我在NVIDIA论坛上发布了这个帖子，我想我会多点帮忙。
我在尝试扩展代码以执行多个案例时遇到了麻烦。我一直在开发最常见的案例，现在是测试时间，我需要确保它适用于不同的情况。目前我的内核是在一个循环中执行的（有原因我们没有做一个内核调用来完成整个事情。）来计算矩阵行的值。最常见的情况是512列乘512行。我需要考虑尺寸为512 x 512,1024 x 512,512 x 1024和其他组合的基质，但最大的将是1024 x 1024矩阵。我一直在使用一个相当简单的内核调用：
launchKernel<<<1,512>>>(................)
此内核适用于常见的512x512和512 x 1024（分别为列，行）情况，但不适用于1024 x 512情况。这种情况需要1024个线程才能执行。在我天真的时候，我一直在尝试使用不同版本的简单内核调用来启动1024个线程。
launchKernel<<<2,512>>>(................) // 2 blocks with 512 threads each ???
launchKernel<<<1,1024>>>(................) // 1 block with 1024 threads ???
我相信我的问题与我对线程和块的理解不足有关
这是deviceQuery的输出，你可以看到我最多可以有1024个线程
C:\ProgramData\NVIDIA Corporation\NVIDIA GPU Computing SDK 4.1\C\bin\win64\Release\deviceQuery.exe Starting...
CUDA Device Query (Runtime API) version (CUDART static linking)
Found 2 CUDA Capable device(s)
Device 0: "Tesla C2050"
CUDA Driver Version / Runtime Version 4.2 / 4.1
CUDA Capability Major/Minor version number: 2.0
Total amount of global memory: 2688 MBytes (2818572288 bytes)
(14) Multiprocessors x (32) CUDA Cores/MP: 448 CUDA Cores
GPU Clock Speed: 1.15 GHz
Memory Clock rate: 1500.00 Mhz
Memory Bus Width: 384-bit
L2 Cache Size: 786432 bytes
Max Texture Dimension Size (x,y,z) 1D=(65536), 2D=(65536,65535), 3D=(2048,2048,2048)
Max Layered Texture Size (dim) x layers 1D=(16384) x 2048, 2D=(16384,16384) x 2048
Total amount of constant memory: 65536 bytes
Total amount of shared memory per block: 49152 bytes
Total number of registers available per block: 32768
Warp size: 32
Maximum number of threads per block: 1024
Maximum sizes of each dimension of a block: 1024 x 1024 x 64
Maximum sizes of each dimension of a grid: 65535 x 65535 x 65535
Maximum memory pitch: 2147483647 bytes
Texture alignment: 512 bytes
Concurrent copy and execution: Yes with 2 copy engine(s)
Run time limit on kernels: Yes
Integrated GPU sharing Host Memory: No
Support host page-locked memory mapping: Yes
Concurrent kernel execution: Yes
Alignment requirement for Surfaces: Yes
Device has ECC support enabled: Yes
Device is using TCC driver mode: No
Device supports Unified Addressing (UVA): No
Device PCI Bus ID / PCI location ID: 40 / 0
Compute Mode:
< Default (multiple host threads can use ::cudaSetDevice() with device simultaneously) >
Device 1: "Quadro 600"
CUDA Driver Version / Runtime Version 4.2 / 4.1
CUDA Capability Major/Minor version number: 2.1
Total amount of global memory: 1024 MBytes (1073741824 bytes)
( 2) Multiprocessors x (48) CUDA Cores/MP: 96 CUDA Cores
GPU Clock Speed: 1.28 GHz
Memory Clock rate: 800.00 Mhz
Memory Bus Width: 128-bit
L2 Cache Size: 131072 bytes
Max Texture Dimension Size (x,y,z) 1D=(65536), 2D=(65536,65535), 3D=(2048,2048,2048)
Max Layered Texture Size (dim) x layers 1D=(16384) x 2048, 2D=(16384,16384) x 2048
Total amount of constant memory: 65536 bytes
Total amount of shared memory per block: 49152 bytes
Total number of registers available per block: 32768
Warp size: 32
Maximum number of threads per block: 1024
Maximum sizes of each dimension of a block: 1024 x 1024 x 64
Maximum sizes of each dimension of a grid: 65535 x 65535 x 65535
