I have two pointers in memory and I want to swap it atomically but atomic operation in CUDA support only int types. There is a way to do the following swap?
classA* a1 = malloc(...);
classA* a2 = malloc(...);
atomicSwap(a1,a2);
When writing device-side code...
While CUDA provides atomics, they can't cover multiple (possibly remote) memory locations at once.
To perform this swap, you will need to "protect" access to both these values with something like mutex, and have whoever wants to write values to them take a hold of the mutex for the duration of the critical section (like in C++'s host-side std::lock_guard). This can be done using CUDA's actual atomic facilities, e.g. compare-and-swap, and is the subject of this question:
Implementing a critical section in CUDA
A caveat to the above is mentioned by #RobertCrovella: If you can make do with, say, a pair of 32-bit offsets rather than a 64-bit pointer, then if you were to store them in a 64-bit aligned struct, you could use compare-and-exchange on the whole struct to implement an atomic swap of the whole struct.
... but is it really device side code?
Your code actually doesn't look like something one would run on the device: Memory allocation is usually (though not always) done from the host side before you launch your kernel and do actual work. If you could make sure these alterations only happen on the host side (think CUDA events and callbacks), and that device-side code will not be interfered with by them - you can just use your plain vanilla C++ facilities for concurrent programming (like lock_guard I mentioned above).
I managed to have the needed behaviour, it is not atomic swap but still safe. The context was a monotonic Linked List working both on CPU and GPU:
template<typename T>
union readablePointer
{
T* ptr;
unsigned long long int address;
};
template<typename T>
struct LinkedList
{
struct Node
{
T value;
readablePointer<Node> previous;
};
Node start;
Node end;
int size;
__host__ __device__ void initialize()
{
size = 0;
start.previous.ptr = nullptr;
end.previous.ptr = &start;
}
__host__ __device__ void push_back(T value)
{
Node* node = nullptr;
malloc(&node, sizeof(Node));
readablePointer<Node> nodePtr;
nodePtr.ptr = node;
nodePtr.ptr->value = value;
#ifdef __CUDA_ARCH__
nodePtr.ptr->previous.address = atomicExch(&end.previous.address, nodePtr.address);
atomicAdd(&size,1);
#else
nodePtr.ptr->previous.address = end.previous.address;
end.previous.address = nodePtr.address;
size += 1;
#endif
}
__host__ __device__ T pop_back()
{
assert(end.previous.ptr != &start);
readablePointer<Node> lastNodePtr;
lastNodePtr.ptr = nullptr;
#ifdef __CUDA_ARCH__
lastNodePtr.address = atomicExch(&end.previous.address,end.previous.ptr->previous.address);
atomicSub(&size,1);
#else
lastNodePtr.address = end.previous.address;
end.previous.address = end.previous.ptr->previous.address;
size -= 1;
#endif
T toReturn = lastNodePtr.ptr->value;
free(lastNodePtr.ptr);
return toReturn;
}
__host__ __device__ void clear()
{
while(size > 0)
{
pop_back();
}
}
};
This question already has an answer here:
Summing the rows of a matrix (stored in either row-major or column-major order) in CUDA
(1 answer)
Closed 5 years ago.
I declared two GPU memory pointers, and allocated the GPU memory, transfer data and launch the kernel in the main:
// declare GPU memory pointers
char * gpuIn;
char * gpuOut;
// allocate GPU memory
cudaMalloc(&gpuIn, ARRAY_BYTES);
cudaMalloc(&gpuOut, ARRAY_BYTES);
// transfer the array to the GPU
cudaMemcpy(gpuIn, currIn, ARRAY_BYTES, cudaMemcpyHostToDevice);
// launch the kernel
role<<<dim3(1),dim3(40,20)>>>(gpuOut, gpuIn);
// copy back the result array to the CPU
cudaMemcpy(currOut, gpuOut, ARRAY_BYTES, cudaMemcpyDeviceToHost);
cudaFree(gpuIn);
cudaFree(gpuOut);
And this is my code inside the kernel:
__global__ void role(char * gpuOut, char * gpuIn){
int idx = threadIdx.x;
int idy = threadIdx.y;
char live = '0';
char dead = '.';
char f = gpuIn[idx][idy];
if(f==live){
gpuOut[idx][idy]=dead;
}
else{
gpuOut[idx][idy]=live;
}
}
But here are some errors, I think here are some errors on the pointers. Any body can give a help?
