[CUDA] GPU并行计算两一维数组之和

书指[GPU高性能编程CUDA实战].(桑德斯).聂雪军等.扫描版.pdf

一.多线程块并行计算 

设一维数组大小为N。

#include "cuda_runtime.h"  
#include "cublas_v2.h"  

#include   
#include   

#define N 1024

using namespace std;

__global__ void init(int *a, int *b, int *c)
{
	int tid = blockIdx.x;
	if (tid < N)
	{
		a[tid] = tid;
		b[tid] = tid * tid;
	}
}

__global__ void add(int *a, int *b, int *c)
{
	int tid = blockIdx.x;
	if (tid < N)
	{
		c[tid] = a[tid] + b[tid];
	}
}

int main()
{
	int a[N], b[N], c[N];
	int *dev_a, *dev_b, *dev_c;


	cudaMalloc((void **)&dev_a, N * sizeof(int));
	cudaMalloc((void **)&dev_b, N * sizeof(int));
	cudaMalloc((void **)&dev_c, N * sizeof(int));

	
	init << > > (dev_a, dev_b, dev_c);
	add << > > (dev_a, dev_b, dev_c);

	cudaMemcpy(a, dev_a, N * sizeof(int), cudaMemcpyDeviceToHost);
	cudaMemcpy(b, dev_b, N * sizeof(int), cudaMemcpyDeviceToHost);
	cudaMemcpy(c, dev_c, N * sizeof(int), cudaMemcpyDeviceToHost);


	cudaFree(dev_a);
	cudaFree(dev_b);
	cudaFree(dev_c);

	printf("%d+%d=%d\n", a[100], b[100], c[100]);
	

	getchar();
	return 0;
}

二.多线程并行计算

Nvidia 1050Ti 4G的每个线程块最大只能有1024个线程。设一维数组大小为N。

1.当一维数组大小小于等于1024时

这种情况下只需要使用一个线程块中的N个线程计算。详见书P44。

#include "cuda_runtime.h"  
#include "cublas_v2.h"  

#include   
#include   

#define N 1024

using namespace std;

__global__ void init(int *a, int *b, int *c)
{
	int i = threadIdx.x;

	if(i < N)
	{
		a[i] = i;
		b[i] = i * i;
	}
		
}

__global__ void add(int *a, int *b, int *c)
{
	int i = threadIdx.x;

	if(i < N)
	{
		c[i] = a[i] + b[i];
	}
}

int main()
{
	int a[N], b[N], c[N];
	int *dev_a, *dev_b, *dev_c;

	cudaMalloc((void **)&dev_a, N * sizeof(int));
	cudaMalloc((void **)&dev_b, N * sizeof(int));
	cudaMalloc((void **)&dev_c, N * sizeof(int));

	init<<<1, N>>>(dev_a, dev_b, dev_c);
	add <<<1, N>>>(dev_a, dev_b, dev_c);

	cudaMemcpy(a, dev_a, N * sizeof(int), cudaMemcpyDeviceToHost);
	cudaMemcpy(b, dev_b, N * sizeof(int), cudaMemcpyDeviceToHost);
	cudaMemcpy(c, dev_c, N * sizeof(int), cudaMemcpyDeviceToHost);

	cudaFree(dev_a);
	cudaFree(dev_b);
	cudaFree(dev_c);

	printf("%d + %d = %d\n", a[100], b[100], c[100]);


	getchar();
	return 0;
}

2.当一维数组大小大于1024时

设每个线程块只使用512个线程，则这种情况下需要使用(N-511)/512个线程块。详见书P45。

#include "cuda_runtime.h"  
#include "cublas_v2.h"  

#include   
#include   

#define N 2000

using namespace std;

__global__ void init(int *a, int *b, int *c)
{
	int i = threadIdx.x + blockIdx.x * blockDim.x;

	if(i < N)
	{
		a[i] = i;
		b[i] = i * i;
	}
		
}

__global__ void add(int *a, int *b, int *c)
{
	int i = threadIdx.x + blockIdx.x * blockDim.x;

	if(i < N)
	{
		c[i] = a[i] + b[i];
	}
}

int main()
{
	int a[N], b[N], c[N];
	int *dev_a, *dev_b, *dev_c;

	cudaMalloc((void **)&dev_a, N * sizeof(int));
	cudaMalloc((void **)&dev_b, N * sizeof(int));
	cudaMalloc((void **)&dev_c, N * sizeof(int));

	init << <(N + 511) / 512, 512 >> > (dev_a, dev_b, dev_c);
	add << <(N + 511) / 512, 512 >> > (dev_a, dev_b, dev_c);

	cudaMemcpy(a, dev_a, N * sizeof(int), cudaMemcpyDeviceToHost);
	cudaMemcpy(b, dev_b, N * sizeof(int), cudaMemcpyDeviceToHost);
	cudaMemcpy(c, dev_c, N * sizeof(int), cudaMemcpyDeviceToHost);

	cudaFree(dev_a);
	cudaFree(dev_b);
	cudaFree(dev_c);

	printf("%d + %d = %d\n", a[100], b[100], c[100]);

	getchar();
	return 0;
}