using constant memory prints address instead of value in cuda - cuda

I am trying to use the constant memory in the code with constant memory assigned value from kernel not using cudacopytosymbol.
#include <iostream>
using namespace std;
#define N 10
//__constant__ int constBuf_d[N];
__constant__ int *constBuf;
__global__ void foo( int *results )
{
int tdx = threadIdx.x;
int idx = blockIdx.x * blockDim.x + tdx;
if( idx < N )
{
constBuf[idx]=1;
results[idx] = constBuf[idx];
}
}
// main routine that executes on the host
int main(int argc, char* argv[])
{
int *results_h = new int[N];
int *results_d;
cudaMalloc((void **)&results_d, N*sizeof(int));
foo <<< 1, 10 >>> ( results_d );
cudaMemcpy(results_h, results_d, N*sizeof(int), cudaMemcpyDeviceToHost);
for( int i=0; i < N; ++i )
printf("%i ", results_h[i] );
delete(results_h);
}
output shows
6231808 6226116 0 0 0 0 0 0 0 0
I want the program to print the value assigned to constant memory through the kenel in the code.

Constant memory is, as the name implies, constant/read-only with respect to device code. What you are trying to do is illegal and can't be made to work.
To set values in constant memory, you currently have two choices:
set the value from host code via the cudaMemcpyToSymbol API call (or its equivalents)
use static initialisation at compile time
In the latter case something like this would work:
__constant__ int constBuf[N] = { 16, 2, 77, 40, 12, 3, 5, 3, 6, 6 };
__global__ void foo( int *results )
{
int tdx = threadIdx.x;
int idx = blockIdx.x * blockDim.x + tdx;
if( tdx < N )
{
results[idx] = constBuf[tdx]; // Note changes here!
}
}

Related

CUDA syncthreads() block other threads

#define TS 32
int num_devices = 0;
__global__ void shared_kernel(float* A, float* B, float* C, int M, int N, int K) {
int global_col = blockDim.x * blockIdx.x + threadIdx.x;
int global_row = blockDim.y * blockIdx.y + threadIdx.y;
int local_col = threadIdx.x;
int local_row = threadIdx.y;
if (global_row >= M || global_col >= N) return;
__shared__ float Asub[TS][TS];
__shared__ float Bsub[TS][TS];
const int num_tiles = K / TS;
float acc = 0;
for(int t = 0; t < num_tiles; t++){
const int t_row = TS * t + local_row;
const int t_col = TS * t + local_col;
Asub[local_row][local_col] = A[global_row * K + t_col];
Bsub[local_row][local_col] = B[t_row * N + global_col];
__syncthreads();
printf("[DEBUG] first sync threads, global_row: %d, global_col: %d\n", global_row, global_col);
for (int k = 0; k < K; ++k) {
acc += Asub[local_row][k] * Bsub[k][local_col];
}
__syncthreads();
printf("[DEBUG] second sync threads, global_row: %d, global_col: %d\n", global_row, global_col);
}
C[global_row * N + global_col] = acc;
}
static float *a_d, *b_d, *c_d;
void mat_mul(float *A, float *B, float *C, int M, int N, int K) {
cudaMemcpy(a_d, A, M * K * sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(b_d, B, K * N * sizeof(float), cudaMemcpyHostToDevice);
dim3 blockDim(TS, TS);
dim3 gridDim(M/TS, N/TS);
shared_kernel<<<gridDim, blockDim>>>(a_d, b_d, c_d, M, N, K);
cudaMemcpy(C, c_d, M * N * sizeof(float), cudaMemcpyDeviceToHost);
cudaDeviceSynchronize();
}
void mat_mul_init(float *A, float *B, float *C, int M, int N, int K) {
cudaGetDeviceCount(&num_devices);
cudaSetDevice(0);
cudaMalloc(&a_d, M * K * sizeof(float));
cudaMalloc(&b_d, K * N * sizeof(float));
cudaMalloc(&c_d, M * N * sizeof(float));
}
Above example is a matrix multiplication with shared memory.
I ran above kernel with dim3 blockDim(TS, TS) and dim3 gridDim(M/TS, N/TS) and M, N, K = 128.
I checked that float * C has zero value after launching kernel. Also, I found that only few of global_row are printed(from 37 to 81) after first __syncthreads(), and there is no printf DEBUG message after the second __syncthreads().
I suspect that __syncthreads() is causing the problem, but I don't know how to fix it. My code is almost the same as other matrix multiplication code in other site.
Would you give me some hint how to solve this?
Any time you are having trouble with a CUDA code, I recommend using proper CUDA error checking and run your code with compute-sanitizer or cuda-memcheck. For this type of analysis, it will be easier if you don't use in-kernel printf.
If you did that, you would see output like this:
========= Invalid __shared__ read of size 4
========= at 0x000002f0 in shared_kernel(float*, float*, float*, int, int, int)
========= by thread (0,2,0) in block (0,1,0)
========= Address 0x00002000 is out of bounds
========= Saved host backtrace up to driver entry point at kernel launch time
... (and more output)
So from that, we can see that your kernel is making invalid __shared__ read operations. Where is that happening in your kernel? You could use the methodology here to identify a specific line of code. However this is a fairly simple kernel, and there is only one line that is reading from shared memory, it is here:
for (int k = 0; k < K; ++k) {
acc += Asub[local_row][k] * Bsub[k][local_col]; // shared reads here
A quick inspection will show that if you let this loop iterate over a range of K=128, then you will index out of bounds here:
for (int k = 0; k < K; ++k) {
acc += Asub[local_row][k] * Bsub[k][local_col];
^ ^
when k is greater than 31, because this would exceed your shared array dimensions:
#define TS 32
__shared__ float Asub[TS][TS];
__shared__ float Bsub[TS][TS];
I'm not going to bother writing a fixed kernel/code for you, because as you've already pointed out, this topic is covered in many other places, and a canonical example is already provided in the programming guide.
FWIW, if i change your for-loop to this:
for (int k = 0; k < TS; ++k) {
then the run-time errors go away for me. cuda-memcheck reports no errors.

