cudaMalloc and cudaMemcpy not working on kernel call - memory

I have an array already initialized that I am trying to use in each thread of the kernel call (each thread uses a different part of the array so there are no dependencies). I create the array and save memory on the device using cudaMalloc and the array is copied from host to device using cudaMemcpy.
I pass the pointer returned by cudaMalloc to the kernel call to be used by each thread.
int SIZE = 100;
int* data = new int[SIZE];
int* d_data = 0;
cutilSafeCall( cudaMalloc(&d_data, SIZE * sizeof(int)) );
for (int i = 0; i < SIZE; i++)
data[i] = i;
cutilSafeCall( cudaMemcpy(d_data, data, SIZE * sizeof(int), cudaMemcpyHostToDevice) );
This code was taken from here.
For the kernel call.
kernel<<<blocks, threads>>> (results, d_data);
I keep track of the results from each thread by using the struct Result. The next code works without errors.
__global__ void mainKernel(Result res[], int* data){
int x = data[0];
}
But when I assign that value to res:
__global__ void mainKernel(Result res[], int* data){
int threadId = (blockIdx.x * blockDim.x) + threadIdx.x;
int x = data[0];
res[threadId].x = x;
}
An error is raised:
cudaSafeCall() Runtime API error in file , line 355 : an illegal memory access was encountered.
The same error appears with any operation involving the use of that pointer
__global__ void mainKernel(Result res[], int* data){
int threadId = (blockIdx.x * blockDim.x) + threadIdx.x;
int x = data[0];
if (x > 10)
res[threadId].x = 5;
}
There is no problem with the definition of res. Assigning any other value to res[threadId].x does not give me any error.
This is the output of running cuda-memcheck:
========= Invalid __global__ read of size 4
========= at 0x00000150 in mainKernel(Result*, int*)
========= by thread (86,0,0) in block (49,0,0)
========= Address 0x13024c0000 is out of bounds
========= Saved host backtrace up to driver entry point at kernel launch time
========= Host Frame:/usr/lib/x86_64-linux-gnu/libcuda.so.1 (cuLaunchKernel + 0x2cd) [0x150d6d]
========= Host Frame:./out [0x2cc4b]
========= Host Frame:./out [0x46c23]
========= Host Frame:./out [0x3e37]
========= Host Frame:./out [0x3ca1]
========= Host Frame:./out [0x3cd6]
========= Host Frame:./out [0x39e9]
========= Host Frame:/lib/x86_64-linux-gnu/libc.so.6 (__libc_start_main + 0xf5) [0x21ec5]
========= Host Frame:./out [0x31b9]
EDIT:
This is an example of the full code:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <iostream>
#include <assert.h>
typedef struct
{
int x,y,z;
} Result;
__global__ void mainKernel(Result pResults[], int* dataimage)
{
int threadId = (blockIdx.x * blockDim.x) + threadIdx.x;
int xVal = dataimage[0];
if (xVal > 10)
pResults[threadId].x = 5;
}
int main (int argc, char** argv)
{
int NUM_THREADS = 5*5;
int SIZE = 100;
int* data = new int[SIZE];
int* d_data = 0;
cutilSafeCall( cudaMalloc(&d_data, SIZE * sizeof(int)) );
for (int i = 0; i < SIZE; i++)
data[i] = i;
cutilSafeCall( cudaMemcpy(d_data, data, SIZE * sizeof(int), cudaMemcpyHostToDevice) );
unsigned int GPU_ID = 1; // not actually :-)
// unsigned int GPU_ID = cutGetMaxGflopsDeviceId() ;
cudaSetDevice(GPU_ID);
Result * results_GPU = 0;
cutilSafeCall( cudaMalloc( &results_GPU, NUM_THREADS * sizeof(Result)) );
Result * results_CPU = 0;
cutilSafeCall( cudaMallocHost( &results_CPU, NUM_THREADS * sizeof(Result)) );
mainKernel<<<5,5>>> ( results_GPU, d_data );
cudaThreadSynchronize();
cutilSafeCall( cudaMemcpy(results_CPU, results_GPU, NUM_THREADS * sizeof(Result),cudaMemcpyDeviceToHost) );
cutilSafeCall(cudaFree(results_GPU));
cutilSafeCall(cudaFreeHost(results_CPU));
cudaThreadExit();
} // ()

Your problem lies in this sequence of calls:
cutilSafeCall( cudaMalloc(&d_data, SIZE * sizeof(int)) );
for (int i = 0; i < SIZE; i++)
data[i] = i;
cutilSafeCall( cudaMemcpy(d_data, data, SIZE * sizeof(int), cudaMemcpyHostToDevice) );
unsigned int GPU_ID = 1;
cudaSetDevice(GPU_ID);
Result * results_GPU = 0;
cutilSafeCall( cudaMalloc( &results_GPU, NUM_THREADS * sizeof(Result)) );
Result * results_CPU = 0;
cutilSafeCall( cudaMallocHost( &results_CPU, NUM_THREADS * sizeof(Result)) );
mainKernel<<<5,5>>> ( results_GPU, d_data );
What is effectively happening is that you are allocating d_data and running your kernel on different GPUs, and d_data is not valid on the GPU you are launching the kernel on.