The key concept is the storage order of multidimensional arrays in memory -- this is well described here. A useful abstraction is to define a simple class which encapsulates a pointer to a multidimensional array stored in linear memory and provides an operator which gives something like the usual a[i][j] style access. Your code could be modified something like this:
template<typename T>
struct array2d
{
T* p;
size_t lda;
__device__ __host__
array2d(T* _p, size_t _lda) : p(_p), lda(_lda) {};
__device__ __host__
T& operator()(size_t i, size_t j) {
return p[j + i * lda];
}
__device__ __host__
const T& operator()(size_t i, size_t j) const {
return p[j + i * lda];
}
};
__global__ void role(array2d<char> gpuOut, array2d<char> gpuIn){
int idx = threadIdx.x;
int idy = threadIdx.y;
char live = '0';
char dead = '.';
char f = gpuIn(idx,idy);
if(f==live){
gpuOut(idx,idy)=dead;
}
else{
gpuOut(idx,idy)=live;
}
}
int main()
{
const int rows = 5, cols = 6;
const size_t ARRAY_BYTES = sizeof(char) * size_t(rows * cols);
// declare GPU memory pointers
char * gpuIn;
char * gpuOut;
char currIn[rows][cols], currOut[rows][cols];
// allocate GPU memory
cudaMalloc(&gpuIn, ARRAY_BYTES);
cudaMalloc(&gpuOut, ARRAY_BYTES);
// transfer the array to the GPU
cudaMemcpy(gpuIn, currIn, ARRAY_BYTES, cudaMemcpyHostToDevice);
// launch the kernel
role<<<dim3(1),dim3(rows,cols)>>>(array2d<char>(gpuOut, cols), array2d<char>(gpuIn, cols));
// copy back the result array to the CPU
cudaMemcpy(currOut, gpuOut, ARRAY_BYTES, cudaMemcpyDeviceToHost);
cudaFree(gpuIn);
cudaFree(gpuOut);
return 0;
}
The important point here is that a two dimensional C or C++ array stored in linear memory can be addressed as col + row * number of cols. The class in the code above is just a convenient way of expressing this.
First off, here is some code:
int main()
{
int days[] = {1,2,3,4,5};
int *ptr = days;
printf("%u\n", sizeof(days));
printf("%u\n", sizeof(ptr));
return 0;
}
Is there a way to find out the size of the array that ptr is pointing to (instead of just giving its size, which is four bytes on a 32-bit system)?
No, you can't. The compiler doesn't know what the pointer is pointing to. There are tricks, like ending the array with a known out-of-band value and then counting the size up until that value, but that's not using sizeof().
Another trick is the one mentioned by Zan, which is to stash the size somewhere. For example, if you're dynamically allocating the array, allocate a block one int bigger than the one you need, stash the size in the first int, and return ptr+1 as the pointer to the array. When you need the size, decrement the pointer and peek at the stashed value. Just remember to free the whole block starting from the beginning, and not just the array.
The answer is, "No."
What C programmers do is store the size of the array somewhere. It can be part of a structure, or the programmer can cheat a bit and malloc() more memory than requested in order to store a length value before the start of the array.
For dynamic arrays (malloc or C++ new) you need to store the size of the array as mentioned by others or perhaps build an array manager structure which handles add, remove, count, etc. Unfortunately C doesn't do this nearly as well as C++ since you basically have to build it for each different array type you are storing which is cumbersome if you have multiple types of arrays that you need to manage.
For static arrays, such as the one in your example, there is a common macro used to get the size, but it is not recommended as it does not check if the parameter is really a static array. The macro is used in real code though, e.g. in the Linux kernel headers although it may be slightly different than the one below:
#if !defined(ARRAY_SIZE)
#define ARRAY_SIZE(x) (sizeof((x)) / sizeof((x)[0]))
#endif
int main()
{
int days[] = {1,2,3,4,5};
int *ptr = days;
printf("%u\n", ARRAY_SIZE(days));
printf("%u\n", sizeof(ptr));
return 0;
}
You can google for reasons to be wary of macros like this. Be careful.
If possible, the C++ stdlib such as vector which is much safer and easier to use.