Thrust/CUDA replicate an array multiple times combined with the values of another array

Let's say I have two arrays
A = {1, 2, 3}
and
B = {10,20,30,40,50}
I want to generate a new array which would have a size of
sizeof(A) * sizeof(B)
I want to replicate B sizeof(A) times, and on each repetition i, the resultant array should have A[i] added to it. So the result would be something like
{11,21,31,41,51,12,22,32,42,52,13,23,33,43,53}
This task can be interpreted as a 2-dimensional problem where the output array can be treated as a matrix of dimensions sizeof(A) times sizeof(B). In this way, we can use 2D CUDA indexing to achieve the desired functionality. A sample CUDA C++ code of this 2D implementation is shown below:
#include <iostream>
#include <cuda_runtime.h>
#include <cassert>
using namespace std;
__global__ void kernel_replicate(int* a, int* b, int* c, int alen, int blen, int clen)
{
const int ai = blockIdx.x * blockDim.x + threadIdx.x;
const int bi = blockIdx.y * blockDim.y + threadIdx.y;
if(ai<alen && bi<blen)
{
const int ci = ai * blen + bi;
c[ci] = a[ai] + b[bi];
}
}
void replicate_device(int* a, int* b, int* c, int alen, int blen, int clen)
{
dim3 block(16,16);
dim3 grid;
grid.x = (alen + block.x - 1) / block.x;
grid.y = (blen + block.y - 1) / block.y;
kernel_replicate<<<grid, block>>>(a,b,c,alen,blen,clen);
assert(cudaSuccess == cudaDeviceSynchronize());
}
void replicate(int* a, int* b, int* c, int alen, int blen, int clen)
{
int *ad, *bd, *cd;
size_t abytes = alen * sizeof(int);
size_t bbytes = blen * sizeof(int);
size_t cbytes = clen * sizeof(int);
cudaMalloc(&ad, abytes);
cudaMalloc(&bd, bbytes);
cudaMalloc(&cd, cbytes);
cudaMemcpy(ad,a, abytes, cudaMemcpyHostToDevice);
cudaMemcpy(bd,b, bbytes, cudaMemcpyHostToDevice);
replicate_device(ad,bd,cd, alen,blen,clen);
cudaMemcpy(c,cd, cbytes, cudaMemcpyDeviceToHost);
cudaFree(ad);
cudaFree(bd);
cudaFree(cd);
}
int main()
{
const int alen = 3;
const int blen = 5;
const int clen = alen * blen;
int A[alen] = {1,2,3};
int B[blen] = {10,20,30,40,50};
int C[clen] = {0};
replicate(A,B,C,alen, blen, clen);
for(int i=0; i<alen; i++)
{
cout<<A[i]<<" ";
}
cout<<endl;
for(int i=0; i<blen; i++)
{
cout<<B[i]<<" ";
}
cout<<endl;
for(int i=0; i<clen; i++)
{
cout<<C[i]<<" ";
}
cout<<endl;
return 0;
}

Is one CUDA block dimension faster than the other?