In detail, because you call cudaMalloc for d_data before cudaSetDevice, you are allocating d_data on the default device, and then explicitly allocating results_GPU and running the kernel on device 1. Clearly device 1 and the default device are not the same GPU (enumeration of devices usually starts at 0 in the runtime API).
If you change the code like this:
unsigned int GPU_ID = 1;
cutilSafeCall(cudaSetDevice(GPU_ID));
cutilSafeCall( cudaMalloc(&d_data, SIZE * sizeof(int)) );
for (int i = 0; i < SIZE; i++)
data[i] = i;
cutilSafeCall( cudaMemcpy(d_data, data, SIZE * sizeof(int), cudaMemcpyHostToDevice) );
Result * results_GPU = 0;
cutilSafeCall( cudaMalloc( &results_GPU, NUM_THREADS * sizeof(Result)) );
Result * results_CPU = 0;
cutilSafeCall( cudaMallocHost( &results_CPU, NUM_THREADS * sizeof(Result)) );
mainKernel<<<5,5>>> ( results_GPU, d_data );
i.e. select the non-default device before any allocations are made, the problem should disappear. The reason this doesn't happen with your very simple kernel:
__global__ void mainKernel(Result res[], int* data){
int x = data[0];
}
is simply that the CUDA compiler performs very aggressive optimisations by default, and because the result of the read of data[0] isn't actually used, the entire read can be optimised away and you are left with an empty stub kernel which doesn't do anything. Only when the result of the load from memory is used in a memory write will the code not be optimised away during compilation. You can confirm this yourself by dissassembling the code emitted by the compiler, if you are curious.
Note that there are ways to make this work on multi-GPU systems which supported it, via peer-to-peer access, but that must be explicitly configured in your code for that facility to be used.

Related

How to discover physical address corresponding to PCIe device memory?

I'm trying to access a PCIe device memory from a user space program. I open the file: /sys/bus/pci/devices/0000:3b:00.0/resource0 and then I call mmap that will return a virtual address.
When writing at this virtual address (VA) the MMU will translate it to a physical address (PA), the memory controller will convert the write to the PA into a TLP to request a write to the PCIe device. (AFAIU)
How can I get the physical address that is being used? I had a look to /proc//maps and I see that there is an address that coincides with the PCIe bar0 address (0xa0000000).
But this address seems too low, it overlaps with DDR memory.
I also tried this program to convert VA to PA but it doesn't seem to give sensible results for such mapping:
virt2phys$ cat v2p.c
#define _XOPEN_SOURCE 700
#include <fcntl.h> /* open */
#include <stdint.h> /* uint64_t */
#include <stdio.h> /* printf */
#include <stdlib.h> /* size_t */
#include <unistd.h> /* pread, sysconf */
typedef struct {
uint64_t pfn : 55;
unsigned int soft_dirty : 1;
unsigned int file_page : 1;
unsigned int swapped : 1;
unsigned int present : 1;
} PagemapEntry;
/* Parse the pagemap entry for the given virtual address.
