C++ CUDA：不使用'；我不适合各种尺寸

我在做一个三维拉普拉斯算子。我的代码在大小N=32时成功，但在大小N=64或N=128时，我得到了一些不正确的结果：

#include <iostream>
#include <sys/time.h>
#include <cuda.h>
#include <ctime>
#include"res3dcb.cuh"
#include <math.h>
using namespace std;

// Let's start the main program.
int main(void) {

// Choice of N.
int N;
cout<<"Choose matrix dimension (32, 64 or 128)"<<endl;
cin>>N;
int size=(N+2)*(N+2)*(N+2)*sizeof(float);

// Variable statement.
struct timeval t1, t2;
float *x_d, *y_d; 
float *x,*y; 
float gflops;
float NumOps;


//Init x and y.
x = new float[size];
y = new float[size];

for (int i=1;i<N+1;i++)
    for (int j=1;j<N+1;j++) 
        for (int k=1;k<N+1;k++) { 
            x[i*(N+2)*(N+2)+j*(N+2)+k]=1;
        }

// Shadow cases.
for (int i=1;i<N+1;i++) {
    for (int j=1;j<N+1;j++) { 
        x[i*(N+2)*(N+2)+j*(N+2)]=x[i*(N+2)*(N+2)+j*(N+2)+1]; 
        x[i*(N+2)*(N+2)+j*(N+2)+N+1]=x[i*(N+2)*(N+2)+j*(N+2)+N];
    }

for (int k=0;k<N+2;k++) { 
    x[i*(N+2)*(N+2)+k]=x[i*(N+2)*(N+2)+(N+2)+k]; 
    x[i*(N+2)*(N+2)+(N+1)*(N+2)+k]=x[i*(N+2)*(N+2)+N*(N+2)+k];}
}

for (int j=0;j<N+2;j++) 
    for (int k=0;k<N+2;k++) {
        x[(N+2)*j+k]=x[(N+2)*(N+2)+(N+2)*j+k];
        x[(N+1)*(N+2)*(N+2)+(N+2)*j+k]=x[(N+2)*(N+2)*N+(N+2)*j+k];
    }

// Display of initial matrix.
int id_stage=-2;
while (id_stage!=-1) {
    cout<<"Which initial matrix's stage do you want to display? (-1 if you don't want to diplay another one)"<<endl;
cin>>id_stage;
cout<<endl;

if (id_stage != -1) {
    cout<<"Etage "<<id_stage<<" du cube :"<<endl;
    for (int j=0;j<N+2;j++) {
        cout<<"| ";
        for (int k=0;k<N+2;k++) {cout<<x[id_stage*(N+2)*(N+2)+j*(N+2)+k]<<" ";}
        cout<<"|"<<endl;
    }
    cout<<endl;
    }
}


// CPU to GPU.
cudaMalloc( (void**) & x_d, size);
cudaMalloc( (void**) & y_d, size);

cudaMemcpy(x_d, x, size, cudaMemcpyHostToDevice) ;
cudaMemcpy(y_d, y, size, cudaMemcpyHostToDevice) ;

// Solver parameters.
dim3 dimGrid(N/32, N/8, N/8);
dim3 dimBlock(16, 8, 8);


// Solver loop.
gettimeofday(&t1, 0);
res3d<<<dimGrid, dimBlock>>>(x_d, y_d, N); 
cudaDeviceSynchronize();
gettimeofday(&t2, 0);
double time = (1000000.0*(t2.tv_sec-t1.tv_sec) + t2.tv_usec-t1.tv_usec)/1000000.0;


// Power calculation.
NumOps=(1.0e-9)*N*N*N*7;
gflops = ( NumOps / (time));

// GPU to CPU.
cudaMemcpy(y, y_d, size, cudaMemcpyDeviceToHost);
cudaFree(x_d);
cudaFree(y_d);

// Display of final matrix.
id_stage=-2;
while (id_stage!=-1) {
    cout<<"Which output's stage do you want to display? (-1 if you don't want to diplay another one)"<<endl;
    cin>>id_stage;
    cout<<endl;

if (id_stage != -1) {
    cout<<"Etage "<<id_stage<<" du cube :"<<endl;
    for (int j=0;j<N+2;j++) {
        cout<<"| ";
        for (int k=0;k<N+2;k++) {cout<<y[id_stage*(N+2)*(N+2)+j*(N+2)+k]<<" ";}
        cout<<"|"<<endl;
    }
    cout<<endl;
}
}



cout<<"Time : "<<time<<endl;
cout<<"Gflops/s : "<<gflops<<endl;
}

#包括
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#包括
#包括
#包括“res3dcb.cuh”
#包括
使用名称空间std；
//让我们开始主程序。
内部主（空）{
//选择。
int N；
难道你这里有个错误吗

dim3 dimGrid(N/32, N/8, N/8);
dim3 dimBlock(16, 8, 8);

这应该是：

dim3 dimGrid(N/16, N/8, N/8);
dim3 dimBlock(16, 8, 8);

