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linux - CUDA - 关于 “branch” 和 “divergent branch” 的 Visual Profiler 结果的混淆 (2)

转载 作者：塔克拉玛干 更新时间：2023-11-03 01:01:48 25 4
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我使用 NVIDIA Visual Profiler 来分析我的代码。测试内核是:

//////////////////////////////////////////////////////////////// Group 1
static __global__ void gpu_test_divergency_0(float *a, float *b)
{
    int tid = threadIdx.x + blockIdx.x * blockDim.x;
    if (tid < 0)
    {
         a[tid] = tid;
    }
    else
    {
         b[tid] = tid;
    }
}
static __global__ void gpu_test_divergency_1(float *a, float *b)
{
    int tid = threadIdx.x + blockIdx.x * blockDim.x;
    if (tid == 0)
    {
         a[tid] = tid;
    }
    else
    {
         b[tid] = tid;
    }
}
static __global__ void gpu_test_divergency_2(float *a, float *b)
{
    int tid = threadIdx.x + blockIdx.x * blockDim.x;
    if (tid >= 0)
    {
         a[tid] = tid;
    }
    else
    {
         b[tid] = tid;
    }
}
static __global__ void gpu_test_divergency_3(float *a, float *b)
{
    int tid = threadIdx.x + blockIdx.x * blockDim.x;
    if (tid > 0)
    {
         a[tid] = tid;
    }
    else
    {
         b[tid] = tid;
    }
}
//////////////////////////////////////////////////////////////// Group 2
static __global__ void gpu_test_divergency_4(float *a, float *b)
{
    int tid = threadIdx.x + blockIdx.x * blockDim.x;
    if (tid < 0)
    {
         a[tid] = tid + 1;
    }
    else
    {
         b[tid] = tid + 2;
    }
}
static __global__ void gpu_test_divergency_5(float *a, float *b)
{
    int tid = threadIdx.x + blockIdx.x * blockDim.x;
    if (tid == 0)
    {
         a[tid] = tid + 1;
    }
    else
    {
         b[tid] = tid + 2;
    }
}
static __global__ void gpu_test_divergency_6(float *a, float *b)
{
    int tid = threadIdx.x + blockIdx.x * blockDim.x;
    if (tid >= 0)
    {
         a[tid] = tid + 1;
    }
    else
    {
         b[tid] = tid + 2;
    }
}
static __global__ void gpu_test_divergency_7(float *a, float *b)
{
    int tid = threadIdx.x + blockIdx.x * blockDim.x;
    if (tid > 0)
    {
         a[tid] = tid + 1;
    }
    else
    {
         b[tid] = tid + 2;
    }
}
//////////////////////////////////////////////////////////////// Group 3
static __global__ void gpu_test_divergency_8(float *a, float *b)
{
    int tid = threadIdx.x + blockIdx.x * blockDim.x;
    if (tid < 0)
    {
         a[tid] = tid + 1.0;
    }
    else
    {
         b[tid] = tid + 2.0;
    }
}
static __global__ void gpu_test_divergency_9(float *a, float *b)
{
    int tid = threadIdx.x + blockIdx.x * blockDim.x;
    if (tid == 0)
    {
         a[tid] = tid + 1.0;
    }
    else
    {
         b[tid] = tid + 2.0;
    }
}
static __global__ void gpu_test_divergency_10(float *a, float *b)
{
    int tid = threadIdx.x + blockIdx.x * blockDim.x;
    if (tid >= 0)
    {
         a[tid] = tid + 1.0;
    }
    else
    {
         b[tid] = tid + 2.0;
    }
}
static __global__ void gpu_test_divergency_11(float *a, float *b)
{
    int tid = threadIdx.x + blockIdx.x * blockDim.x;
    if (tid > 0)
    {
         a[tid] = tid + 1.0;
    }
    else
    {
         b[tid] = tid + 2.0;
    }
}

当我使用 <<< 1, 32 >>> 启动测试内核时，我从分析器中得到了这样的结果:

gpu_test_divergency_0 :  Branch Efficiency = 100% branch = 1 divergent branch = 0
gpu_test_divergency_1 :  Branch Efficiency = 100% branch = 1 divergent branch = 0
gpu_test_divergency_2 :  Branch Efficiency = 100% branch = 1 divergent branch = 0
gpu_test_divergency_3 :  Branch Efficiency = 100% branch = 1 divergent branch = 0

gpu_test_divergency_4 :  Branch Efficiency = 100% branch = 3 divergent branch = 0
gpu_test_divergency_5 :  Branch Efficiency = 100% branch = 3 divergent branch = 0
gpu_test_divergency_6 :  Branch Efficiency = 100% branch = 2 divergent branch = 0
gpu_test_divergency_7 :  Branch Efficiency = 100% branch = 3 divergent branch = 0

gpu_test_divergency_8 :  Branch Efficiency = 100% branch = 3 divergent branch = 0
gpu_test_divergency_9 :  Branch Efficiency = 75%  branch = 4 divergent branch = 1
gpu_test_divergency_10 : Branch Efficiency = 100% branch = 2 divergent branch = 0
gpu_test_divergency_11 : Branch Efficiency = 75%  branch = 4 divergent branch = 1