*
* #param[out] entry the parsed entry
* #param[in] pagemap_fd file descriptor to an open /proc/pid/pagemap file
* #param[in] vaddr virtual address to get entry for
* #return 0 for success, 1 for failure
*/
int pagemap_get_entry(PagemapEntry *entry, int pagemap_fd, uintptr_t vaddr)
{
size_t nread;
ssize_t ret;
uint64_t data;
uintptr_t vpn;
vpn = vaddr / sysconf(_SC_PAGE_SIZE);
nread = 0;
while (nread < sizeof(data)) {
ret = pread(pagemap_fd, ((uint8_t*)&data) + nread, sizeof(data) - nread,
vpn * sizeof(data) + nread);
nread += ret;
if (ret <= 0) {
return 1;
}
}
entry->pfn = data & (((uint64_t)1 << 55) - 1);
entry->soft_dirty = (data >> 55) & 1;
entry->file_page = (data >> 61) & 1;
entry->swapped = (data >> 62) & 1;
entry->present = (data >> 63) & 1;
return 0;
}
/* Convert the given virtual address to physical using /proc/PID/pagemap.
*
* #param[out] paddr physical address
* #param[in] pid process to convert for
* #param[in] vaddr virtual address to get entry for
* #return 0 for success, 1 for failure
*/
int virt_to_phys_user(uintptr_t *paddr, pid_t pid, uintptr_t vaddr)
{
char pagemap_file[BUFSIZ];
int pagemap_fd;
snprintf(pagemap_file, sizeof(pagemap_file), "/proc/%ju/pagemap", (uintmax_t)pid);
pagemap_fd = open(pagemap_file, O_RDONLY);
if (pagemap_fd < 0) {
return 1;
}
PagemapEntry entry;
if (pagemap_get_entry(&entry, pagemap_fd, vaddr)) {
return 1;
}
close(pagemap_fd);
*paddr = (entry.pfn * sysconf(_SC_PAGE_SIZE)) + (vaddr % sysconf(_SC_PAGE_SIZE));
return 0;
}
int main(int argc, char **argv)
{
pid_t pid;
uintptr_t vaddr, paddr = 0;
if (argc < 3) {
printf("Usage: %s pid vaddr(in hex)\n", argv[0]);
return EXIT_FAILURE;
}
pid = strtoull(argv[1], NULL, 0);
vaddr = strtoull(argv[2], NULL, 16);
if (virt_to_phys_user(&paddr, pid, vaddr)) {
fprintf(stderr, "error: virt_to_phys_user\n");
return EXIT_FAILURE;
};
printf("0x%jx\n", (uintmax_t)paddr);
return EXIT_SUCCESS;
}

Converting a 2D Canny Edge image to 1D edge pixel array in CUDA - Strange behaviour

I have a CUDA kernel which takes an edge image and processes it to create a smaller, 1D array of the edge pixels. Now here is the strange behaviour. Every time I run the kernel and calculate the number of edge pixels in "d_nlist" (see the code near the printf), I get a greater pixel count each time, even when I use the same image and stop the program completely and re-run. Therefore, each time I run it, it takes longer to run, until eventually, it throws an un-caught exception.
My question is, how can I stop this from happening so that I can get consistent results each time I run the kernel?
My device is a Geforce 620.
Constants:
THREADS_X = 32
THREADS_Y = 4
PIXELS_PER_THREAD = 4
MAX_QUEUE_LENGTH = THREADS_X * THREADS_Y * PIXELS_PER_THREAD
IMG_WIDTH = 256
IMG_HEIGHT = 256
IMG_SIZE = IMG_WIDTH * IMG_HEIGHT
BLOCKS_X = IMG_WIDTH / (THREADS_X * PIXELS_PER_THREAD)
BLOCKS_Y = IMG_HEIGHT / THREADS_Y
The kernel is as follows:
__global__ void convert2DEdgeImageTo1DArray( unsigned char const * const image,
unsigned int* const list, int* const glob_index ) {
unsigned int const x = blockIdx.x * THREADS_X*PIXELS_PER_THREAD + threadIdx.x;
unsigned int const y = blockIdx.y * THREADS_Y + threadIdx.y;
volatile int qindex = -1;
volatile __shared__ int sh_qindex[THREADS_Y];
volatile __shared__ int sh_qstart[THREADS_Y];
sh_qindex[threadIdx.y] = -1;
// Start by making an array
volatile __shared__ unsigned int sh_queue[MAX_QUEUE_LENGTH];
// Fill the queue
for(int i=0; i<PIXELS_PER_THREAD; i++)
{
int const xx = i*THREADS_X + x;
// Read one image pixel from global memory
unsigned char const pixel = image[y*IMG_WIDTH + xx];