I have a simple CUDA code that assigns the values of an NxN matrix A to matrix B. In one case, I declare block sizes block(1,32) and have each thread loop over the entries in the first matrix dimension. In the second case,
I declare block sizes block(32,1) and have each thread loop over entries in the
second matrix dimension.
Is there some really obvious reason why, in my code below, threads that loop over the stride 1 memory are significantly slower than those that the loop over stride N memory? I would have thought it was the other way around (if there is any difference at all).
Am I missing something really obvious (a bug, perhaps)?
The complete code is below.
#include <stdio.h>
#include <sys/time.h>
__global__ void addmat_x(int m, int n, int* A, int *B)
{
int idx, ix;
int iy = threadIdx.y + blockIdx.y*blockDim.y;
if (iy < n)
for(ix = 0; ix < m; ix++) {
idx = iy*m + ix; /* iy*m is constant */
B[idx] = A[idx];
}
}
__global__ void addmat_y(int m, int n, int* A, int *B)
{
int ix = threadIdx.x + blockIdx.x*blockDim.x;
int idx, iy;
if (ix < m)
for(iy = 0; iy < n; iy++) {
idx = iy*m + ix;
B[idx] = A[idx];
}
}
double cpuSecond()
{
struct timeval tp;
gettimeofday(&tp,NULL);
return (double) tp.tv_sec + (double)tp.tv_usec*1e-6;
}
int main(int argc, char** argv)
{
int *A, *B;
int *dev_A, *dev_B;
size_t m, n, nbytes;
double etime, start;
m = 1 << 14;
n = 1 << 14;
nbytes = m*n*sizeof(int);
A = (int*) malloc(nbytes);
B = (int*) malloc(nbytes);
memset(A,0,nbytes);
cudaMalloc((void**) &dev_A, nbytes);
cudaMalloc((void**) &dev_B, nbytes);
cudaMemcpy(dev_A, A, nbytes, cudaMemcpyHostToDevice);
#if 1
/* One thread per row */
dim3 block(1,32);
dim3 grid(1,(n+block.y-1)/block.y);
start = cpuSecond();
addmat_x<<<grid,block>>>(m,n,dev_A, dev_B);
#else
/* One thread per column */
dim3 block(32,1);
dim3 grid((m+block.x-1)/block.x,1);
start = cpuSecond();
addmat_y<<<grid,block>>>(m,n,dev_A, dev_B);
#endif
cudaDeviceSynchronize();
etime = cpuSecond() - start;
printf("GPU Kernel %10.3g (s)\n",etime);
cudaFree(dev_A);
cudaFree(dev_B);
free(A);
free(B);
cudaDeviceReset();
}
Lets compare the global memory indexing generated by each thread, in each case.
addmat_x:
Your block dimension is (1,32). This means 1 thread wide in x, 32 threads "long" in y. The threadId.x value for each thread will be 0. The threadIdx.y value for the threads in the warp will range from 0 to 31, as you move from thread to thread in the warp. With that, let's inspect your creation of idx in that kernel:
m = 1 << 14;
...
int iy = threadIdx.y + blockIdx.y*blockDim.y;
idx = iy*m + ix;
let's choose the first block, whose blockIdx.y is 0. Then:
idx = threadIdx.y*(1<<14) + ix;
For the first loop iteration, ix is 0. The idx values generated by each thread will be:
threadIdx.y: | idx:
0 0
1 (1<<14)
2 2*(1<<14)
...
31 31*(1<<14)
For a given loop iteration, the distance from the load or store index from one thread to the next will be 1<<14. i.e. not adjacent. Scattered.
addmat_y:
Your block dimension is (32,1). This means 32 threads wide in x, 1 thread "long" in y. The threadIdx.y value for each thread will be 0. The threadIdx.x value for the threads in the warp will range from 0 to 31, as you move from thread to thread. Now let's inspect your creation of idx in that kernel:
m = 1 << 14;
...
int ix = threadIdx.x + blockIdx.x*blockDim.x;
idx = iy*m + ix;
Let's choose the first block, whose blockIdx.x is 0. Then:
idx = iy*m + threadIdx.x;
For the first loop iteration, iy is 0, so we simply have:
idx = threadIdx.x;
This generates the following index pattern across the warp:
threadIdx.x: | idx:
0 0
1 1
2 2
...
31 31
These indices are adjacent, it is not a scattered load or store, the addresses will coalesce nicely, and this represents "efficient" use of global memory. It will perform faster than the first case.

how to set values in constant structure in cuda [duplicate]

