C++ Cuda performance for double pointers
0
I finally succeded in saving in memory a double pointer in order to use it in cuda.(The code below), but i see that is less performent than if i would flatten the matrix,which is not that great.
Some suggestions to save some time/memory?
I really want to use dynamic 2d array.
#include "cuda_runtime.h"
#include "device_launch_parameters.h"
#include <stdlib.h>
#include <cstdio>
__global__ void fct(int **dev_c)
{
int y = threadIdx.x;
int x = threadIdx.y;
dev_c[y][x] = 3;
}
int main(void)
{
//Output Array
int **cc = new int*[2];
for (int i = 0; i < 2; i++)cc[i] = new int[2];
//Host Array
int ** h_c = (int **)malloc(2 * sizeof(int *));
for (int i = 0; i < 2; i++) {
cudaMalloc((void**)&h_c[i], 2 * sizeof(int));
}
//Devie array
int ** d_c;
cudaMalloc((void **)&d_c, 2 * sizeof(int *));
cudaMemcpy(d_c, h_c, 2 * sizeof(int *), cudaMemcpyHostToDevice);
dim3 d(2, 2);
fct << <1, d >> > (d_c);
for (int i = 0; i < 2; i++) {
cudaMemcpy(cc[i], h_c[i], 2 * sizeof(int), cudaMemcpyDeviceToHost);
}
for (int i = 0; i < 2; i++) {
for (int j = 0; j < 2; j++) {
printf("(%d,%d):%dn", i, j, cc[i][j]);
}
}
int x;
std::cin >> x;
delete h_c;
delete d_c;
}
c++ pointers cuda
asked Dec 30 '18 at 14:34
Vlad ConstantinescuVlad Constantinescu
344
add a comment |
0
I finally succeded in saving in memory a double pointer in order to use it in cuda.(The code below), but i see that is less performent than if i would flatten the matrix,which is not that great.
Some suggestions to save some time/memory?
I really want to use dynamic 2d array.
#include "cuda_runtime.h"
#include "device_launch_parameters.h"
#include <stdlib.h>
#include <cstdio>
__global__ void fct(int **dev_c)
{
int y = threadIdx.x;
int x = threadIdx.y;
dev_c[y][x] = 3;
}
int main(void)
{
//Output Array
int **cc = new int*[2];
for (int i = 0; i < 2; i++)cc[i] = new int[2];
//Host Array
int ** h_c = (int **)malloc(2 * sizeof(int *));
for (int i = 0; i < 2; i++) {
cudaMalloc((void**)&h_c[i], 2 * sizeof(int));
}
//Devie array
int ** d_c;
cudaMalloc((void **)&d_c, 2 * sizeof(int *));
cudaMemcpy(d_c, h_c, 2 * sizeof(int *), cudaMemcpyHostToDevice);
dim3 d(2, 2);
fct << <1, d >> > (d_c);
for (int i = 0; i < 2; i++) {
cudaMemcpy(cc[i], h_c[i], 2 * sizeof(int), cudaMemcpyDeviceToHost);
}
for (int i = 0; i < 2; i++) {
for (int j = 0; j < 2; j++) {
printf("(%d,%d):%dn", i, j, cc[i][j]);
}
}
int x;
std::cin >> x;
delete h_c;
delete d_c;
}
c++ pointers cuda
asked Dec 30 '18 at 14:34
Vlad ConstantinescuVlad Constantinescu
344
If you know the width of your 2D matrix at compile time, it's possible to use doubly-subscripted access (even in device code) while still maintaining the performance benefits of indexed vs. double-pointer access. This answer discusses various methods including the known-width approach. If you do not know the width of your 2D matrix at compile time, I'm not aware of any method to do doubly-subscripted access without 2 pointer dereferences per access.
– Robert Crovella
Dec 30 '18 at 16:31
Thank you! I will keep in mind this but sadly for this project i need that array to be dynamic.
– Vlad Constantinescu
Dec 30 '18 at 23:50
add a comment |
0
0
0
I finally succeded in saving in memory a double pointer in order to use it in cuda.(The code below), but i see that is less performent than if i would flatten the matrix,which is not that great.
Some suggestions to save some time/memory?