unsigned int const queue_val = (y << 16) + xx;
if(pixel)
{
do {
qindex++;
sh_qindex[threadIdx.y] = qindex;
sh_queue[threadIdx.y*THREADS_X*PIXELS_PER_THREAD + qindex] = queue_val;
} while (sh_queue[threadIdx.y*THREADS_X*PIXELS_PER_THREAD + qindex] != queue_val);
}
// Reload index from smem (last thread to write to smem will have updated it)
qindex = sh_qindex[threadIdx.y];
}
// Let thread 0 reserve the space required in the global list
__syncthreads();
if(threadIdx.x == 0 && threadIdx.y == 0)
{
// Find how many items are stored in each list
int total_index = 0;
#pragma unroll
for(int i=0; i<THREADS_Y; i++)
{
sh_qstart[i] = total_index;
total_index += (sh_qindex[i] + 1u);
}
// Calculate the offset in the global list
unsigned int global_offset = atomicAdd(glob_index, total_index);
#pragma unroll
for(int i=0; i<THREADS_Y; i++)
{
sh_qstart[i] += global_offset;
}
}
__syncthreads();
// Copy local queues to global queue
for(int i=0; i<=qindex; i+=THREADS_X)
{
if(i + threadIdx.x > qindex)
break;
unsigned int qvalue = sh_queue[threadIdx.y*THREADS_X*PIXELS_PER_THREAD + i + threadIdx.x];
list[sh_qstart[threadIdx.y] + i + threadIdx.x] = qvalue;
}
}
The following is the method which calls the kernel:
void call2DTo1DKernel(unsigned char const * const h_image)
{
// Device side allocation
unsigned char *d_image = NULL;
unsigned int *d_list = NULL;
int h_nlist, *d_nlist = NULL;
cudaMalloc((void**)&d_image, sizeof(unsigned char)*IMG_SIZE);
cudaMalloc((void**)&d_list, sizeof(unsigned int)*IMG_SIZE);
cudaMalloc((void**)&d_nlist, sizeof(int));
// Time measurement initialization
cudaEvent_t start, stop, startio, stopio;
cudaEventCreate(&start);
cudaEventCreate(&stop);
cudaEventCreate(&startio);
cudaEventCreate(&stopio);
// Start timer w/ io
cudaEventRecord(startio,0);
// Copy image data to device
cudaMemcpy((void*)d_image, (void*)h_image, sizeof(unsigned char)*IMG_SIZE, cudaMemcpyHostToDevice);
// Start timer
cudaEventRecord(start,0);
// Kernel call
// Phase 1 : Convert 2D binary image to 1D pixel array
dim3 dimBlock1(THREADS_X, THREADS_Y);
dim3 dimGrid1(BLOCKS_X, BLOCKS_Y);
convert2DEdgeImageTo1DArray<<<dimGrid1, dimBlock1>>>(d_image, d_list, d_nlist);
// Stop timer
cudaEventRecord(stop,0);
cudaEventSynchronize(stop);
// Stop timer w/ io
cudaEventRecord(stopio,0);
cudaEventSynchronize(stopio);
// Time measurement
cudaEventElapsedTime(&et,start,stop);
cudaEventElapsedTime(&etio,startio,stopio);
// Time measurement deinitialization
cudaEventDestroy(start);
cudaEventDestroy(stop);
cudaEventDestroy(startio);
cudaEventDestroy(stopio);
// Get list size
cudaMemcpy((void*)&h_nlist, (void*)d_nlist, sizeof(int), cudaMemcpyDeviceToHost);
// Report on console
printf("%d pixels processed...\n", h_nlist);
// Device side dealloc
cudaFree(d_image);
cudaFree(d_space);
cudaFree(d_list);
cudaFree(d_nlist);
}
Thank you very much in advance for your help everyone.
As a preamble, let me suggest some troubleshooting steps that are useful:
instrument your code with proper cuda error checking
run your code with cuda-memcheck e.g. cuda-memcheck ./myapp
If you do the above steps, you'll find that your kernel is failing, and the failures have to do with global writes of size 4. So that will focus your attention on the last segment of your kernel, beginning with the comment // Copy local queues to global queue
Regarding your code, then, you have at least 2 problems:
The addressing/indexing in your final segment of your kernel, where you are writing the individual queues out to global memory, is messed up. I'm not going to try and debug this for you.