此外，如评论中所述，您在此处过度分配内存：

x = new float[size];
y = new float[size];

因为
size
是以字节计算的，而不是以元素计算的。
我发现了错误。DimGrid和DimBlock没有错，因为我在x轴上耕作

错误是我在全局内核中的“if”。下面是一个性能更好、结果正确的算法：

#include <assert.h>
#include <stdio.h>
#include <iostream>
#include <sys/time.h>
#include <cuda.h>
#include <ctime>
#include <math.h>
#include"reslap3D.cu"
using namespace std;

// Let's start the main program.
int main(void) {

// Variable statement.
struct timeval t1, t2;
float gflops;
float NumOps;
double time;
long int N=128;
int size=(N+2);
int size3=size*size*size*sizeof(float);
float *x = new float[size3];
float *y = new float[size3];
float *d_x;
float *d_y;

//Init x.
for (int i=1;i<N+1;i++) 
    for (int j=1;j<N+1;j++) 
        for (int k=1;k<N+1;k++)
            x[size*size*i+size*j+k]=cos(k);

// Shadow cells.
for (int i=1;i<N+1;i++) {
    for (int j=1;j<N+1;j++) { x[i*(N+2)*(N+2)+j*(N+2)]=x[i*(N+2)*(N+2)+j*(N+2)+1]; x[i*(N+2)*(N+2)+j*(N+2)+N+1]=x[i*(N+2)*(N+2)+j*(N+2)+N];}

    for (int k=0;k<N+2;k++) { x[i*(N+2)*(N+2)+k]=x[i*(N+2)*(N+2)+(N+2)+k]; x[i*(N+2)*(N+2)+(N+1)*(N+2)+k]=x[i*(N+2)*(N+2)+N*(N+2)+k];}
}

// CPU to GPU.
cudaMalloc((void **) &d_x, size3);
cudaMalloc((void **) &d_y, size3);
cudaMemcpy(d_x, x, size3, cudaMemcpyHostToDevice);
cudaMemcpy(d_y, y, size3, cudaMemcpyHostToDevice);

// Solver parameters.
dim3 dimBlock(2, 2, 64);
dim3 dimGrid(64, 64);

// Solver loop.
gettimeofday(&t1, 0);
kernel1 <<<dimGrid, dimBlock>>> (d_x, d_y, size, N);
cudaDeviceSynchronize();
gettimeofday(&t2, 0);
time = (1000000.0*(t2.tv_sec-t1.tv_sec) + t2.tv_usec-t1.tv_usec)/1000000.0;

// GPU to CPU.  
cudaMemcpy(y, d_y, size3, cudaMemcpyDeviceToHost);
cudaFree(d_x);
cudaFree(d_y);

// Power calculation.
NumOps=(1.0e-9)*N*N*N*7;
gflops = ( NumOps / (time));

// Display of final matrix.
int id_stage=-2;
while (id_stage!=-1) {
    cout<<"Which output's stage do you want to display? (-1 if you don't want to diplay another one)"<<endl;
    cin>>id_stage;
    cout<<endl;

    if (id_stage != -1) {
        cout<<"Stage "<<id_stage<<" of cube :"<<endl;
        for (int j=0;j<N+2;j++) {
            cout<<"| ";
            for (int k=0;k<N+2;k++) {cout<<y[id_stage*(N+2)*(N+2)+j*(N+2)+k]<<" ";}
            cout<<"|"<<endl;
        }
        cout<<endl;
    }
}

// Display of performances.
cout<<"Time : "<<time<<endl;
cout<<"Gflops/s : "<<gflops<<endl;      
}

}
你说的“结果是假的”是什么意思？数据结构是否包含意外数据？哪一个？在哪个阶段？事实上，我的结果应该只有0。当N=32时是这样，但如果N=64或128，则会出现一些1和2，如下所示：。N=128的示例听起来像是代码审查的问题，但事实并非如此。当您为x和y调用new时，为什么只为malloc指定有意义的size？
#include <assert.h>
#include <stdio.h>
#include <iostream>
#include <sys/time.h>
#include <cuda.h>
#include <ctime>
#include <math.h>
#include"reslap3D.cu"
using namespace std;

// Let's start the main program.
int main(void) {

// Variable statement.
struct timeval t1, t2;
float gflops;
float NumOps;
double time;
long int N=128;
int size=(N+2);
int size3=size*size*size*sizeof(float);
float *x = new float[size3];
float *y = new float[size3];
float *d_x;
float *d_y;

//Init x.
for (int i=1;i<N+1;i++) 
    for (int j=1;j<N+1;j++) 
        for (int k=1;k<N+1;k++)
            x[size*size*i+size*j+k]=cos(k);

