Skip to content
New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

hw08/cuda #9

Open
wants to merge 1 commit into
base: main
Choose a base branch
from
Open
Show file tree
Hide file tree
Changes from all commits
Commits
File filter

Filter by extension

Filter by extension

Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
2 changes: 1 addition & 1 deletion CMakeLists.txt
Original file line number Diff line number Diff line change
Expand Up @@ -6,7 +6,7 @@ if (NOT CMAKE_BUILD_TYPE)
set(CMAKE_BUILD_TYPE Release)
endif()
# 如果需要指定显卡版本号的话:
# set(CMAKE_CUDA_ARCHITECTURES 52)
set(CMAKE_CUDA_ARCHITECTURES 70)

project(hellocmake LANGUAGES CXX CUDA)

Expand Down
20 changes: 11 additions & 9 deletions main.cu
Original file line number Diff line number Diff line change
Expand Up @@ -8,18 +8,18 @@

// 这是基于“边角料法”的,请把他改成基于“网格跨步循环”的:10 分
__global__ void fill_sin(int *arr, int n) {
int i = blockIdx.x * blockDim.x + threadIdx.x;
if (i < n) return;
arr[i] = sinf(i);
// 基于网格跨步循环,自动根据n指定blockDim和gridDim
for (size_t i = blockDim.x * blockIdx.x + threadIdx.x; i < n; i += blockDim.x * gridDim.x)
arr[i] = sinf(i);
}

__global__ void filter_positive(int *counter, int *res, int const *arr, int n) {
int i = blockIdx.x * blockDim.x + threadIdx.x;
if (i < n) return;
if (i >= n) return;
if (arr[i] >= 0) {
// 这里有什么问题?请改正:10 分
int loc = *counter;
*counter += 1;
// 会有其他线程干扰对counter+1的过程,应替代为原子操作
int loc = atomicAdd(counter, 1);
res[loc] = n;
}
}
Expand All @@ -31,20 +31,22 @@ int main() {
std::vector<int, CudaAllocator<int>> counter(1);

// fill_sin 改成“网格跨步循环”以后,这里三重尖括号里的参数如何调整?10 分
fill_sin<<<n / 1024, 1024>>>(arr.data(), n);
fill_sin<<<128, 1024>>>(arr.data(), n);

// 这里的“边角料法”对于不是 1024 整数倍的 n 会出错,为什么?请修复:10 分
filter_positive<<<n / 1024, 1024>>>(counter.data(), res.data(), arr.data(), n);
// 出错:若n = 1025,n / 1024 = 1,总共会启动1*1024=1024个线程,会漏掉1个元素
filter_positive<<<(n + 1024 - 1)/ 1024, 1024>>>(counter.data(), res.data(), arr.data(), n);

// 这里 CPU 访问数据前漏了一步什么操作?请补上:10 分
cudaDeviceSynchronize();

if (counter[0] <= n / 50) {
printf("Result too short! %d <= %d\n", counter[0], n / 50);
return -1;
}
for (int i = 0; i < counter[0]; i++) {
if (res[i] < 0) {
printf("Wrong At %d: %f < 0\n", i, res[i]);
printf("Wrong At %d: %d < 0\n", i, res[i]);
return -1; // 突然想起了ICPC有一年队名叫“蓝翔WA掘机”的,笑不活了:)
}
}
Expand Down