You are not initializing your d_nlist variable to zero. Therefore when you do an atomic add to it, you are adding your values to a junk value, which will tend to increase as you repeat the process.
Here's some code which has the problems removed, (I did not try to sort out your queue copy code) and error checking added. It produces repeatable results for me:
$ cat t216.cu
#include <stdio.h>
#include <stdlib.h>
#define THREADS_X 32
#define THREADS_Y 4
#define PIXELS_PER_THREAD 4
#define MAX_QUEUE_LENGTH (THREADS_X*THREADS_Y*PIXELS_PER_THREAD)
#define IMG_WIDTH 256
#define IMG_HEIGHT 256
#define IMG_SIZE (IMG_WIDTH*IMG_HEIGHT)
#define BLOCKS_X (IMG_WIDTH/(THREADS_X*PIXELS_PER_THREAD))
#define BLOCKS_Y (IMG_HEIGHT/THREADS_Y)
#define cudaCheckErrors(msg) \
do { \
cudaError_t __err = cudaGetLastError(); \
if (__err != cudaSuccess) { \
fprintf(stderr, "Fatal error: %s (%s at %s:%d)\n", \
msg, cudaGetErrorString(__err), \
__FILE__, __LINE__); \
fprintf(stderr, "*** FAILED - ABORTING\n"); \
exit(1); \
} \
} while (0)
__global__ void convert2DEdgeImageTo1DArray( unsigned char const * const image,
unsigned int* const list, int* const glob_index ) {
unsigned int const x = blockIdx.x * THREADS_X*PIXELS_PER_THREAD + threadIdx.x;
unsigned int const y = blockIdx.y * THREADS_Y + threadIdx.y;
volatile int qindex = -1;
volatile __shared__ int sh_qindex[THREADS_Y];
volatile __shared__ int sh_qstart[THREADS_Y];
sh_qindex[threadIdx.y] = -1;
// Start by making an array
volatile __shared__ unsigned int sh_queue[MAX_QUEUE_LENGTH];
// Fill the queue
for(int i=0; i<PIXELS_PER_THREAD; i++)
{
int const xx = i*THREADS_X + x;
// Read one image pixel from global memory
unsigned char const pixel = image[y*IMG_WIDTH + xx];
unsigned int const queue_val = (y << 16) + xx;
if(pixel)
{
do {
qindex++;
sh_qindex[threadIdx.y] = qindex;
sh_queue[threadIdx.y*THREADS_X*PIXELS_PER_THREAD + qindex] = queue_val;
} while (sh_queue[threadIdx.y*THREADS_X*PIXELS_PER_THREAD + qindex] != queue_val);
}
// Reload index from smem (last thread to write to smem will have updated it)
qindex = sh_qindex[threadIdx.y];
}
// Let thread 0 reserve the space required in the global list
__syncthreads();
if(threadIdx.x == 0 && threadIdx.y == 0)
{
// Find how many items are stored in each list
int total_index = 0;
#pragma unroll
for(int i=0; i<THREADS_Y; i++)
{
sh_qstart[i] = total_index;
total_index += (sh_qindex[i] + 1u);
}
// Calculate the offset in the global list
unsigned int global_offset = atomicAdd(glob_index, total_index);
#pragma unroll
for(int i=0; i<THREADS_Y; i++)
{
sh_qstart[i] += global_offset;
}
}
__syncthreads();
// Copy local queues to global queue
/*
for(int i=0; i<=qindex; i+=THREADS_X)
{
if(i + threadIdx.x > qindex)
break;
unsigned int qvalue = sh_queue[threadIdx.y*THREADS_X*PIXELS_PER_THREAD + i + threadIdx.x];
list[sh_qstart[threadIdx.y] + i + threadIdx.x] = qvalue;
}
*/
}
void call2DTo1DKernel(unsigned char const * const h_image)
{
// Device side allocation
unsigned char *d_image = NULL;
unsigned int *d_list = NULL;
int h_nlist=0, *d_nlist = NULL;