Maximum memory pitch: 2147483647 bytes
Texture alignment: 512 bytes
Concurrent copy and execution: Yes with 1 copy engine(s)
Run time limit on kernels: Yes
Integrated GPU sharing Host Memory: No
Support host page-locked memory mapping: Yes
Concurrent kernel execution: Yes
Alignment requirement for Surfaces: Yes
Device has ECC support enabled: No
Device is using TCC driver mode: No
Device supports Unified Addressing (UVA): No
Device PCI Bus ID / PCI location ID: 15 / 0
Compute Mode:
< Default (multiple host threads can use ::cudaSetDevice() with device simultaneously) >
deviceQuery, CUDA Driver = CUDART, CUDA Driver Version = 4.2, CUDA Runtime Version = 4.1, NumDevs = 2, Device = Tesla C2050, Device = Quadro 600
我只使用Tesla C2050设备 这是我的内核的剥离版本，因此您可以了解它正在做什么。
#define twoPi 6.283185307179586
#define speed_of_light 3.0E8
#define MaxSize 999
__global__ void calcRx4CPP4
(
const float *array1,
const double *array2,
const float scalar1,
const float scalar2,
const float scalar3,
const float scalar4,
const float scalar5,
const float scalar6,
const int scalar7,
const int scalar8,
float *outputArray1,
float *outputArray2)
{
float scalar9;
int idx;
double scalar10;
double scalar11;
float sumReal, sumImag;
float real, imag;
float coeff1, coeff2, coeff3, coeff4;
sumReal = 0.0;
sumImag = 0.0;
// kk loop 1 .. 512 (scalar7)
idx = (blockIdx.x * blockDim.x) + threadIdx.x;
/* Declare the shared memory parameters */
__shared__ float SharedArray1[MaxSize];
__shared__ double SharedArray2[MaxSize];
/* populate the arrays on shared memory */
SharedArray1[idx] = array1[idx]; // first 512 elements
SharedArray2[idx] = array2[idx];
if (idx+blockDim.x < MaxSize){
SharedArray1[idx+blockDim.x] = array1[idx+blockDim.x];
SharedArray2[idx+blockDim.x] = array2[idx+blockDim.x];
}
__syncthreads();
// input scalars used here.
scalar10 = ...;
scalar11 = ...;
for (int kk = 0; kk < scalar8; kk++)
{
/* some calculations */
// SharedArray1, SharedArray2 and scalar9 used here
sumReal = ...;
sumImag = ...;
}
/* calculation of the exponential of a complex number */
real = ...;
imag = ...;
coeff1 = (sumReal * real);
coeff2 = (sumReal * imag);
coeff3 = (sumImag * real);
coeff4 = (sumImag * imag);
outputArray1[idx] = (coeff1 - coeff4);
outputArray2[idx] = (coeff2 + coeff3);
}
因为我的每个块的最大线程数是1024，我以为我能够继续使用简单的内核启动，我错了吗？
如何使用1024个线程成功启动每个内核？
答案 0 :(得分：5)
您不希望改变每个块的线程数。您应该使用CUDA占用计算器为您的内核获得每个块的最佳线程数。获得该数字后，只需启动获取所需线程总数所需的块数。如果给定案例所需的线程数并不总是每个块的多个线程，则可以在内核顶部添加代码以中止不需要的线程。 （if () return;）。然后，您可以使用额外的参数向内核传递矩阵的维度，或者使用x和y网格维度，具体取决于内核中需要哪些信息（我还没有研究过）。
我的猜测是你遇到1024个线程的问题是，即使你的GPU支持一个块中的那么多线程，还有另一个限制因素，你可以在每个块中拥有的线程数基于内核中的资源使用情况。限制因素可以是共享存储器或寄存器使用。占用计算器将告诉您哪些信息仅在您想要优化内核时才重要。
答案 1 :(得分：3)
如果您使用一个包含1024个线程的块，则会遇到问题，因为MaxSize只有999，导致数据错误。
让我们为最后一个帖子＃1023模拟它
__shared__ float SharedArray1[999];
__shared__ double SharedArray2[999];
/* populate the arrays on shared memory */
SharedArray1[1023] = array1[1023];
SharedArray2[1023] = array2[1023];
if (2047 < MaxSize)
{
SharedArray1[2047] = array1[2047];
SharedArray2[2047] = array2[2047];
}
__syncthreads();
如果您现在在计算中使用所有这些元素，那么这应该不起作用。 （您的计算代码未显示为假设）