cudaMemcpy2D与cudaMemcpy的复制时间问题

2009-12-25 01:35:11 1k 阅读 AI 写同款 · GEO 优化AI 选题数据监测一键分发 ›

CUDA显存管理核心：cudaMalloc、cudaMemcpy与cudaFree详解在GPU并行计算与深度学习领域，高效管理设备显存是提升性能的关键基础。其核心原理在于CPU（主机）与GPU（设备）拥有独立的内存空间，数据必须通过特定的API在两者间进行搬运。CUDA运行时API为此提供了基石性的显存管理工具，它们直接决定了程序的正确性与执行效率。掌握这些底层技术，对于优化模型训练、解决OOM（Out Of Memory）错误以及进行深度性能调优具有重要价值。本文将以向量加法为例，深入剖析cudaMalloc、cudaMemcpy和cudaFree这三个核心API的功能、使用场景与常见陷阱阅读详情

1、[/size] float *ac, *bc, *cc;
clock_t start, endTime;
size_t pitch_a, pitch_b, pitch_c;
int newn = ((n + BLOCK_SIZE - 1) / BLOCK_SIZE) * BLOCK_SIZE;

start = clock();

cudaMallocPitch((void**) &ac, &pitch_a, sizeof(float) * newn, newn);
cudaMallocPitch((void**) &bc, &pitch_b, sizeof(float) * newn, newn);
cudaMallocPitch((void**) &cc, &pitch_c, sizeof(float) * newn, newn);

cudaMemset(ac, 0, pitch_a * newn);
cudaMemset(bc, 0, pitch_b * newn);

cudaMemcpy2D(ac, pitch_a, a, sizeof(float) * lda, sizeof(float) * n, n, cudaMemcpyHostToDevice);
cudaMemcpy2D(bc, pitch_b, b, sizeof(float) * ldb, sizeof(float) * n, n, cudaMemcpyHostToDevice);

2、// allocate host memory for matrices A and B
unsigned int size_A = WA * HA;
unsigned int mem_size_A = sizeof(float) * size_A;

float* h_A = (float*) malloc(mem_size_A);
unsigned int size_B = WB * HB;
unsigned int mem_size_B = sizeof(float) * size_B;
float* h_B = (float*) malloc(mem_size_B);

// initialize host memory
randomInit(h_A, size_A);
randomInit(h_B, size_B);

// allocate device memory
float* d_A;
cutilSafeCall(cudaMalloc((void**) &d_A, mem_size_A));
float* d_B;
cutilSafeCall(cudaMalloc((void**) &d_B, mem_size_B));

// copy host memory to device
cutilSafeCall(cudaMemcpy(d_A, h_A, mem_size_A,
cudaMemcpyHostToDevice) );
cutilSafeCall(cudaMemcpy(d_B, h_B, mem_size_B,
cudaMemcpyHostToDevice) );

cudaMemcpy2D花的时间比cudaMemcpy多了将近一倍。。。不懂这是为什么？然后我把第一个程序中的cudaMemcpy2D换成cudaMemcpy后花的时间还是不变？？

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