I am trying to use the constant memory in the code with constant memory assigned value from kernel not using cudacopytosymbol.
#include <iostream>
using namespace std;
#define N 10
//__constant__ int constBuf_d[N];
__constant__ int *constBuf;
__global__ void foo( int *results )
{
int tdx = threadIdx.x;
int idx = blockIdx.x * blockDim.x + tdx;
if( idx < N )
{
constBuf[idx]=1;
results[idx] = constBuf[idx];
}
}
// main routine that executes on the host
int main(int argc, char* argv[])
{
int *results_h = new int[N];
int *results_d;
cudaMalloc((void **)&results_d, N*sizeof(int));
foo <<< 1, 10 >>> ( results_d );
cudaMemcpy(results_h, results_d, N*sizeof(int), cudaMemcpyDeviceToHost);
for( int i=0; i < N; ++i )
printf("%i ", results_h[i] );
delete(results_h);
}
output shows
6231808 6226116 0 0 0 0 0 0 0 0
I want the program to print the value assigned to constant memory through the kenel in the code.
Constant memory is, as the name implies, constant/read-only with respect to device code. What you are trying to do is illegal and can't be made to work.
To set values in constant memory, you currently have two choices:
set the value from host code via the cudaMemcpyToSymbol API call (or its equivalents)
use static initialisation at compile time
In the latter case something like this would work:
__constant__ int constBuf[N] = { 16, 2, 77, 40, 12, 3, 5, 3, 6, 6 };
__global__ void foo( int *results )
{
int tdx = threadIdx.x;
int idx = blockIdx.x * blockDim.x + tdx;
if( tdx < N )
{
results[idx] = constBuf[tdx]; // Note changes here!
}
}

