2013-03-15 66 views
4

以下是CUDA编程示例,它基本上是C但内部带有NVidia CUDA函数。我一直试图解释这个代码示例,并找出它正在尝试做什么。我的问题是这个程序编译得很好,但是它需要什么参数?例如,这CUDA程序正在在Linux模拟器但是一旦运行./program返回运行:C/CUDA程序输出

用法:./program数 分段故障

哪些程序输入参数。谢谢。

#include <assert.h> 
#include <stdio.h> 

//#define N 100000 

__host__ void saxpy_host(int length, float alpha, float * x, float * y) 
{ 
    for (int i = 0; i < length; ++i) 
    y[i] = alpha*x[i] + y[i]; 
} 

__global__ void saxpy (int length, float alpha, float * x, float * y) 
{ 
    int i; 
    i = blockIdx.x*blockDim.x + threadIdx.x; 
    if (i < length) y[i] = alpha*x[i]+y[i]; 
    __syncthreads(); 
} 

int main(int argc, char* argv[]) { 

if (argc != 2) { 
    printf("Usage: %s number\n", argv[0]); 
    return -1; 
} 

int N = atoi(argv[1]); 

// host data 
float alpha = 0.5; 
float x[N], xback[N]; 
float y[N], yback[N]; 

int size; 
int i; 
int blocks; 

// determining size 
size = sizeof(float)*N; 


// device data 
float * dxp, * dyp; 


// fill host data 
for (i = 0; i < N; i++) { 
    x[i] = (float) (rand() % 128); 
    y[i] = (float) (rand() % 256); 
} 


// Allocating and Moving data to device 
cudaMalloc((void**) &dxp, size); 
cudaMalloc((void**) &dyp, size); 
cudaMemcpy (dxp, x, size, cudaMemcpyHostToDevice); 
cudaMemcpy (dyp, y, size, cudaMemcpyHostToDevice); 

// size of thread blocks 
blocks = (N + 31)/32; 
saxpy <<< blocks, 32 >>> (N, alpha, dxp, dyp); 

// bring back data 
cudaMemcpy (xback, dxp, size, cudaMemcpyDeviceToHost); 
cudaMemcpy (yback, dyp, size, cudaMemcpyDeviceToHost); 

// Calculating host SAXPY 
saxpy_host (N, alpha, (float *) &x, (float *) &y); 

// checking computation on host matches computation on GPU 
for (i = 0; i < N; i++) { 
    assert (yback[i] == y[i]) ; 
    //printf ("%i %f %f \n", i, yback[i], y[i]); 
} 


// free device data 
cudaFree(dxp); cudaFree(dyp); 

return 0; 
} 

回答

3
int N = atoi(argv[1]); 

该方案需要一个单一的整数作为命令行参数。 (例如,将它称为./program 5

然后计算一个SAXPY(一个源于早期BLAS实现的旧术语,但它卡住了,它意味着“单个(精确的,又称浮点数)实数alpha x加y” )与尺寸为N的矢量。

+0

任何想法是什么int值。将值设置为100000会返回以下错误:“”int main(int,char **):声明'yback [i] == y [i]'失败。中止“”一个非常大的数字会导致错误“”Segmentation Fault“” – 2013-03-15 20:08:30

+0

它运行的值是小的 - '5'还是'32',可能? – us2012 2013-03-15 20:12:22

+0

不,它错误与相同断言失败错误 – 2013-03-15 20:13:57