有谁知道cub :: scan支持的最大大小是多少？我的核心转储输入大小超过5亿。我想确保我没有做错任何事......
这是我的代码：
#define CUB_STDERR
#include <stdio.h>
#include "cub/util_allocator.cuh"
#include "cub/device/device_scan.cuh"
#include <sys/time.h>
using namespace cub;
bool g_verbose = false; // Whether to display input/output to console
CachingDeviceAllocator g_allocator(true); // Caching allocator for device memory
typedef int mytype;
/**
* Solve inclusive-scan problem
*/
static void solve(mytype *h_in, mytype *h_cpu, int n)
{
mytype inclusive = 0;
for (int i = 0; i < n; ++i) {
inclusive += h_in[i];
h_cpu[i] = inclusive;
}
}
static int compare(mytype *h_cpu, mytype *h_o, int n)
{
for (int i = 0; i < n; i++) {
if (h_cpu[i] != h_o[i]) {
return i + 1;
}
}
return 0;
}
/**
* Main
*/
int main(int argc, char** argv)
{
cudaSetDevice(0);
struct timeval start, end;
int num_items = 1073741824;
const int repetitions = 5;
mytype *h_in, *h_out, *h_cpu;
const int size = num_items * sizeof(mytype);
// Allocate host arrays
h_in = (mytype *)malloc(size);
h_out = (mytype *)malloc(size);
h_cpu = (mytype *)malloc(size);
// Initialize problem and solution
for (int i = 0; i < num_items; i++) {
h_in[i] = i;
h_out[i] = 0;
h_cpu[i] = 0;
}
solve(h_in, h_cpu, num_items);
// Allocate problem device arrays
mytype *d_in = NULL;
CubDebugExit(g_allocator.DeviceAllocate((void**)&d_in, sizeof(mytype) * num_items));
// Initialize device input
CubDebugExit(cudaMemcpy(d_in, h_in, sizeof(mytype) * num_items, cudaMemcpyHostToDevice));
// Allocate device output array
mytype *d_out = NULL;
CubDebugExit(g_allocator.DeviceAllocate((void**)&d_out, sizeof(mytype) * num_items));
// Allocate temporary storage
void *d_temp_storage = NULL;
size_t temp_storage_bytes = 0;
CubDebugExit(DeviceScan::InclusiveSum(d_temp_storage, temp_storage_bytes, d_in, d_out, num_items));
CubDebugExit(g_allocator.DeviceAllocate(&d_temp_storage, temp_storage_bytes));
// Run
gettimeofday(&start, NULL);
for (long i = 0; i < repetitions; i++)
DeviceScan::InclusiveSum(d_temp_storage, temp_storage_bytes, d_in, d_out, num_items);
cudaThreadSynchronize();
gettimeofday(&end, NULL);
double ctime = end.tv_sec + end.tv_usec / 1000000.0 - start.tv_sec - start.tv_usec / 1000000.0;
cudaMemcpy(h_out, d_out, sizeof(mytype) * num_items, cudaMemcpyDeviceToHost);
int cmp = compare(h_cpu, h_out, num_items);
printf("%d\t", num_items);
if (!cmp)
printf("\t%7.4fs \n", ctime);
printf("\n");
if (h_in) delete[] h_in;
if (h_out) delete[] h_out;
if (h_cpu) delete[] h_cpu;
if (d_in) CubDebugExit(g_allocator.DeviceFree(d_in));
if (d_out) CubDebugExit(g_allocator.DeviceFree(d_out));
if (d_temp_storage) CubDebugExit(g_allocator.DeviceFree(d_temp_storage));
printf("\n\n");
return 0;
}
答案 0 :(得分：2)
问题在于：
const int size = num_items * sizeof(mytype);
可以通过将其更改为：
来修复const size_t size = num_items * sizeof(mytype);
代码中num_items的值超过10亿。当我们将它乘以sizeof(mytype)时，我们将它乘以4，结果超过40亿。此值无法存储在int变量中。如果您尝试使用它，那么您的后续主机代码将做坏事。这个问题（核心转储）实际上与CUDA无关。如果删除了所有CUB元素，代码将进行核心转储。