cudaMalloc((void**)&d_image, sizeof(unsigned char)*IMG_SIZE);
cudaMalloc((void**)&d_list, sizeof(unsigned int)*IMG_SIZE);
cudaMalloc((void**)&d_nlist, sizeof(int));
cudaCheckErrors("cudamalloc fail");
// Time measurement initialization
cudaEvent_t start, stop, startio, stopio;
cudaEventCreate(&start);
cudaEventCreate(&stop);
cudaEventCreate(&startio);
cudaEventCreate(&stopio);
float et, etio;
// Start timer w/ io
cudaEventRecord(startio,0);
cudaMemcpy(d_nlist, &h_nlist, sizeof(int), cudaMemcpyHostToDevice);
// Copy image data to device
cudaMemcpy((void*)d_image, (void*)h_image, sizeof(unsigned char)*IMG_SIZE, cudaMemcpyHostToDevice);
cudaCheckErrors("cudamemcpy 1");
// Start timer
cudaEventRecord(start,0);
// Kernel call
// Phase 1 : Convert 2D binary image to 1D pixel array
dim3 dimBlock1(THREADS_X, THREADS_Y);
dim3 dimGrid1(BLOCKS_X, BLOCKS_Y);
convert2DEdgeImageTo1DArray<<<dimGrid1, dimBlock1>>>(d_image, d_list, d_nlist);
cudaDeviceSynchronize();
cudaCheckErrors("kernel fail");
// Stop timer
cudaEventRecord(stop,0);
cudaEventSynchronize(stop);
// Stop timer w/ io
cudaEventRecord(stopio,0);
cudaEventSynchronize(stopio);
// Time measurement
cudaEventElapsedTime(&et,start,stop);
cudaEventElapsedTime(&etio,startio,stopio);
// Time measurement deinitialization
cudaEventDestroy(start);
cudaEventDestroy(stop);
cudaEventDestroy(startio);
cudaEventDestroy(stopio);
// Get list size
cudaMemcpy((void*)&h_nlist, (void*)d_nlist, sizeof(int), cudaMemcpyDeviceToHost);
cudaCheckErrors("cudaMemcpy 2");
// Report on console
printf("%d pixels processed...\n", h_nlist);
// Device side dealloc
cudaFree(d_image);
// cudaFree(d_space);
cudaFree(d_list);
cudaFree(d_nlist);
}
int main(){
unsigned char *image;
image = (unsigned char *)malloc(IMG_SIZE * sizeof(unsigned char));
if (image == 0) {printf("malloc fail\n"); return 0;}
for (int i =0 ; i<IMG_SIZE; i++)
image[i] = rand()%2;
call2DTo1DKernel(image);
call2DTo1DKernel(image);
call2DTo1DKernel(image);
call2DTo1DKernel(image);
call2DTo1DKernel(image);
cudaCheckErrors("some error");
return 0;
}
$ nvcc -arch=sm_20 -O3 -o t216 t216.cu
$ ./t216
32617 pixels processed...
32617 pixels processed...
32617 pixels processed...
32617 pixels processed...
32617 pixels processed...
$ ./t216
32617 pixels processed...
32617 pixels processed...
32617 pixels processed...
32617 pixels processed...
32617 pixels processed...
$

cudaFree is not freeing memory

The code below calculates the dot product of two vectors a and b. The correct result is 8192. When I run it for the first time the result is correct. Then when I run it for the second time the result is the previous result + 8192 and so on:
1st iteration: result = 8192
2nd iteration: result = 8192 + 8192
3rd iteration: result = 8192 + 8192
and so on.
I checked by printing it on screen and the device variable dev_c is not freed. What's more writing to it causes something like a sum, the result beeing the previous value plus the new one being written to it. I guess that could be something with the atomicAdd() operation, but nonetheless cudaFree(dev_c) should erase it after all.