// Shadow cells.
for (int i=1;i<N+1;i++) {
    for (int j=1;j<N+1;j++) { x[i*(N+2)*(N+2)+j*(N+2)]=x[i*(N+2)*(N+2)+j*(N+2)+1]; x[i*(N+2)*(N+2)+j*(N+2)+N+1]=x[i*(N+2)*(N+2)+j*(N+2)+N];}

    for (int k=0;k<N+2;k++) { x[i*(N+2)*(N+2)+k]=x[i*(N+2)*(N+2)+(N+2)+k]; x[i*(N+2)*(N+2)+(N+1)*(N+2)+k]=x[i*(N+2)*(N+2)+N*(N+2)+k];}
}

// CPU to GPU.
cudaMalloc((void **) &d_x, size3);
cudaMalloc((void **) &d_y, size3);
cudaMemcpy(d_x, x, size3, cudaMemcpyHostToDevice);
cudaMemcpy(d_y, y, size3, cudaMemcpyHostToDevice);

// Solver parameters.
dim3 dimBlock(2, 2, 64);
dim3 dimGrid(64, 64);

// Solver loop.
gettimeofday(&t1, 0);
kernel1 <<<dimGrid, dimBlock>>> (d_x, d_y, size, N);
cudaDeviceSynchronize();
gettimeofday(&t2, 0);
time = (1000000.0*(t2.tv_sec-t1.tv_sec) + t2.tv_usec-t1.tv_usec)/1000000.0;

// GPU to CPU.  
cudaMemcpy(y, d_y, size3, cudaMemcpyDeviceToHost);
cudaFree(d_x);
cudaFree(d_y);

// Power calculation.
NumOps=(1.0e-9)*N*N*N*7;
gflops = ( NumOps / (time));

// Display of final matrix.
int id_stage=-2;
while (id_stage!=-1) {
    cout<<"Which output's stage do you want to display? (-1 if you don't want to diplay another one)"<<endl;
    cin>>id_stage;
    cout<<endl;

    if (id_stage != -1) {
        cout<<"Stage "<<id_stage<<" of cube :"<<endl;
        for (int j=0;j<N+2;j++) {
            cout<<"| ";
            for (int k=0;k<N+2;k++) {cout<<y[id_stage*(N+2)*(N+2)+j*(N+2)+k]<<" ";}
            cout<<"|"<<endl;
        }
        cout<<endl;
    }
}

// Display of performances.
cout<<"Time : "<<time<<endl;
cout<<"Gflops/s : "<<gflops<<endl;      
}

#define D(x,y,z) size*size*(x)+size*(y)+z
__global__ void kernel1(float *x, float *y, int size, int N)
{
__shared__ float sdata0[4][4][66];
__shared__ float sdata64[4][4][66];

int c0 = blockIdx.x*blockDim.x + threadIdx.x+1;
int c1 = blockIdx.y*blockDim.y + threadIdx.y+1;
int c2 = threadIdx.z+1;
int i = threadIdx.x+1, j = threadIdx.y+1, k = threadIdx.z+1;

if (threadIdx.x == 0) 
{ sdata0[i-1][j][k] = x[D(c0-1,c1,c2)];     
  sdata64[i-1][j][k] = x[D(c0-1,c1,c2+64)]; 
}
if (threadIdx.x == 1) 
{ sdata0[i+1][j][k] = x[D(c0+1,c1,c2)];     
  sdata64[i+1][j][k] = x[D(c0+1,c1,c2+64)]; 
}

if (threadIdx.y == 0) 
{ sdata0[i][j-1][k] = x[D(c0,c1-1,c2)]; 
  sdata64[i][j-1][k] = x[D(c0,c1-1,c2+64)]; 
}
if (threadIdx.y == 1) 
{ sdata0[i][j+1][k] = x[D(c0,c1+1,c2)]; 
  sdata64[i][j+1][k] = x[D(c0,c1+1,c2+64)]; 
}

if (threadIdx.z == 0) 
{ sdata0[i][j][k-1] = x[D(c0,c1,c2-1)]; 
  sdata64[i][j][k-1] = x[D(c0,c1,c2+63)]; 
}
if (threadIdx.z == 63) 
{ sdata0[i][j][k+1] = x[D(c0,c1,c2+1)]; 
  sdata64[i][j][k+1] = x[D(c0,c1,c2+65)]; 
}


sdata0[i][j][k] = x[D(c0,c1,c2)];

sdata64[i][j][k] = x[D(c0,c1,c2+64)];

__syncthreads();

             y[D(c0, c1, c2)] = sdata0[i+1][j][k]
                      + sdata0[i-1][j][k]
                          + sdata0[i][j+1][k] 
                      + sdata0[i][j-1][k]
                      + sdata0[i][j][k+1] 
                      + sdata0[i][j][k-1]

                      - 6 * sdata0[i][j][k];

             y[D(c0, c1, c2+64)] = sdata64[i+1][j][k]
                      + sdata64[i-1][j][k]
                          + sdata64[i][j+1][k] 
                      + sdata64[i][j-1][k]
                      + sdata64[i][j][k+1] 
                      + sdata64[i][j][k-1]

                      - 6 * sdata64[i][j][k];