I really want to use dynamic 2d array.
#include "cuda_runtime.h"
#include "device_launch_parameters.h"
#include <stdlib.h>
#include <cstdio>
__global__ void fct(int **dev_c)
{
int y = threadIdx.x;
int x = threadIdx.y;
dev_c[y][x] = 3;
}
int main(void)
{
//Output Array
int **cc = new int*[2];
for (int i = 0; i < 2; i++)cc[i] = new int[2];
//Host Array
int ** h_c = (int **)malloc(2 * sizeof(int *));
for (int i = 0; i < 2; i++) {
cudaMalloc((void**)&h_c[i], 2 * sizeof(int));
}
//Devie array
int ** d_c;
cudaMalloc((void **)&d_c, 2 * sizeof(int *));
cudaMemcpy(d_c, h_c, 2 * sizeof(int *), cudaMemcpyHostToDevice);
dim3 d(2, 2);
fct << <1, d >> > (d_c);
for (int i = 0; i < 2; i++) {
cudaMemcpy(cc[i], h_c[i], 2 * sizeof(int), cudaMemcpyDeviceToHost);
}
for (int i = 0; i < 2; i++) {
for (int j = 0; j < 2; j++) {
printf("(%d,%d):%dn", i, j, cc[i][j]);
}
}
int x;
std::cin >> x;
delete h_c;
delete d_c;
}
c++ pointers cuda
asked Dec 30 '18 at 14:34
Vlad ConstantinescuVlad Constantinescu
344
I finally succeded in saving in memory a double pointer in order to use it in cuda.(The code below), but i see that is less performent than if i would flatten the matrix,which is not that great.
Some suggestions to save some time/memory?
I really want to use dynamic 2d array.
#include "cuda_runtime.h"
#include "device_launch_parameters.h"
#include <stdlib.h>
#include <cstdio>
__global__ void fct(int **dev_c)
{
int y = threadIdx.x;
int x = threadIdx.y;
dev_c[y][x] = 3;
}
int main(void)
{
//Output Array
int **cc = new int*[2];
for (int i = 0; i < 2; i++)cc[i] = new int[2];
//Host Array
int ** h_c = (int **)malloc(2 * sizeof(int *));
for (int i = 0; i < 2; i++) {
cudaMalloc((void**)&h_c[i], 2 * sizeof(int));
}
//Devie array
int ** d_c;
cudaMalloc((void **)&d_c, 2 * sizeof(int *));
cudaMemcpy(d_c, h_c, 2 * sizeof(int *), cudaMemcpyHostToDevice);
dim3 d(2, 2);
fct << <1, d >> > (d_c);
for (int i = 0; i < 2; i++) {
cudaMemcpy(cc[i], h_c[i], 2 * sizeof(int), cudaMemcpyDeviceToHost);
}
for (int i = 0; i < 2; i++) {
for (int j = 0; j < 2; j++) {
printf("(%d,%d):%dn", i, j, cc[i][j]);
}
}
int x;
std::cin >> x;
delete h_c;
delete d_c;
}
c++ pointers cuda
c++ pointers cuda
asked Dec 30 '18 at 14:34
Vlad ConstantinescuVlad Constantinescu
344
asked Dec 30 '18 at 14:34
Vlad ConstantinescuVlad Constantinescu
344
asked Dec 30 '18 at 14:34
Vlad ConstantinescuVlad Constantinescu
344
asked Dec 30 '18 at 14:34
Vlad ConstantinescuVlad Constantinescu
344
asked Dec 30 '18 at 14:34
Vlad ConstantinescuVlad Constantinescu
344
344
If you know the width of your 2D matrix at compile time, it's possible to use doubly-subscripted access (even in device code) while still maintaining the performance benefits of indexed vs. double-pointer access. This answer discusses various methods including the known-width approach. If you do not know the width of your 2D matrix at compile time, I'm not aware of any method to do doubly-subscripted access without 2 pointer dereferences per access.
– Robert Crovella
Dec 30 '18 at 16:31
Thank you! I will keep in mind this but sadly for this project i need that array to be dynamic.
