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我有一个内核，它计算总和。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*。然而，在分析内核时，使用了 34 个寄存器。我需要减少到 30 个寄存器以允许执行 1024 个线程。

任何人都可以看到有什么问题吗？

__global__ void sum_kernel(float* values, float bk_size, int start_idx, int end_idx, int resolution, float* avgs){

    // Allocate shared memory (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;
}
  • 假设每个指针有 2 个寄存器，每个 unsigned int 有 1 个寄存器，参数存储在常量内存中。
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1 回答 1

6

您无法根据声明变量的数量来估计使用的寄存器数量。编译器可以使用寄存器进行地址计算或存储您未明确声明的临时变量等。

例如，我已经反汇编了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;
    }
}

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，没有明确声明的变量。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互换的，用于存储常量、内存地址和全局内存值。

于 2013-10-20T18:20:52.677 回答