There is a clean solution with C++ templates, without using sizeof(). The following getSize() function returns the size of any static array:
#include <cstddef>
template<typename T, size_t SIZE>
size_t getSize(T (&)[SIZE]) {
return SIZE;
}
Here is an example with a foo_t structure:
#include <cstddef>
template<typename T, size_t SIZE>
size_t getSize(T (&)[SIZE]) {
return SIZE;
}
struct foo_t {
int ball;
};
int main()
{
foo_t foos3[] = {{1},{2},{3}};
foo_t foos5[] = {{1},{2},{3},{4},{5}};
printf("%u\n", getSize(foos3));
printf("%u\n", getSize(foos5));
return 0;
}
Output:
3
5
As all the correct answers have stated, you cannot get this information from the decayed pointer value of the array alone. If the decayed pointer is the argument received by the function, then the size of the originating array has to be provided in some other way for the function to come to know that size.
Here's a suggestion different from what has been provided thus far,that will work: Pass a pointer to the array instead. This suggestion is similar to the C++ style suggestions, except that C does not support templates or references:
#define ARRAY_SZ 10
void foo (int (*arr)[ARRAY_SZ]) {
printf("%u\n", (unsigned)sizeof(*arr)/sizeof(**arr));
}
But, this suggestion is kind of silly for your problem, since the function is defined to know exactly the size of the array that is passed in (hence, there is little need to use sizeof at all on the array). What it does do, though, is offer some type safety. It will prohibit you from passing in an array of an unwanted size.
int x[20];
int y[10];
foo(&x); /* error */
foo(&y); /* ok */
If the function is supposed to be able to operate on any size of array, then you will have to provide the size to the function as additional information.
For this specific example, yes, there is, IF you use typedefs (see below). Of course, if you do it this way, you're just as well off to use SIZEOF_DAYS, since you know what the pointer is pointing to.
If you have a (void *) pointer, as is returned by malloc() or the like, then, no, there is no way to determine what data structure the pointer is pointing to and thus, no way to determine its size.
#include <stdio.h>
#define NUM_DAYS 5
typedef int days_t[ NUM_DAYS ];
#define SIZEOF_DAYS ( sizeof( days_t ) )
int main() {
days_t days;
days_t *ptr = &days;
printf( "SIZEOF_DAYS: %u\n", SIZEOF_DAYS );
printf( "sizeof(days): %u\n", sizeof(days) );
printf( "sizeof(*ptr): %u\n", sizeof(*ptr) );
printf( "sizeof(ptr): %u\n", sizeof(ptr) );
return 0;
}
Output:
SIZEOF_DAYS: 20
sizeof(days): 20
sizeof(*ptr): 20
sizeof(ptr): 4
There is no magic solution. C is not a reflective language. Objects don't automatically know what they are.
But you have many choices:
Obviously, add a parameter
Wrap the call in a macro and automatically add a parameter
Use a more complex object. Define a structure which contains the dynamic array and also the size of the array. Then, pass the address of the structure.
You can do something like this:
int days[] = { /*length:*/5, /*values:*/ 1,2,3,4,5 };
int *ptr = days + 1;
printf("array length: %u\n", ptr[-1]);
return 0;
My solution to this problem is to save the length of the array into a struct Array as a meta-information about the array.
#include <stdio.h>
#include <stdlib.h>
struct Array
{
int length;
double *array;
};
typedef struct Array Array;
Array* NewArray(int length)
{
/* Allocate the memory for the struct Array */
Array *newArray = (Array*) malloc(sizeof(Array));
/* Insert only non-negative length's*/
newArray->length = (length > 0) ? length : 0;
newArray->array = (double*) malloc(length*sizeof(double));
return newArray;
}
void SetArray(Array *structure,int length,double* array)
{
structure->length = length;
structure->array = array;
}
void PrintArray(Array *structure)
{
if(structure->length > 0)
{
int i;
printf("length: %d\n", structure->length);
for (i = 0; i < structure->length; i++)
printf("%g\n", structure->array[i]);
}
else
printf("Empty Array. Length 0\n");
}
int main()
{
int i;
Array *negativeTest, *days = NewArray(5);
double moreDays[] = {1,2,3,4,5,6,7,8,9,10};
for (i = 0; i < days->length; i++)
days->array[i] = i+1;
PrintArray(days);
SetArray(days,10,moreDays);
PrintArray(days);
negativeTest = NewArray(-5);
PrintArray(negativeTest);
return 0;
}
But you have to care about set the right length of the array you want to store, because the is no way to check this length, like our friends massively explained.
