2013-10-20 32 views
2

我有一个内核,它计算总和。如果我通过内核计算所声明的变量数,我会假设每个内核共有5个寄存器*。但是,在分析内核时,会使用34个寄存器。我需要下到30个寄存器来允许执行1024个线程。CUDA - 内核使用比预期更多的寄存器?

任何人都可以看到什么是错的?

__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; 
} 
  • 假设每个指针2个寄存器,每个无符号整型1个寄存器,存储在常数存储参数。

回答

6

根据声明变量的数量,您无法估计已用寄存器的数量。该编译器可以使用的寄存器进行地址的计算或存储你没有明确声明等

例如临时变量,我拆开你的内核函数的第一部分,即

__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; 
    } 
} 

具有以下导致

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代码段,没有显式地声明的变量。从反汇编代码可以看出,编译器使用了3寄存器,即R0,R1R2。这些寄存器在功能上可互换,并用于存储常量,内存地址和全局内存值。

+0

小修正:谓词不存储在R寄存器中,但是谓词寄存器(在这种情况下为P0)。 – njuffa

+0

@njuffa非常感谢。我立即修复了答案的最后一句。 – JackOLantern