– Vlad Constantinescu
Dec 30 '18 at 23:50
add a comment |
If you know the width of your 2D matrix at compile time, it's possible to use doubly-subscripted access (even in device code) while still maintaining the performance benefits of indexed vs. double-pointer access. This answer discusses various methods including the known-width approach. If you do not know the width of your 2D matrix at compile time, I'm not aware of any method to do doubly-subscripted access without 2 pointer dereferences per access.
– Robert Crovella
Dec 30 '18 at 16:31
Thank you! I will keep in mind this but sadly for this project i need that array to be dynamic.
– Vlad Constantinescu
Dec 30 '18 at 23:50
If you know the width of your 2D matrix at compile time, it's possible to use doubly-subscripted access (even in device code) while still maintaining the performance benefits of indexed vs. double-pointer access. This answer discusses various methods including the known-width approach. If you do not know the width of your 2D matrix at compile time, I'm not aware of any method to do doubly-subscripted access without 2 pointer dereferences per access.
– Robert Crovella
Dec 30 '18 at 16:31
If you know the width of your 2D matrix at compile time, it's possible to use doubly-subscripted access (even in device code) while still maintaining the performance benefits of indexed vs. double-pointer access. This answer discusses various methods including the known-width approach. If you do not know the width of your 2D matrix at compile time, I'm not aware of any method to do doubly-subscripted access without 2 pointer dereferences per access.
– Robert Crovella
Dec 30 '18 at 16:31
Thank you! I will keep in mind this but sadly for this project i need that array to be dynamic.
– Vlad Constantinescu
Dec 30 '18 at 23:50
Thank you! I will keep in mind this but sadly for this project i need that array to be dynamic.
– Vlad Constantinescu
Dec 30 '18 at 23:50
add a comment |
1 Answer
1
active
oldest
votes
0
You may actually want to use flattened matrix with some pointer tricks:
int main() {
const int size = 10;
auto arr = new int*[size];
arr[0] = new int[size * size];
for(int i = 1; i < size; i++) {
arr[i] = arr[0] + (i * size);
}
}
This way, you can still access the matrix with arr[x][y] syntax, but the actual memory is contiguous (which is not only faster to allocate*, but faster to access, given cache pre-fetching memory around the one you desire to use).
*It is faster to allocate size * size memory once, rather than allocating size times size elements.
Side note: using delete on a malloced memory is undefined behaviour. Don't mix new/new + delete/delete with malloc + free.
answered Dec 30 '18 at 14:41
FureeishFureeish
3,27321029
Thank you! Still, if i wanna save that arr in cuda memory i can do it like a simple vector?Or i have to alloc every row?
– Vlad Constantinescu
Dec 30 '18 at 15:33
@VladConstantinescu What do you mean bysimple vector?
– Fureeish
Dec 30 '18 at 15:34
I mean , 1d array ( int *a = new int[size])
– Vlad Constantinescu
Dec 30 '18 at 15:37
You can pass this matrix as a 1d array simply by passing*arr(orarr[0]), if that's what you are asking. It's still a little unclear. Do you want to pass the whole matrix as a single 1d array? Or you want to pass single rows as 1d arrays?
– Fureeish
Dec 30 '18 at 15:49
i want to pass to the kernel function(fct) the whole 2d matrix. There is a way that i will still be able to use the syntax matrix[i][j] inside the function ?
– Vlad Constantinescu
Dec 30 '18 at 15:54
|
show 2 more comments
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
You may actually want to use flattened matrix with some pointer tricks:
int main() {
const int size = 10;
auto arr = new int*[size];
arr[0] = new int[size * size];
for(int i = 1; i < size; i++) {
arr[i] = arr[0] + (i * size);
}
}
This way, you can still access the matrix with arr[x][y] syntax, but the actual memory is contiguous (which is not only faster to allocate*, but faster to access, given cache pre-fetching memory around the one you desire to use).
*It is faster to allocate size * size memory once, rather than allocating size times size elements.
Side note: using delete on a malloced memory is undefined behaviour. Don't mix new/new + delete/delete with malloc + free.
answered Dec 30 '18 at 14:41
FureeishFureeish
3,27321029
Thank you! Still, if i wanna save that arr in cuda memory i can do it like a simple vector?Or i have to alloc every row?