#define N 8192
#define THREADS_PER_BLOCK 512
#define NUMBER_OF_BLOCKS (N/THREADS_PER_BLOCK)
#include <stdio.h>
__global__ void dot( int *a, int *b, int *c ) {
__shared__ int temp[THREADS_PER_BLOCK];
int index = threadIdx.x + blockIdx.x * blockDim.x;
temp[threadIdx.x] = a[index] * b[index];
__syncthreads();
if( 0 == threadIdx.x ) {
int sum = 0;
for( int i= 0; i< THREADS_PER_BLOCK; i++ ){
sum += temp[i];
}
atomicAdd(c,sum);
}
}
int main( void ) {
int *a, *b, *c;
int *dev_a, *dev_b, *dev_c;
int size = N * sizeof( int);
cudaMalloc( (void**)&dev_a, size );
cudaMalloc( (void**)&dev_b, size );
cudaMalloc( (void**)&dev_c, sizeof(int));
a = (int*)malloc(size);
b = (int*)malloc(size);
c = (int*)malloc(sizeof(int));
for(int i = 0 ; i < N ; i++){
a[i] = 1;
b[i] = 1;
}
cudaMemcpy( dev_a, a, size, cudaMemcpyHostToDevice);
cudaMemcpy( dev_b, b, size, cudaMemcpyHostToDevice);
dot<<< N/THREADS_PER_BLOCK,THREADS_PER_BLOCK>>>( dev_a, dev_b, dev_c);
cudaMemcpy( c, dev_c, sizeof(int) , cudaMemcpyDeviceToHost);
printf("Dot product = %d\n", *c);
cudaFree(dev_a);
cudaFree(dev_b);
cudaFree(dev_c);
free(a);
free(b);
free(c);
return 0;
}
cudaFree doesn't erase anything, it simply returns memory to a pool to be re-allocated. cudaMalloc doesn't guarantee the value of memory that has been allocated. You need to initialize memory (both global and shared) that your program uses, in order to have consistent results. The same is true for malloc and free, by the way.
From the documentation of cudaMalloc();
The memory is not cleared.
That means that dev_c is not initialized, and your atomicAdd(c,sum); will add to any random value that happens to be stored in memory at the returned position.

Cuda-memcheck not reporting out of bounds shared memory access

I am runnig the follwoing code using shared memory:
__global__ void computeAddShared(int *in , int *out, int sizeInput){
//not made parameters gidata and godata to emphasize that parameters get copy of address and are different from pointers in host code
extern __shared__ float temp[];
int tid = blockIdx.x * blockDim.x + threadIdx.x;
int ltid = threadIdx.x;
temp[ltid] = 0;
while(tid < sizeInput){
temp[ltid] += in[tid];
tid+=gridDim.x * blockDim.x; // to handle array of any size
}
__syncthreads();
int offset = 1;
while(offset < blockDim.x){
if(ltid % (offset * 2) == 0){
temp[ltid] = temp[ltid] + temp[ltid + offset];
}
__syncthreads();
offset*=2;
}
if(ltid == 0){
out[blockIdx.x] = temp[0];
}
}
int main(){
int size = 16; // size of present input array. Changes after every loop iteration
int cidata[] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
/*FILE *f;
f = fopen("invertedList.txt" , "w");
a[0] = 1 + (rand() % 8);
fprintf(f, "%d,",a[0]);
for( int i = 1 ; i< N; i++){
a[i] = a[i-1] + (rand() % 8) + 1;
fprintf(f, "%d,",a[i]);
}
fclose(f);*/
int* gidata;
int* godata;
cudaMalloc((void**)&gidata, size* sizeof(int));
cudaMemcpy(gidata,cidata, size * sizeof(int), cudaMemcpyHostToDevice);
int TPB = 4;
int blocks = 10; //to get things kicked off
cudaEvent_t start, stop;
cudaEventCreate(&start);
cudaEventCreate(&stop);
cudaEventRecord(start, 0);
while(blocks != 1 ){
if(size < TPB){
TPB = size; // size is 2^sth
}
blocks = (size+ TPB -1 ) / TPB;
cudaMalloc((void**)&godata, blocks * sizeof(int));
computeAddShared<<<blocks, TPB,TPB>>>(gidata, godata,size);
cudaFree(gidata);
gidata = godata;
size = blocks;
}
//printf("The error by cuda is %s",cudaGetErrorString(cudaGetLastError()));
cudaEventRecord(stop, 0);
cudaEventSynchronize(stop);
float elapsedTime;
cudaEventElapsedTime(&elapsedTime , start, stop);
printf("time is %f ms", elapsedTime);
int *output = (int*)malloc(sizeof(int));
cudaMemcpy(output, gidata, sizeof(int), cudaMemcpyDeviceToHost);
//Cant free either earlier as both point to same location
cudaError_t chk = cudaFree(godata);
if(chk!=0){
printf("First chk also printed error. Maybe error in my logic\n");
}
printf("The error by threadsyn is %s", cudaGetErrorString(cudaGetLastError()));
printf("The sum of the array is %d\n", output[0]);
getchar();
return 0;
}
Clearly, the first while loop in computeAddShared is causing out of bounds error because I am allocating 4 bytes to shared memory. Why does cudamemcheck not catch this. Below is the output of cuda-memcheck
========= CUDA-MEMCHECK
time is 12.334816 msThe error by threadsyn is no errorThe sum of the array is 13
6
========= ERROR SUMMARY: 0 errors
Shared memory allocation granularity. The Hardware undoubtedly has a page size for allocations (probably the same as the L1 cache line side). With only 4 threads per block, there will "accidentally" be enough shared memory in a single page to let you code work. If you used a sensible number of threads block (ie. a round multiple of the warp size) the error would be detected because there would not be enough allocated memory.

