我熟悉一个配备Pascal P100 GPU + Nvlink的新集群。我写了一个ping-pong程序来测试gpu< - > gpu和gpu< - > cpu带宽和点对点访问。 (我知道cuda样本包含这样的程序,但我想自己做以便更好地理解。)Nvlink带宽看似合理(双向约35 GB / s,理论最大值为40)。然而,在调试乒乓球时,我发现了一些奇怪的行为。
首先,无论我指定什么cudaMemcpyKind,cudaMemcpyAsync都会成功,例如,如果cudaMemcpyAsync正在将内存从主机复制到设备,即使我将cudaMemcpyDeviceToHost作为类型传递,它也会成功。
其次,当主机内存未被页面锁定时,cudaMemcpyAsync会执行以下操作:
这种行为是期待的吗?我已经包含了一个在我的系统上演示它的最小工作示例代码(示例不是乒乓应用程序,它所做的只是使用各种参数测试cudaMemcpyAsync)。
P100s启用了UVA,因此我认为cudaMemcpyAsync只是推断src和dst指针的位置并忽略cudaMemcpyKind参数。但是,我不确定为什么cudaMemcpyAsync无法为非页面锁定的主机内存引发错误。我的印象是严格的禁忌。
#include <stdio.h>
#include <cuda_runtime.h>
#include <stdlib.h>
#define gpuErrchk(ans) { gpuAssert((ans), __FILE__, __LINE__); }
inline void gpuAssert(cudaError_t code, const char *file, int line, bool abort=true)
{
if (code != cudaSuccess)
{
fprintf(stderr,"GPUassert: %s %s %d\n", cudaGetErrorString(code), file, line);
if (abort) exit(code);
}
}
__global__ void checkDataDevice( int* current, int* next, int expected_current_val, int n )
{
int tid = threadIdx.x + blockIdx.x*blockDim.x;
for( int i = tid; i < n; i += blockDim.x*gridDim.x )
{
if( current[i] != expected_current_val )
printf( "Error on device: expected = %d, current[%d] = %d\n"
, expected_current_val
, i
, current[i] );
// Increment the data so the next copy is properly tested
next[i] = current[i] + 1;
}
}
void checkDataHost( int* current, int* next, int expected_current_val, int n )
{
for( int i = 0; i < n; i++ )
{
if( current[i] != expected_current_val )
printf( "Error on host: expected = %d, current[%d] = %d\n"
, expected_current_val
, i
, current[i] );
// Increment the data so the next copy is properly tested
next[i] = current[i] + 1;
}
}
int main( int argc, char** argv )
{
bool pagelocked = true;
// invoking the executable with any additional argument(s) will turn off page locked memory, i.e.,
// Run with pagelocked memory: ./a.out
// Run with ordinary malloc'd memory: ./a.out jkfdlsja
if( argc > 1 )
pagelocked = false;
int copybytes = 1e8; // Ok to use int instead of size_t for 1e8.
cudaStream_t* stream = (cudaStream_t*)malloc( sizeof(cudaStream_t) );
cudaStreamCreate( stream );
int* srcHost;
int* dstHost;
int* srcDevice;
int* dstDevice;
cudaMalloc( (void**)&srcDevice, copybytes );
cudaMalloc( (void**)&dstDevice, copybytes );
if( pagelocked )
{
printf( "Using page locked memory\n" );
cudaMallocHost( (void**)&srcHost, copybytes );
cudaMallocHost( (void**)&dstHost, copybytes );
}
else
{
printf( "Using non page locked memory\n" );
srcHost = (int*)malloc( copybytes );
dstHost = (int*)malloc( copybytes );
}
for( int i = 0; i < copybytes/sizeof(int); i++ )
srcHost[i] = 1;
cudaMemcpyKind kinds[4];
kinds[0] = cudaMemcpyHostToDevice;
kinds[1] = cudaMemcpyDeviceToHost;
kinds[2] = cudaMemcpyHostToHost;
kinds[3] = cudaMemcpyDeviceToDevice;
// Test cudaMemcpyAsync in both directions,
// iterating through all "cudaMemcpyKinds" to verify
// that they don't matter.