– Vlad Constantinescu
Dec 30 '18 at 15:33
@VladConstantinescu What do you mean bysimple vector?
– Fureeish
Dec 30 '18 at 15:34
I mean , 1d array ( int *a = new int[size])
– Vlad Constantinescu
Dec 30 '18 at 15:37
You can pass this matrix as a 1d array simply by passing*arr(orarr[0]), if that's what you are asking. It's still a little unclear. Do you want to pass the whole matrix as a single 1d array? Or you want to pass single rows as 1d arrays?
– Fureeish
Dec 30 '18 at 15:49
i want to pass to the kernel function(fct) the whole 2d matrix. There is a way that i will still be able to use the syntax matrix[i][j] inside the function ?
– Vlad Constantinescu
Dec 30 '18 at 15:54
|
show 2 more comments
0
You may actually want to use flattened matrix with some pointer tricks:
int main() {
const int size = 10;
auto arr = new int*[size];
arr[0] = new int[size * size];
for(int i = 1; i < size; i++) {
arr[i] = arr[0] + (i * size);
}
}
This way, you can still access the matrix with arr[x][y] syntax, but the actual memory is contiguous (which is not only faster to allocate*, but faster to access, given cache pre-fetching memory around the one you desire to use).
*It is faster to allocate size * size memory once, rather than allocating size times size elements.
Side note: using delete on a malloced memory is undefined behaviour. Don't mix new/new + delete/delete with malloc + free.
answered Dec 30 '18 at 14:41
FureeishFureeish
3,27321029
Thank you! Still, if i wanna save that arr in cuda memory i can do it like a simple vector?Or i have to alloc every row?
– Vlad Constantinescu
Dec 30 '18 at 15:33
@VladConstantinescu What do you mean bysimple vector?
– Fureeish
Dec 30 '18 at 15:34
I mean , 1d array ( int *a = new int[size])
– Vlad Constantinescu
Dec 30 '18 at 15:37
You can pass this matrix as a 1d array simply by passing*arr(orarr[0]), if that's what you are asking. It's still a little unclear. Do you want to pass the whole matrix as a single 1d array? Or you want to pass single rows as 1d arrays?
– Fureeish
Dec 30 '18 at 15:49
i want to pass to the kernel function(fct) the whole 2d matrix. There is a way that i will still be able to use the syntax matrix[i][j] inside the function ?
– Vlad Constantinescu
Dec 30 '18 at 15:54
|
show 2 more comments
0
0
0
You may actually want to use flattened matrix with some pointer tricks:
int main() {
const int size = 10;
auto arr = new int*[size];
arr[0] = new int[size * size];
for(int i = 1; i < size; i++) {
arr[i] = arr[0] + (i * size);
}
}
This way, you can still access the matrix with arr[x][y] syntax, but the actual memory is contiguous (which is not only faster to allocate*, but faster to access, given cache pre-fetching memory around the one you desire to use).
*It is faster to allocate size * size memory once, rather than allocating size times size elements.
Side note: using delete on a malloced memory is undefined behaviour. Don't mix new/new + delete/delete with malloc + free.
answered Dec 30 '18 at 14:41
FureeishFureeish
3,27321029
You may actually want to use flattened matrix with some pointer tricks:
int main() {
const int size = 10;
auto arr = new int*[size];
arr[0] = new int[size * size];
for(int i = 1; i < size; i++) {
arr[i] = arr[0] + (i * size);
}
}
This way, you can still access the matrix with arr[x][y] syntax, but the actual memory is contiguous (which is not only faster to allocate*, but faster to access, given cache pre-fetching memory around the one you desire to use).
*It is faster to allocate size * size memory once, rather than allocating size times size elements.
Side note: using delete on a malloced memory is undefined behaviour. Don't mix new/new + delete/delete with malloc + free.
answered Dec 30 '18 at 14:41
FureeishFureeish
3,27321029
edited Dec 30 '18 at 15:22
answered Dec 30 '18 at 14:41
FureeishFureeish
3,27321029
answered Dec 30 '18 at 14:41
FureeishFureeish
3,27321029
answered Dec 30 '18 at 14:41
FureeishFureeish
3,27321029
3,27321029
Thank you! Still, if i wanna save that arr in cuda memory i can do it like a simple vector?Or i have to alloc every row?
