 6; a[7]= 7;
float* b = (float*)(malloc(8*sizeof(float)));
setIndexedGridInfo(a,b);
The code segment at System.cu:
void
setIndexedGridInfo(float* a, float*b)
{
thrust::device_ptr<float> d_oldData(a);
thrust::device_ptr<float> d_newData(b);
float c = 0.0;
thrust::for_each(
thrust::make_zip_iterator(thrust::make_tuple(d_oldData,d_newData)),
thrust::make_zip_iterator(thrust::make_tuple(d_oldData+8,d_newData+8)),
grid_functor(c));
}
grid_functor is defined in _kernel.cu
struct grid_functor
{
float a;
__host__ __device__
grid_functor(float grid_Info) : a(grid_Info) {}
template <typename Tuple>
__device__
void operator()(Tuple t)
{
volatile float data = thrust::get<0>(t);
float pos = data + 0.1;
thrust::get<1>(t) = pos;
}
};
I also get these on the Output window (I use Visual Studio):
First-chance exception at 0x000007fefdc7cacd in Particles.exe:
Microsoft C++ exception: cudaError_enum at memory location
0x0029eb60.. First-chance exception at 0x000007fefdc7cacd in
smokeParticles.exe: Microsoft C++ exception:
thrust::system::system_error at memory location 0x0029ecf0.. Unhandled
exception at 0x000007fefdc7cacd in Particles.exe: Microsoft C++
exception: thrust::system::system_error at memory location
0x0029ecf0..
What is causing the problem?
You are trying to use host memory pointers in functions expecting pointers in device memory. This code is the problem:
float* a= (float*)(malloc(8*sizeof(float)));
a[0]= 0; a[1]= 1; a[2]= 2; a[3]= 3; a[4]= 4; a[5]= 5; a[6]= 6; a[7]= 7;
float* b = (float*)(malloc(8*sizeof(float)));
setIndexedGridInfo(a,b);
.....
thrust::device_ptr<float> d_oldData(a);
thrust::device_ptr<float> d_newData(b);
The thrust::device_ptr is intended for "wrapping" a device memory pointer allocated with the CUDA API so that thrust can use it. You are trying to treat a host pointer directly as a device pointer. That is illegal. You could modify your setIndexedGridInfo function like this:
void setIndexedGridInfo(float* a, float*b, const int n)
{
thrust::device_vector<float> d_oldData(a,a+n);
thrust::device_vector<float> d_newData(b,b+n);
float c = 0.0;
thrust::for_each(
thrust::make_zip_iterator(thrust::make_tuple(d_oldData.begin(),d_newData.begin())),
thrust::make_zip_iterator(thrust::make_tuple(d_oldData.end(),d_newData.end())),
grid_functor(c));
}
The device_vector constructor will allocate device memory and then copy the contents of your host memory to the device. That should fix the error you are seeing, although I am not sure what you are trying to do with the for_each iterator and whether the functor you have wrttien is correct.
Edit:
Here is a complete, compilable, runnable version of your code:
#include <cstdlib>
#include <cstdio>
#include <thrust/device_vector.h>
#include <thrust/for_each.h>
#include <thrust/copy.h>
struct grid_functor
{
float a;
__host__ __device__
grid_functor(float grid_Info) : a(grid_Info) {}
template <typename Tuple>
__device__
void operator()(Tuple t)
{
volatile float data = thrust::get<0>(t);
float pos = data + 0.1f;
thrust::get<1>(t) = pos;
}
};
void setIndexedGridInfo(float* a, float*b, const int n)
{
thrust::device_vector<float> d_oldData(a,a+n);
thrust::device_vector<float> d_newData(b,b+n);
float c = 0.0;
thrust::for_each(
thrust::make_zip_iterator(thrust::make_tuple(d_oldData.begin(),d_newData.begin())),
thrust::make_zip_iterator(thrust::make_tuple(d_oldData.end(),d_newData.end())),
grid_functor(c));
thrust::copy(d_newData.begin(), d_newData.end(), b);
}
int main(void)
{
const int n = 8;
float* a= (float*)(malloc(n*sizeof(float)));
a[0]= 0; a[1]= 1; a[2]= 2; a[3]= 3; a[4]= 4; a[5]= 5; a[6]= 6; a[7]= 7;
float* b = (float*)(malloc(n*sizeof(float)));
setIndexedGridInfo(a,b,n);
for(int i=0; i<n; i++) {
fprintf(stdout, "%d (%f,%f)\n", i, a[i], b[i]);
}
return 0;
}
I can compile and run this code on an OS 10.6.8 host with CUDA 4.1 like this:
$ nvcc -Xptxas="-v" -arch=sm_12 -g -G thrustforeach.cu
./thrustforeach.cu(18): Warning: Cannot tell what pointer points to, assuming global memory space
./thrustforeach.cu(20): Warning: Cannot tell what pointer points to, assuming global memory space
./thrustforeach.cu(18): Warning: Cannot tell what pointer points to, assuming global memory space
./thrustforeach.cu(20): Warning: Cannot tell what pointer points to, assuming global memory space
ptxas info : Compiling entry function '_ZN6thrust6detail7backend4cuda6detail23launch_closure_by_valueINS2_18for_each_n_closureINS_12zip_iteratorINS_5tupleINS0_15normal_iteratorINS_10device_ptrIfEEEESB_NS_9null_typeESC_SC_SC_SC_SC_SC_SC_EEEEi12grid_functorEEEEvT_' for 'sm_12'
ptxas info : Used 14 registers, 160+0 bytes lmem, 16+16 bytes smem, 4 bytes cmem[1]
ptxas info : Compiling entry function '_ZN6thrust6detail7backend4cuda6detail23launch_closure_by_valueINS2_18for_each_n_closureINS_12zip_iteratorINS_5tupleINS0_15normal_iteratorINS_10device_ptrIfEEEESB_NS_9null_typeESC_SC_SC_SC_SC_SC_SC_EEEEj12grid_functorEEEEvT_' for 'sm_12'
ptxas info : Used 14 registers, 160+0 bytes lmem, 16+16 bytes smem, 4 bytes cmem[1]
$ ./a.out
0 (0.000000,0.100000)
1 (1.000000,1.100000)
2 (2.000000,2.100000)
3 (3.000000,3.100000)
4 (4.000000,4.100000)
5 (5.000000,5.100000)
6 (6.000000,6.100000)
7 (7.000000,7.100000)