当我修改上面的代码行，并编译正确的GPU（例如我的情况下为-arch=sm_35，或者对于Titan X GPU进行-arch=sm_52）时，我得到正确答案（和没有seg fault / core dump）。
通常，追逐seg故障/核心转储类型错误时的正确起点是识别此错误来自主机代码，您应该尝试本地化确切的源代码行那就是产生这个错误。这可以通过在代码中放入许多printf语句来简单/繁琐地完成，直到您确定代码行，之后您没有看到任何printf输出，或者使用主机代码调试器，例如gdb在linux上。
另请注意，编写的代码在主机上需要略多于12GB的内存，而GPU上的内存略多于8GB，因此只能在此类设置中正常运行。
供参考，这里是固定代码（基于OP发布的here）：
#define CUB_STDERR
#include <stdio.h>
#include "cub/util_allocator.cuh"
#include "cub/device/device_scan.cuh"
#include <sys/time.h>
using namespace cub;
bool g_verbose = false; // Whether to display input/output to console
CachingDeviceAllocator g_allocator(true); // Caching allocator for device memory
typedef int mytype;
/**
* Solve inclusive-scan problem
*/
static void solve(mytype *h_in, mytype *h_cpu, int n)
{
mytype inclusive = 0;
for (int i = 0; i < n; ++i) {
inclusive += h_in[i];
h_cpu[i] = inclusive;
}
}
static int compare(mytype *h_cpu, mytype *h_o, int n)
{
for (int i = 0; i < n; i++) {
if (h_cpu[i] != h_o[i]) {
return i + 1;
}
}
return 0;
}
/**
* Main
*/
int main(int argc, char** argv)
{
cudaSetDevice(0);
struct timeval start, end;
int num_items = 1073741824;
const int repetitions = 5;
mytype *h_in, *h_out, *h_cpu;
const size_t size = num_items * sizeof(mytype);
// Allocate host arrays
h_in = (mytype *)malloc(size);
h_out = (mytype *)malloc(size);
h_cpu = (mytype *)malloc(size);
// Initialize problem and solution
for (int i = 0; i < num_items; i++) {
h_in[i] = i;
h_out[i] = 0;
h_cpu[i] = 0;
}
solve(h_in, h_cpu, num_items);
// Allocate problem device arrays
mytype *d_in = NULL;
CubDebugExit(g_allocator.DeviceAllocate((void**)&d_in, sizeof(mytype) * num_items));
// Initialize device input
CubDebugExit(cudaMemcpy(d_in, h_in, sizeof(mytype) * num_items, cudaMemcpyHostToDevice));
// Allocate device output array
mytype *d_out = NULL;
CubDebugExit(g_allocator.DeviceAllocate((void**)&d_out, sizeof(mytype) * num_items));
// Allocate temporary storage
void *d_temp_storage = NULL;
size_t temp_storage_bytes = 0;
CubDebugExit(DeviceScan::InclusiveSum(d_temp_storage, temp_storage_bytes, d_in, d_out, num_items));
CubDebugExit(g_allocator.DeviceAllocate(&d_temp_storage, temp_storage_bytes));
// Run
gettimeofday(&start, NULL);
for (long i = 0; i < repetitions; i++)
DeviceScan::InclusiveSum(d_temp_storage, temp_storage_bytes, d_in, d_out, num_items);
cudaThreadSynchronize();
gettimeofday(&end, NULL);
double ctime = end.tv_sec + end.tv_usec / 1000000.0 - start.tv_sec - start.tv_usec / 1000000.0;
cudaMemcpy(h_out, d_out, sizeof(mytype) * num_items, cudaMemcpyDeviceToHost);
int cmp = compare(h_cpu, h_out, num_items);
printf("%d\t", num_items);
if (!cmp)
printf("\t%7.4fs \n", ctime);
printf("\n");
if (h_in) delete[] h_in;
if (h_out) delete[] h_out;
if (h_cpu) delete[] h_cpu;
if (d_in) CubDebugExit(g_allocator.DeviceFree(d_in));
if (d_out) CubDebugExit(g_allocator.DeviceFree(d_out));
if (d_temp_storage) CubDebugExit(g_allocator.DeviceFree(d_temp_storage));
printf("\n\n");
return 0;
}