当我使用 <<< 1, 64 >>> 启动测试内核时，我从分析器中得到了这样的结果:

gpu_test_divergency_0 :  Branch Efficiency = 100% branch = 2 divergent branch = 0
gpu_test_divergency_1 :  Branch Efficiency = 100% branch = 2 divergent branch = 0
gpu_test_divergency_2 :  Branch Efficiency = 100% branch = 2 divergent branch = 0
gpu_test_divergency_3 :  Branch Efficiency = 100% branch = 2 divergent branch = 0

gpu_test_divergency_4 :  Branch Efficiency = 100% branch = 6 divergent branch = 0
gpu_test_divergency_5 :  Branch Efficiency = 100% branch = 6 divergent branch = 0
gpu_test_divergency_6 :  Branch Efficiency = 100% branch = 4 divergent branch = 0
gpu_test_divergency_7 :  Branch Efficiency = 100% branch = 5 divergent branch = 0

gpu_test_divergency_8 :  Branch Efficiency = 100%  branch = 6 divergent branch = 0
gpu_test_divergency_9 :  Branch Efficiency = 85.7% branch = 7 divergent branch = 1
gpu_test_divergency_10 : Branch Efficiency = 100%  branch = 4 divergent branch = 0
gpu_test_divergency_11 : Branch Efficiency = 83.3% branch = 6 divergent branch = 1

我在 Linux 上使用 CUDA Capability 2.0 和 NVIDIA Visual Profiler v4.2 的“GeForce GTX 570”。根据文件:

“分支”-“执行内核的线程采用的分支数。如果 warp 中至少有一个线程采用分支，则此计数器将递增 1。”

“发散分支” - “一个 warp 中发散分支的数量。如果 warp 中至少有一个线程通过数据依赖项发散(即遵循不同的执行路径)，则此计数器将递增 1条件分支。”

但我对结果真的很困惑。为什么每个测试组的“分支”数量不同？为什么只有第三个测试组似乎有正确的“发散分支”？

@JackOLantern:我在 Release模式下编译。我按照你的方法拆了它。 “gpu_test_divergency_4”的结果和你的完全一样，但是“gpu_test_divergency_0”的结果不同:

    Function : _Z21gpu_test_divergency_0PfS_
/*0000*/     /*0x00005de428004404*/     MOV R1, c [0x1] [0x100];
/*0008*/     /*0x94001c042c000000*/     S2R R0, SR_CTAid_X;
/*0010*/     /*0x84009c042c000000*/     S2R R2, SR_Tid_X;
/*0018*/     /*0x20009ca320044000*/     IMAD R2, R0, c [0x0] [0x8], R2;
/*0020*/     /*0xfc21dc23188e0000*/     ISETP.LT.AND P0, pt, R2, RZ, pt;
/*0028*/     /*0x0920de0418000000*/     I2F.F32.S32 R3, R2;
/*0030*/     /*0x9020204340004000*/     @!P0 ISCADD R0, R2, c [0x0] [0x24], 0x2;
/*0038*/     /*0x8020804340004000*/     @P0 ISCADD R2, R2, c [0x0] [0x20], 0x2;
/*0040*/     /*0x0000e08590000000*/     @!P0 ST [R0], R3;
/*0048*/     /*0x0020c08590000000*/     @P0 ST [R2], R3;
/*0050*/     /*0x00001de780000000*/     EXIT;

我想，正如您所说，转换指令(在本例中为 I2F)不会添加额外的分支。

但我看不到这些反汇编代码与 Profiler 结果之间的关系。我从另一篇文章(https://devtalk.nvidia.com/default/topic/463316/branch-divergent-branches/)了解到，发散分支是根据 SM 上的实际线程(warp)运行情况计算的。所以我估计我们不能仅仅根据这些反汇编代码来推导出每次实际运行的分支发散。我对吗？

最佳答案

跟进 - 使用 VOTE Intrinsics 检查线程分歧

我认为检查 warps 内线程分歧的最佳方法是使用投票内在函数，尤其是 __ballot 和 __popc 内在函数。关于 __ballot 和 __popc 的很好的解释可以在 Shane Cook，CUDA Programming，Morgan Kaufmann 的书中找到。

__ballot的原型(prototype)如下

unsigned int __ballot(int predicate);

如果谓词非零，__ballot 返回一个设置了第 N 位的值，其中 N 是 threadIdx.x.