Get statistics for a list of numbers using GPU

I have several lists of numbers on a file . For example,
.333, .324, .123 , .543, .00054
.2243, .333, .53343 , .4434
Now, I want to get the number of times each number occurs using the GPU. I believe this will be faster to do on the GPU than the CPU because each thread can process one list. What data structure should I use on the GPU to easily get the above counts. For example , for the above, the answer will look as follows:
.333 = 2 times in entire file
.324 = 1 time
etc..
I looking for a general solution. Not one that works only on devices with specific compute capability
Just writing kernel suggested by Pavan to see if I have implemented it efficiently:
int uniqueEle = newend.valiter – d_A;
int* count;
cudaMalloc((void**)&count, uniqueEle * sizeof(int)); // stores the count of each unique element
int TPB = 256;
int blocks = uniqueEle + TPB -1 / TPB;
//Cast d_I to raw pointer called d_rawI
launch<<<blocks,TPB>>>(d_rawI,count,uniqueEle);
__global__ void launch(int *i, int* count, int n){
int id = blockDim.x * blockIdx.x + threadIdx.x;
__shared__ int indexes[256];
if(id < n ){
indexes[threadIdx.x] = i[id];
//as occurs between two blocks
if(id % 255 == 0){
count[indexes] = i[id+1] - i[id];
}
}
__syncthreads();
if(id < ele - 1){
if(threadIdx.x < 255)
count[id] = indexes[threadIdx.x+1] – indexes[threadIdx.x];
}
}
Question: how to modify this kernel so that it handles arrays of arbitrary size. I.e , handle the condition when the total number of threads < number of elements
Here is how I would do the code in matlab
A = [333, .324, .123 , .543, .00054 .2243, .333, .53343 , .4434];
[values, locations] = unique(A); % Find unique values and their locations
counts = diff([0, locations]); % Find the count based on their locations
There is no easy way to do this in plain cuda, but you can use existing libraries to do this.
1) Thrust
It is also being shipped with CUDA toolkit from CUDA 4.0.
The matlab code can be roughly translated into thrust by using the following functions. I am not too proficient with thrust, but I am just trying to give you an idea on what routines to look at.
float _A[] = {.333, .324, .123 , .543, .00054 .2243, .333, .53343 , .4434};
int _I[] = {0, 1, 2, 3, 4, 5, 6, 7, 8};
float *A, *I;
// Allocate memory on device and cudaMempCpy values from _A to A and _I to I
int num = 9;
// Values vector
thrust::device_vector<float>d_A(A, A+num);
// Need to sort to get same values together
thrust::stable_sort(d_A, d_A+num);
// Vector containing 0 to num-1
thrust::device_vector<int>d_I(I, I+num);
// Find unique values and elements
thrust::device_vector<float>d_Values(num), d_Locations(num), d_counts(num);
// Find unique elements
thrust::device_vector<float>::iterator valiter;
thrust::device_vector<int>::iterator idxiter;
thrust::pair<valiter, idxiter> new_end;
new_end = thrust::unique_by_key(d_A, d_A+num, d_I, d_Values, d_Locations);
You now have the locations of the first instance of each unique value. You can now launch a kernel to find the differences between adjacent elements from 0 to new_end in d_Locations. Subtract the final value from num to get the count for final location.
EDIT (Adding code that was provided over chat)
Here is how the difference code needs to be done
#define MAX_BLOCKS 65535
#define roundup(A, B) = (((A) + (B) - 1) / (B))
int uniqueEle = newend.valiter – d_A;
int* count;
cudaMalloc((void**)&count, uniqueEle * sizeof(int));
int TPB = 256;
int num_blocks = roundup(uniqueEle, TPB);
int blocks_y = roundup(num_blocks, MAX_BLOCKS);
int blocks_x = roundup(num_blocks, blocks_y);
dim3 blocks(blocks_x, blocks_y);
kernel<<<blocks,TPB>>>(d_rawI, count, uniqueEle);
__global__ void kernel(float *i, int* count, int n)
{
int tx = threadIdx.x;
int bid = blockIdx.y * gridDim.x + blockIdx.x;
int id = blockDim.x * bid + tx;
__shared__ int indexes[256];
if (id < n) indexes[tx] = i[id];
__syncthreads();
if (id < n - 1) {
if (tx < 255) count[id] = indexes[tx + 1] - indexes[tx];
else count[id] = i[id + 1] - indexes[tx];
}
if (id == n - 1) count[id] = n - indexes[tx];
return;
}
2) ArrayFire
This is an easy to use, free array based library.
You can do the following in ArrayFire.
using namespace af;
float h_A[] = {.333, .324, .123 , .543, .00054 .2243, .333, .53343 , .4434};
int num = 9;
// Transfer data to device
array A(9, 1, h_A);
array values, locations, original;
// Find the unique values and locations
setunique(values, locations, original, A);
// Locations are 0 based, add 1.
// Add *num* at the end to find count of last value.
array counts = diff1(join(locations + 1, num));
Disclosure: I work for AccelerEyes, that develops this software.
To answer the latest addenum to this question - the diff kernel which would complete the thrust method proposed by Pavan could look something like this:
template<int blcksz>
__global__ void diffkernel(const int *i, int* count, const int n) {
int id = blockDim.x * blockIdx.x + threadIdx.x;
int strd = blockDim.x * gridDim.x;
int nmax = blcksz * ((n/blcksz) + ((n%blcksz>0) ? 1 : 0));
__shared__ int indices[blcksz+1];
for(; id<nmax; id+=strd) {
// Data load
indices[threadIdx.x] = (id < n) ? i[id] : n;
if (threadIdx.x == (blcksz-1))
indices[blcksz] = ((id+1) < n) ? i[id+1] : n;
__syncthreads();
// Differencing calculation
int diff = indices[threadIdx.x+1] - indices[threadIdx.x];
// Store
if (id < n) count[id] = diff;
__syncthreads();
}
}
here is a solution:
__global__ void counter(float* a, int* b, int N)
{
int idx = blockIdx.x*blockDim.x+threadIdx.x;
if(idx < N)
{
float my = a[idx];
int count = 0;
for(int i=0; i < N; i++)
{
if(my == a[i])
count++;
}
b[idx]=count;
}
}
int main()
{
int threads = 9;
int blocks = 1;
int N = blocks*threads;
float* h_a;
int* h_b;
float* d_a;
int* d_b;
h_a = (float*)malloc(N*sizeof(float));
h_b = (int*)malloc(N*sizeof(int));
cudaMalloc((void**)&d_a,N*sizeof(float));
cudaMalloc((void**)&d_b,N*sizeof(int));
h_a[0]= .333f;
h_a[1]= .324f;
h_a[2]= .123f;
h_a[3]= .543f;
h_a[4]= .00054f;
h_a[5]= .2243f;
h_a[6]= .333f;
h_a[7]= .53343f;
h_a[8]= .4434f;
cudaMemcpy(d_a,h_a,N*sizeof(float),cudaMemcpyHostToDevice);
counter<<<blocks,threads>>>(d_a,d_b,N);
cudaMemcpy(h_b,d_b,N*sizeof(int),cudaMemcpyDeviceToHost);
for(int i=0; i < N; i++)
{
printf("%f = %d times\n",h_a[i],h_b[i]);
}
cudaFree(d_a);
cudaFree(d_b);
free(h_a);
free(h_b);
getchar();
return 0;
}