This is how I personally do it in my code. I like to keep it as simple as possible while still able to get values that I need.
typedef struct intArr {
int size;
int* arr;
} intArr_t;
int main() {
intArr_t arr;
arr.size = 6;
arr.arr = (int*)malloc(sizeof(int) * arr.size);
for (size_t i = 0; i < arr.size; i++) {
arr.arr[i] = i * 10;
}
return 0;
}
No, you can't use sizeof(ptr) to find the size of array ptr is pointing to.
Though allocating extra memory(more than the size of array) will be helpful if you want to store the length in extra space.
int main()
{
int days[] = {1,2,3,4,5};
int *ptr = days;
printf("%u\n", sizeof(days));
printf("%u\n", sizeof(ptr));
return 0;
}
Size of days[] is 20 which is no of elements * size of it's data type.
While the size of pointer is 4 no matter what it is pointing to.
Because a pointer points to other element by storing it's address.
In strings there is a '\0' character at the end so the length of the string can be gotten using functions like strlen. The problem with an integer array, for example, is that you can't use any value as an end value so one possible solution is to address the array and use as an end value the NULL pointer.
#include <stdio.h>
/* the following function will produce the warning:
* ‘sizeof’ on array function parameter ‘a’ will
* return size of ‘int *’ [-Wsizeof-array-argument]
*/
void foo( int a[] )
{
printf( "%lu\n", sizeof a );
}
/* so we have to implement something else one possible
* idea is to use the NULL pointer as a control value
* the same way '\0' is used in strings but this way
* the pointer passed to a function should address pointers
* so the actual implementation of an array type will
* be a pointer to pointer
*/
typedef char * type_t; /* line 18 */
typedef type_t ** array_t;
int main( void )
{
array_t initialize( int, ... );
/* initialize an array with four values "foo", "bar", "baz", "foobar"
* if one wants to use integers rather than strings than in the typedef
* declaration at line 18 the char * type should be changed with int
* and in the format used for printing the array values
* at line 45 and 51 "%s" should be changed with "%i"
*/
array_t array = initialize( 4, "foo", "bar", "baz", "foobar" );
int size( array_t );
/* print array size */
printf( "size %i:\n", size( array ));
void aprint( char *, array_t );
/* print array values */
aprint( "%s\n", array ); /* line 45 */
type_t getval( array_t, int );
/* print an indexed value */
int i = 2;
type_t val = getval( array, i );
printf( "%i: %s\n", i, val ); /* line 51 */
void delete( array_t );
/* free some space */
delete( array );
return 0;
}
/* the output of the program should be:
* size 4:
* foo
* bar
* baz
* foobar
* 2: baz
*/
#include <stdarg.h>
#include <stdlib.h>
array_t initialize( int n, ... )
{
/* here we store the array values */
type_t *v = (type_t *) malloc( sizeof( type_t ) * n );
va_list ap;
va_start( ap, n );
int j;
for ( j = 0; j < n; j++ )
v[j] = va_arg( ap, type_t );
va_end( ap );
/* the actual array will hold the addresses of those
* values plus a NULL pointer
*/
array_t a = (array_t) malloc( sizeof( type_t *) * ( n + 1 ));
a[n] = NULL;
for ( j = 0; j < n; j++ )
a[j] = v + j;
return a;
}
int size( array_t a )
{
int n = 0;
while ( *a++ != NULL )
n++;
return n;
}
void aprint( char *fmt, array_t a )
{
while ( *a != NULL )
printf( fmt, **a++ );
}
type_t getval( array_t a, int i )
{
return *a[i];
}
void delete( array_t a )
{
free( *a );
free( a );
}
#include <stdio.h>
#include <string.h>
#include <stddef.h>
#include <stdlib.h>
#define array(type) struct { size_t size; type elem[0]; }
void *array_new(int esize, int ecnt)
{
size_t *a = (size_t *)malloc(esize*ecnt+sizeof(size_t));
if (a) *a = ecnt;
return a;
}
#define array_new(type, count) array_new(sizeof(type),count)
#define array_delete free
#define array_foreach(type, e, arr) \
for (type *e = (arr)->elem; e < (arr)->size + (arr)->elem; ++e)
int main(int argc, char const *argv[])
{
array(int) *iarr = array_new(int, 10);
array(float) *farr = array_new(float, 10);
array(double) *darr = array_new(double, 10);
array(char) *carr = array_new(char, 11);
