C++ CUDA-内核使用的寄存器比预期的多?
我有一个计算和的内核。如果我通过内核计算声明的变量数量,我会假设每个内核总共有5个寄存器*。但是,在分析内核时,使用了34个寄存器。我需要使用30个寄存器,以允许执行1024个线程 有人能看出哪里出了问题吗C++ CUDA-内核使用的寄存器比预期的多?,c++,cuda,C++,Cuda,我有一个计算和的内核。如果我通过内核计算声明的变量数量,我会假设每个内核总共有5个寄存器*。但是,在分析内核时,使用了34个寄存器。我需要使用30个寄存器,以允许执行1024个线程 有人能看出哪里出了问题吗 __global__ void sum_kernel(float* values, float bk_size, int start_idx, int end_idx, int resolution, float* avgs){ // Allocate shared memory
__global__ void sum_kernel(float* values, float bk_size, int start_idx, int end_idx, int resolution, float* avgs){
// Allocate shared memory (assuming a maximum of 1024 threads).
__shared__ float sums[1024];
// Boundary check.
if(blockIdx.x == 0){
avgs[blockIdx.x] = values[start_idx];
return;
}
else if(blockIdx.x == resolution-1) {
avgs[blockIdx.x] = values[start_idx+(end_idx-start_idx)-1];
return;
}
else if(blockIdx.x > resolution -2){
return;
}
// Iteration index calculation.
unsigned int idx_prev = floor((blockIdx.x + 0) * bk_size) + 1;
unsigned int from = idx_prev + threadIdx.x*(bk_size / blockDim.x);
unsigned int to = from + (bk_size / blockDim.x);
to = (to < (end_idx-start_idx))? to : (end_idx-start_idx);
// Partial average calculation using shared memory.
sums[threadIdx.x] = 0;
for (from; from < to; from++)
{
sums[threadIdx.x] += values[from+start_idx];
}
__syncthreads();
// Addition of partial sums.
if(threadIdx.x != 0) return;
from = 1;
for(from; from < 1024; from++)
{
sum += sums[from];
}
avgs[blockIdx.x] = sum;
}
\uuuuu全局\uuuuu无效和\uu内核(浮点*值、浮点bk大小、整数开始\uIDX、整数结束\uIDX、整数分辨率、浮点*平均值){
//分配共享内存(假设最多1024个线程)。
__共享浮点数和[1024];
//边界检查。
if(blockIdx.x==0){
avgs[blockIdx.x]=值[start_idx];
返回;
}
else if(blockIdx.x==分辨率-1){
avgs[blockIdx.x]=值[start_idx+(end_idx-start_idx)-1];
返回;
}
else if(blockIdx.x>分辨率-2){
返回;
}
//迭代指数计算。
无符号整数idx_prev=地板((块idx.x+0)*bk_大小)+1;
unsigned int from=idx_prev+threadIdx.x*(bk_size/blockDim.x);
无符号int-to=from+(bk_size/blockDim.x);
to=(to<(end_idx-start_idx))?to:(end_idx-start_idx);
//使用共享内存进行部分平均计算。
和[threadIdx.x]=0;
for(from;from
- 假设每个指针有2个寄存器,每个无符号int有1个寄存器,参数存储在常量内存中
__global__ void sum_kernel(float* values, float bk_size, int start_idx, int end_idx, int resolution, float* avgs){
// Boundary check.
if(blockIdx.x == 0){
avgs[blockIdx.x] = values[start_idx];
return;
}
else if(blockIdx.x == resolution-1) {
avgs[blockIdx.x] = values[start_idx+(end_idx-start_idx)-1];
return;
}
else if(blockIdx.x > resolution -2){
return;
}
}
有以下结果
code for sm_20
Function : _Z10sum_kernelPffiiiS_
.headerflags @"EF_CUDA_SM20 EF_CUDA_PTX_SM(EF_CUDA_SM20)"
/*0000*/ MOV R1, c[0x1][0x100]; /* 0x2800440400005de4 */ R1 = [0x1][0x100]
/*0008*/ S2R R2, SR_CTAID.X; /* 0x2c00000094009c04 */ R2 = BlockIdx.x
/*0010*/ MOV R0, c[0x0][0x34]; /* 0x28004000d0001de4 */ R0 = [0x0][0x34]
/*0018*/ ISETP.EQ.AND P0, PT, R2, RZ, PT; /* 0x190e0000fc21dc23 */ if (R2 == 0)
/*0020*/ @P0 BRA 0x78; /* 0x40000001400001e7 */
/*0028*/ MOV R0, c[0x0][0x30]; /* 0x28004000c0001de4 */
/*0030*/ IADD R0, R0, -0x1; /* 0x4800fffffc001c03 */
/*0038*/ ISETP.NE.AND P0, PT, R2, R0, PT; /* 0x1a8e00000021dc23 */
/*0040*/ @P0 EXIT ; /* 0x80000000000001e7 */
/*0048*/ MOV R0, c[0x0][0x2c]; /* 0x28004000b0001de4 */
/*0050*/ ISCADD R2, R2, c[0x0][0x34], 0x2; /* 0x40004000d0209c43 */
/*0058*/ ISCADD R0, R0, c[0x0][0x20], 0x2; /* 0x4000400080001c43 */
/*0060*/ LDU R0, [R0+-0x4]; /* 0x8bfffffff0001c85 */
/*0068*/ ST [R2], R0; /* 0x9000000000201c85 */
/*0070*/ BRA 0x98; /* 0x4000000080001de7 */
/*0078*/ MOV R2, c[0x0][0x28]; /* 0x28004000a0009de4 */
/*0080*/ ISCADD R2, R2, c[0x0][0x20], 0x2; /* 0x4000400080209c43 */
/*0088*/ LDU R2, [R2]; /* 0x8800000000209c85 */ R2 used for addressing and storing gmem data
/*0090*/ ST [R0], R2; /* 0x9000000000009c85 */ R0 used for addressing
/*0098*/ EXIT ; /* 0x8000000000001de7 */
在上面的CUDA代码片段中,没有显式声明的变量。从反汇编代码中可以看出,编译器使用了
3
寄存器,即R0
、R1
和R2
。这些寄存器在功能上是可交换的,用于存储常量、内存地址和全局内存值。小更正:谓词不是存储在R寄存器中,而是存储在谓词寄存器中(在本例中为P0)。@njuffa非常感谢。我已立即修正了答案的最后一句话。