int expected_current_val = 1;
for( int kind = 0; kind<4; kind++ )
{
// Host to device copy
cudaMemcpyAsync( dstDevice
, srcHost
, copybytes
, kinds[kind]
, *stream );
gpuErrchk( cudaDeviceSynchronize() );
checkDataDevice<<<56*8,256>>>( dstDevice
, srcDevice
, expected_current_val
, copybytes/sizeof(int) );
expected_current_val++;
// Device to host copy
cudaMemcpyAsync( dstHost
, srcDevice
, copybytes
, kinds[kind]
, *stream );
gpuErrchk( cudaDeviceSynchronize() );
checkDataHost( dstHost
, srcHost
, expected_current_val
, copybytes/sizeof(int) );
expected_current_val++;
}
free( stream );
cudaFree( srcDevice );
cudaFree( dstDevice );
if( pagelocked )
{
cudaFreeHost( srcHost );
cudaFreeHost( dstHost );
}
else
{
free( srcHost );
free( dstHost );
}
return 0;
}
答案 0 :(得分:5)
如果遇到CUDA代码问题,我强烈建议使用严格(==检查每个调用返回代码)proper CUDA error checking。
您的错误检查存在缺陷,并且这些缺陷会导致您的一些混淆。
首先,在页面锁定的情况下,给定(映射)指针在主机和设备上都是可访问/有效的。因此,每个可能的方向枚举(H2D,D2H,D2D,H2H)都是合法且有效的。因此,不会返回任何错误,并且复制操作成功。
在非页面锁定的情况下,上述情况并非如此,因此一般来说,指示的传输方向更好地匹配隐含的传输方向,如指针所检查的那样。如果没有,cudaMemcpyAsync
将返回错误代码(cudaErrorInvalidValue
== 11)。在您的情况下,您忽略此错误结果。如果你有足够的耐心(如果你只是标记了第一个错误,而不是打印10M +元素中的每个不匹配会更好),你可以通过cuda-memcheck
运行你的代码来证明这一点(另一件好事) 当你遇到CUDA代码时遇到问题),或者只是进行适当,严格的错误检查。
当cudaMemcpyAsync
操作指示失败时,操作未成功完成,因此不会复制数据,并且数据检查表明不匹配。希望现在这并不奇怪,因为预期的复制操作实际上并没有发生(也没有失败&#34;默默地&#34;)。
也许你很困惑,认为在任何类型的异步操作中捕获错误的方法是执行cudaDeviceSynchronize
,然后检查错误。
这对cudaMemcpyAsync
不正确。在调用cudaMemcpyAsync
操作时可以检测到的错误将由调用本身立即返回,并且由于后续CUDA调用(显然),将不会返回 错误是非粘性的。
故事的寓意:
cuda-memcheck
运行代码。这是一个功能齐全的示例,对您的代码进行了一些微不足道的修改,以使输出更加健全&#34;在失败的情况下,证明在失败案例中指出了错误:
$ cat t153.cu
#include <stdio.h>
#include <stdlib.h>
#define gpuErrchk(ans) { gpuAssert((ans), __FILE__, __LINE__); }
inline void gpuAssert(cudaError_t code, const char *file, int line, bool abort=true)
{
if (code != cudaSuccess)
{
fprintf(stderr,"GPUassert: %s %s %d\n", cudaGetErrorString(code), file, line);
if (abort) exit(code);
}
}
__global__ void checkDataDevice( int* current, int* next, int expected_current_val, int n )
{
int tid = threadIdx.x + blockIdx.x*blockDim.x;
for( int i = tid; i < n; i += blockDim.x*gridDim.x )
{
if( current[i] != expected_current_val )
printf( "Error on device: expected = %d, current[%d] = %d\n"
, expected_current_val
, i
, current[i] );
// Increment the data so the next copy is properly tested
next[i] = current[i] + 1;
}
}
void checkDataHost( int* current, int* next, int expected_current_val, int n )
{
for( int i = 0; i < n; i++ )
{
if( current[i] != expected_current_val ){
printf( "Error on host: expected = %d, current[%d] = %d\n"
, expected_current_val
, i
, current[i] );
exit(0);}
// Increment the data so the next copy is properly tested
next[i] = current[i] + 1;
}
}
int main( int argc, char** argv )
{
bool pagelocked = true;
// invoking the executable with any additional argument(s) will turn off page locked memory, i.e.,
// Run with pagelocked memory: ./a.out
// Run with ordinary malloc'd memory: ./a.out jkfdlsja
if( argc > 1 )
pagelocked = false;
int copybytes = 1e8; // Ok to use int instead of size_t for 1e8.