for (int i = 0; i < iarr->size; ++i) {
iarr->elem[i] = i;
farr->elem[i] = i*1.0f;
darr->elem[i] = i*1.0;
carr->elem[i] = i+'0';
}
array_foreach(int, e, iarr) {
printf("%d ", *e);
}
array_foreach(float, e, farr) {
printf("%.0f ", *e);
}
array_foreach(double, e, darr) {
printf("%.0lf ", *e);
}
carr->elem[carr->size-1] = '\0';
printf("%s\n", carr->elem);
return 0;
}
#define array_size 10
struct {
int16 size;
int16 array[array_size];
int16 property1[(array_size/16)+1]
int16 property2[(array_size/16)+1]
} array1 = {array_size, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
#undef array_size
array_size is passing to the size variable:
#define array_size 30
struct {
int16 size;
int16 array[array_size];
int16 property1[(array_size/16)+1]
int16 property2[(array_size/16)+1]
} array2 = {array_size};
#undef array_size
Usage is:
void main() {
int16 size = array1.size;
for (int i=0; i!=size; i++) {
array1.array[i] *= 2;
}
}
Most implementations will have a function that tells you the reserved size for objects allocated with malloc() or calloc(), for example GNU has malloc_usable_size()
However, this will return the size of the reversed block, which can be larger than the value given to malloc()/realloc().
There is a popular macro, which you can define for finding number of elements in the array (Microsoft CRT even provides it OOB with name _countof):
#define countof(x) (sizeof(x)/sizeof((x)[0]))
Then you can write:
int my_array[] = { ... some elements ... };
printf("%zu", countof(my_array)); // 'z' is correct type specifier for size_t
I am having an issue with my OpenCL kernel. The input arguments are corrupt when they are passed to the kernel. What makes this strange is this same exact kernel executes flawlessly on mac osx. Once I started porting my code over to windows (windows 8 64-bit) I started having this issue.
I have provided an example using my camera struct. The x,y,z coordinates are defined as <0,0,200>. However, when they make it to my kernel they show as <0,-0.00132704, -0.00132704>.
I have a kernel that accepts two structs.
typedef struct{
cl_float d;
cl_float3 eye;
cl_float3 lookat;
cl_float3 u;
cl_float3 v;
cl_float3 w;
cl_float3 up;
}rt_cl_camera;
typedef struct {
float r;
float g;
float b;
} rt_cl_rgb;
I have slimmed down my kernel for the sake of testing. After tracking down the issues I noticed that my input paramaters were not coming over correctly. However, I have determined that my output is being passed back correctly.
__kernel void ray_trace_scene( __global rt_cl_rgb* output,
__global rt_cl_camera* camera,
const unsigned int pcount)
{
int pixel = get_global_id(0);
if(pixel < pcount){
output[pixel].r = camera->eye.x;
output[pixel].g = camera->eye.y;
output[pixel].b = camera->eye.z;
}// End Pixel computation
}//End kernel
I am creating my input buffer with the follwoing:
cl_mem cam_input;
cl_uint cam_error;
cam_input = clCreateBuffer(context, CL_MEM_READ_ONLY, sizeof(rt_cl_camera), NULL, &cam_error);
I am also checking to make sure my buffer was created successfully with
if (cam_error != CL_SUCCESS || !cam_input) {
throw std::runtime_error(CLERROR_FAILED_DEVBUFF);
}
I then write my data into my buffer with the following.
cl_uint err = 0;
err = clEnqueueWriteBuffer(commands, cam_input, CL_TRUE, 0, sizeof(rt_cl_camera), cam_ptr, 0, NULL, NULL);
if (err != CL_SUCCESS) {
throw std::runtime_error("Failed to write camera");
}
and finally linking my argument for the appropriate command line slot. Please note that slot zero is being used for my output.
err |= clSetKernelArg(kernel, 1, sizeof(cl_mem), &cam_input);
and checking that everything was successful..
if (err != CL_SUCCESS) {
throw std::runtime_error(CLERROR_FAILED_CMDARGS);
}
I am not receiving any error messages from openCL at any step of the process. Has anyone run into this? Any help is greatly appreciated.
side note - At each step of the way I am printing out my local variables to make sure they are correct and valid before I pass them over to the GPU.
Looks an alignment/packing issue. Try using float4 instead of float3 in the struct, and move float d at the end.