CUDA memory limitations

If I try to send to my CUDA device a struct wich is heavier than the size of memory available, will CUDA give me any kind of warning or error?
I'm asking that because my GPU has 1024 MBytes (1073414144 bytes) Total amount of global memory, but I don't know how I should handle and eventual problem.
That's my code:
#define VECSIZE 2250000
#define WIDTH 1500
#define HEIGHT 1500
// Matrices are stored in row-major order:
// M(row, col) = *(M.elements + row * M.width + col)
struct Matrix
{
int width;
int height;
int* elements;
};
int main()
{
Matrix M;
M.width = WIDTH;
M.height = HEIGHT;
M.elements = (int *) calloc(VECSIZE,sizeof(int));
int row, col;
// define Matrix M
// Matrix generator:
for (int i = 0; i < M.height; i++)
for(int j = 0; j < M.width; j++)
{
row = i;
col = j;
if (i == j)
M.elements[row * M.width + col] = INFINITY;
else
{
M.elements[row * M.width + col] = (rand() % 2); // because 'rand() % 1' just does not seems to work ta all.
if (M.elements[row * M.width + col] == 0) // can't have zero weight.
M.elements[row * M.width + col] = INFINITY;
else if (M.elements[row * M.width + col] == 2)
M.elements[row * M.width + col] = 1;
}
}
// Declare & send device Matrix to Device.
Matrix d_M;
d_M.width = M.width;
d_M.height = M.height;
size_t size = M.width * M.height * sizeof(int);
cudaMalloc(&d_M.elements, size);
cudaMemcpy(d_M.elements, M.elements, size, cudaMemcpyHostToDevice);
int *d_k= (int*) malloc(sizeof(int));
cudaMalloc((void**) &d_k, sizeof (int));
int *d_width=(int*)malloc(sizeof(int));
cudaMalloc((void**) &d_width, sizeof(int));
unsigned int *width=(unsigned int*)malloc(sizeof(unsigned int));
width[0] = M.width;
cudaMemcpy(d_width, width, sizeof(int), cudaMemcpyHostToDevice);
int *d_height=(int*)malloc(sizeof(int));
cudaMalloc((void**) &d_height, sizeof(int));
unsigned int *height=(unsigned int*)malloc(sizeof(unsigned int));
height[0] = M.height;
cudaMemcpy(d_height, height, sizeof(int), cudaMemcpyHostToDevice);
/*
et cetera .. */
While you may not currently be sending enough data to the GPU to max out it's memory, when you do, your cudaMalloc will return the error code cudaErrorMemoryAllocation which as per the cuda api docs, signals that the memory allocation failed. I note that in your example code you are not checking the return values of the cuda calls. These return codes need to be checked to make sure your program is running correctly. The cuda api does not throw exceptions: you must check the return codes. See this article for info on checking the errors and getting meaningful messages about the errors
If you are using cutil.h, then it provides two very useful macros:
CUDA_SAFE_CALL (used while issuing functions like cudaMalloc, cudaMemcpy etc.)
and
CUT_CHECK_ERROR (used after executing a kernel to check for errors in kernel execution).
They take care of the errors, if any, by using the error checking mechanism detailed in the article provided by flipchart.

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