cudaStream_t* stream = (cudaStream_t*)malloc( sizeof(cudaStream_t) );
cudaStreamCreate( stream );
int* srcHost;
int* dstHost;
int* srcDevice;
int* dstDevice;
cudaMalloc( (void**)&srcDevice, copybytes );
cudaMalloc( (void**)&dstDevice, copybytes );
if( pagelocked )
{
printf( "Using page locked memory\n" );
cudaMallocHost( (void**)&srcHost, copybytes );
cudaMallocHost( (void**)&dstHost, copybytes );
}
else
{
printf( "Using non page locked memory\n" );
srcHost = (int*)malloc( copybytes );
dstHost = (int*)malloc( copybytes );
}
for( int i = 0; i < copybytes/sizeof(int); i++ )
srcHost[i] = 1;
cudaMemcpyKind kinds[4];
kinds[0] = cudaMemcpyHostToDevice;
kinds[1] = cudaMemcpyDeviceToHost;
kinds[2] = cudaMemcpyHostToHost;
kinds[3] = cudaMemcpyDeviceToDevice;
// Test cudaMemcpyAsync in both directions,
// iterating through all "cudaMemcpyKinds" to verify
// that they don't matter.
int expected_current_val = 1;
for( int kind = 0; kind<4; kind++ )
{
// Host to device copy
cudaMemcpyAsync( dstDevice
, srcHost
, copybytes
, kinds[kind]
, *stream );
gpuErrchk( cudaDeviceSynchronize() );
checkDataDevice<<<56*8,256>>>( dstDevice
, srcDevice
, expected_current_val
, copybytes/sizeof(int) );
expected_current_val++;
// Device to host copy
cudaMemcpyAsync( dstHost
, srcDevice
, copybytes
, kinds[kind]
, *stream );
gpuErrchk( cudaDeviceSynchronize() );
checkDataHost( dstHost
, srcHost
, expected_current_val
, copybytes/sizeof(int) );
expected_current_val++;
}
free( stream );
cudaFree( srcDevice );
cudaFree( dstDevice );
if( pagelocked )
{
cudaFreeHost( srcHost );
cudaFreeHost( dstHost );
}
else
{
free( srcHost );
free( dstHost );
}
return 0;
}
$ nvcc -arch=sm_61 -o t153 t153.cu
$ cuda-memcheck ./t153 a
========= CUDA-MEMCHECK
Using non page locked memory
========= Program hit cudaErrorInvalidValue (error 11) due to "invalid argument" on CUDA API call to cudaMemcpyAsync.
========= Saved host backtrace up to driver entry point at error
========= Host Frame:/usr/lib/x86_64-linux-gnu/libcuda.so.1 [0x2ef423]
========= Host Frame:./t153 [0x489a3]
========= Host Frame:./t153 [0x2e11]
========= Host Frame:/lib/x86_64-linux-gnu/libc.so.6 (__libc_start_main + 0xf5) [0x21ec5]
========= Host Frame:./t153 [0x2a49]
=========
Error on host: expected = 2, current[0] = 0
========= ERROR SUMMARY: 1 error
$