CUDA编程系列学习-从入门到放弃

   本文主要从代码方面进行CUDA系列知识的学习，基础知识就不一一细讲了，毕竟很多大牛已经介绍得很详细了，，，，

   bong  bong bong 开始吧！

一、整数相加

   由GPU计算两个整数求和，使用一个Block 一个thread就可以解决问题了。

代码如下：

__global__ void add(int *a, int *b, int *c) {
	*c = *a + *b;
}


void firstDemo(){
	//host value of a b c
	int hA, hB, hC;
	hA = 10, hB = 20, hC = 0;
	//device value of a b c 
	int *devA, *devB, *devC;
	//allocate device memery of a b c (use void**)
	cudaMalloc((void**)&devA, sizeof(int));
	cudaMalloc((void**)&devB, sizeof(int));
	cudaMalloc((void**)&devC, sizeof(int));

	//copy host value to device (use hA address &hA)
	cudaMemcpy(devA, &hA, sizeof(int), cudaMemcpyHostToDevice);
	cudaMemcpy(devB, &hB, sizeof(int), cudaMemcpyHostToDevice);  

	//launch add() kernel on GPU 
	add << <1, 1 >> > (devA, devB, devC);
	//return the answer from the device
	cudaMemcpy(&hC, devC, sizeof(int), cudaMemcpyDeviceToHost);
	printf("C=%d\n", hC);

	//Free the device memory
	cudaFree(devA);
	cudaFree(devB);
	cudaFree(devC);
}

二、一维数组相加

   在上面的例子的基础上，将整数改为数组，并且一个Block 里面只设置一个线程（实现Block层次的粗粒度并行），所以kernel函数里面下标的索引就用blockIdx.x。

注意：由整数变为数组后函数参数的变化，是否取地址符

   代码如下：

__global__ void vectorAdd(int *a, int *b, int *c) {
	//use blockidx.x to access block's index
	c[blockIdx.x] = a[blockIdx.x] + b[blockIdx.x];
}
#define N  100 
void secondDemo() {
	//host value of a b c
	int *hA, *hB, *hC;
	size_t size = N * sizeof(int);
	hA = (int*)malloc(size);
	hB = (int*)malloc(size);
	hC = (int*)malloc(size);

	//srand( (unsigned) time(NULL) );
	for (int i = 0; i < N; i++) {
		hA[i] = rand()%20;
		hB[i] = rand() % 20;
	}
	printf("A:\n");
	for (int i = 0; i < N; i++) {
		printf("%5d", hA[i]);
	}
	printf("\nB:\n");
	for (int i = 0; i < N; i++) {
		printf("%5d", hB[i]);
	}
	printf("\n");
	//device value of a b c 
	int *devA, *devB, *devC;
	//allocate device memery of a b c (use void**)
	cudaMalloc((void**)&devA, size);
	cudaMalloc((void**)&devB, size);
	cudaMalloc((void**)&devC, size);

	//copy host value to device (use hA address &hA)
	cudaMemcpy(devA, hA, size, cudaMemcpyHostToDevice);
	cudaMemcpy(devB, hB, size, cudaMemcpyHostToDevice);
	//launch add() kernel on GPU 
	vectorAdd << > > (devA, devB, devC);
	//return the answer from the device
	cudaMemcpy(hC, devC, size, cudaMemcpyDeviceToHost);
	
	printf("C:\n");
	for (int i = 0; i < N; i++) {
		printf("%5d", hC[i]);
	}
	printf("\n");

	//Free the device memory
	cudaFree(devA);
	cudaFree(devB);
	cudaFree(devC);

	free(hA);
	free(hB);
	free(hC);
}

三、一维数组点乘求和-使用共享内存

该例子只用了一个Block，在该Block里面声明 了一块共享内存数组，通过线程索引将乘积保存到对应元素，利用栅栏同步函数等待所有线程计算完成后再用线程0来累加点乘之和。

kernel:

__global__ void dotProduct(int *a, int *b, int *c) {
	__shared__ int shareArray[N];
	shareArray[threadIdx.x] = a[threadIdx.x] * b[threadIdx.x];
	__syncthreads();

	if (0 == threadIdx.x) {
		int sum = 0;
		for (int i = 0; i < N; i++) {
			sum += shareArray[i];
		}
		*c = sum;
	}
}

void thirdDemo_SharMem() {
	//host value of a b c
	int *hA, *hB, *hCone;
	size_t size = N * sizeof(int);
	hA = (int*)malloc(size);
	hB = (int*)malloc(size);
	hCone = (int*)malloc(sizeof(int));

	//srand( (unsigned) time(NULL) );
	for (int i = 0; i < N; i++) {
		hA[i] = i+1;
		hB[i] = 2;
	}
	printf("A:\n");
	for (int i = 0; i < N; i++) {
		printf("%5d", hA[i]);
	}
	printf("\nB:\n");
	for (int i = 0; i < N; i++) {
		printf("%5d", hB[i]);
	}
	printf("\n");
	//device value of a b c 
	int *devA, *devB, *devC;
	//allocate device memery of a b c (use void**)
	cudaMalloc((void**)&devA, size);
	cudaMalloc((void**)&devB, size);
	cudaMalloc((void**)&devC, sizeof(int));

	//copy host value to device (use hA address &hA)
	cudaMemcpy(devA, hA, size, cudaMemcpyHostToDevice);
	cudaMemcpy(devB, hB, size, cudaMemcpyHostToDevice);
	//launch add() kernel on GPU   use one Block
	dotProduct << <1, N >> > (devA, devB, devC);
	//return the answer from the device
	cudaMemcpy(hCone, devC, sizeof(int), cudaMemcpyDeviceToHost);

	printf("C:\n");
	printf("%5d", *hCone);

	//Free the device memory
	cudaFree(devA);
	cudaFree(devB);
	cudaFree(devC);

	free(hA);
	free(hB);
	free(hCone);
}

四：