– Vlad Constantinescu
Dec 30 '18 at 15:33
@VladConstantinescu What do you mean bysimple vector?
– Fureeish
Dec 30 '18 at 15:34
I mean , 1d array ( int *a = new int[size])
– Vlad Constantinescu
Dec 30 '18 at 15:37
You can pass this matrix as a 1d array simply by passing*arr(orarr[0]), if that's what you are asking. It's still a little unclear. Do you want to pass the whole matrix as a single 1d array? Or you want to pass single rows as 1d arrays?
– Fureeish
Dec 30 '18 at 15:49
i want to pass to the kernel function(fct) the whole 2d matrix. There is a way that i will still be able to use the syntax matrix[i][j] inside the function ?
– Vlad Constantinescu
Dec 30 '18 at 15:54
|
show 2 more comments
Thank you! Still, if i wanna save that arr in cuda memory i can do it like a simple vector?Or i have to alloc every row?
– Vlad Constantinescu
Dec 30 '18 at 15:33
@VladConstantinescu What do you mean bysimple vector?
– Fureeish
Dec 30 '18 at 15:34
I mean , 1d array ( int *a = new int[size])
– Vlad Constantinescu
Dec 30 '18 at 15:37
You can pass this matrix as a 1d array simply by passing*arr(orarr[0]), if that's what you are asking. It's still a little unclear. Do you want to pass the whole matrix as a single 1d array? Or you want to pass single rows as 1d arrays?
– Fureeish
Dec 30 '18 at 15:49
i want to pass to the kernel function(fct) the whole 2d matrix. There is a way that i will still be able to use the syntax matrix[i][j] inside the function ?
– Vlad Constantinescu
Dec 30 '18 at 15:54
Thank you! Still, if i wanna save that arr in cuda memory i can do it like a simple vector?Or i have to alloc every row?
– Vlad Constantinescu
Dec 30 '18 at 15:33
Thank you! Still, if i wanna save that arr in cuda memory i can do it like a simple vector?Or i have to alloc every row?
– Vlad Constantinescu
Dec 30 '18 at 15:33
@VladConstantinescu What do you mean by
simple vector?– Fureeish
Dec 30 '18 at 15:34
@VladConstantinescu What do you mean by
simple vector?– Fureeish
Dec 30 '18 at 15:34
I mean , 1d array ( int *a = new int[size])
– Vlad Constantinescu
Dec 30 '18 at 15:37
I mean , 1d array ( int *a = new int[size])
– Vlad Constantinescu
Dec 30 '18 at 15:37
You can pass this matrix as a 1d array simply by passing
*arr (or arr[0]), if that's what you are asking. It's still a little unclear. Do you want to pass the whole matrix as a single 1d array? Or you want to pass single rows as 1d arrays?– Fureeish
Dec 30 '18 at 15:49
You can pass this matrix as a 1d array simply by passing
*arr (or arr[0]), if that's what you are asking. It's still a little unclear. Do you want to pass the whole matrix as a single 1d array? Or you want to pass single rows as 1d arrays?– Fureeish
Dec 30 '18 at 15:49
i want to pass to the kernel function(fct) the whole 2d matrix. There is a way that i will still be able to use the syntax matrix[i][j] inside the function ?
– Vlad Constantinescu
Dec 30 '18 at 15:54
i want to pass to the kernel function(fct) the whole 2d matrix. There is a way that i will still be able to use the syntax matrix[i][j] inside the function ?
– Vlad Constantinescu
Dec 30 '18 at 15:54
|
show 2 more comments
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If you know the width of your 2D matrix at compile time, it's possible to use doubly-subscripted access (even in device code) while still maintaining the performance benefits of indexed vs. double-pointer access. This answer discusses various methods including the known-width approach. If you do not know the width of your 2D matrix at compile time, I'm not aware of any method to do doubly-subscripted access without 2 pointer dereferences per access.
– Robert Crovella
Dec 30 '18 at 16:31
Thank you! I will keep in mind this but sadly for this project i need that array to be dynamic.
– Vlad Constantinescu
Dec 30 '18 at 23:50