另一方面，__popc 返回使用 32 位参数设置的位数。

因此，通过联合使用 __ballot、__popc 和 atomicAdd，可以检查 warp 是否发散。

为此，我设置了如下代码

#include <cuda.h>
#include <stdio.h>
#include <iostream>

#include <cuda.h>
#include <cuda_runtime.h>

__device__ unsigned int __ballot_non_atom(int predicate)
{
    if (predicate != 0) return (1 << (threadIdx.x % 32));
    else return 0;
}

__global__ void gpu_test_divergency_0(unsigned int* d_ballot, int Num_Warps_per_Block)
{
    int tid = threadIdx.x + blockIdx.x * blockDim.x;

    const unsigned int warp_num = threadIdx.x >> 5;

    atomicAdd(&d_ballot[warp_num+blockIdx.x*Num_Warps_per_Block],__popc(__ballot_non_atom(tid > 2)));
    //  atomicAdd(&d_ballot[warp_num+blockIdx.x*Num_Warps_per_Block],__popc(__ballot(tid > 2)));

}

#include <conio.h>

int main(int argc, char *argv[])
{
    unsigned int Num_Threads_per_Block      = 64;
    unsigned int Num_Blocks_per_Grid        = 1;
    unsigned int Num_Warps_per_Block        = Num_Threads_per_Block/32;
    unsigned int Num_Warps_per_Grid         = (Num_Threads_per_Block*Num_Blocks_per_Grid)/32;

    unsigned int* h_ballot = (unsigned int*)malloc(Num_Warps_per_Grid*sizeof(unsigned int));
    unsigned int* d_ballot; cudaMalloc((void**)&d_ballot, Num_Warps_per_Grid*sizeof(unsigned int));

    for (int i=0; i<Num_Warps_per_Grid; i++) h_ballot[i] = 0;

    cudaMemcpy(d_ballot, h_ballot, Num_Warps_per_Grid*sizeof(unsigned int), cudaMemcpyHostToDevice);

    gpu_test_divergency_0<<<Num_Blocks_per_Grid,Num_Threads_per_Block>>>(d_ballot,Num_Warps_per_Block);

    cudaMemcpy(h_ballot, d_ballot, Num_Warps_per_Grid*sizeof(unsigned int), cudaMemcpyDeviceToHost);

    for (int i=0; i<Num_Warps_per_Grid; i++) { 
        if ((h_ballot[i] == 0)||(h_ballot[i] == 32)) std::cout << "Warp " << i << " IS NOT divergent- Predicate true for " << h_ballot[i] << " threads\n";
            else std::cout << "Warp " << i << " IS divergent - Predicate true for " << h_ballot[i] << " threads\n";
    }

    getch();
    return EXIT_SUCCESS;
}

请注意，我现在正在计算能力为 1.2 的卡上运行代码，因此在上面的示例中，我使用的是 __ballot_non_atom，它是 的非固有等效项__ballot，因为 __ballot 仅适用于 >= 2.0 的计算能力。换句话说，如果你有一张计算能力>=2.0的卡，请取消注释内核函数中使用__ballot的指令。

使用上面的代码，您可以通过简单地更改内核函数中的相关谓词来使用上面的所有内核函数。

上一个答案

我在 release 模式下为计算能力 2.0 编译了你的代码，我使用 -keep 来保留中间文件和 cuobjdump 实用程序来生成两个内核的反汇编，即:

static __global__ void gpu_test_divergency_0(float *a, float *b)
{
    int tid = threadIdx.x + blockIdx.x * blockDim.x;
    if (tid < 0) a[tid] = tid;
    else b[tid] = tid;
}

和

static __global__ void gpu_test_divergency_4(float *a, float *b)
{
    int tid = threadIdx.x + blockIdx.x * blockDim.x;
    if (tid < 0) a[tid] = tid + 1;
    else b[tid] = tid + 2;
}

结果如下

gpu_test_divergency_0

/*0000*/        MOV R1, c[0x1][0x100];                 /* 0x2800440400005de4 */
/*0008*/        S2R R0, SR_CTAID.X;                    /* 0x2c00000094001c04 */
/*0010*/        S2R R2, SR_TID.X;                      /* 0x2c00000084009c04 */
/*0018*/        IMAD R2, R0, c[0x0][0x8], R2;          /* 0x2004400020009ca3 */
/*0020*/        ISETP.LT.AND P0, PT, R2, RZ, PT;       /* 0x188e0000fc21dc23 */
/*0028*/        I2F.F32.S32 R0, R2;                    /* 0x1800000009201e04 */
/*0030*/   @!P0 ISCADD R3, R2, c[0x0][0x24], 0x2;      /* 0x400040009020e043 */
/*0038*/    @P0 ISCADD R2, R2, c[0x0][0x20], 0x2;      /* 0x4000400080208043 */
/*0040*/   @!P0 ST [R3], R0;                           /* 0x9000000000302085 */
/*0048*/    @P0 ST [R2], R0;                           /* 0x9000000000200085 */
/*0050*/        EXIT ;                                 /* 0x8000000000001de7 */

和

gpu_test_divergency_4

/*0000*/        MOV R1, c[0x1][0x100];                 /* 0x2800440400005de4 */
/*0008*/        S2R R0, SR_CTAID.X;                    /* 0x2c00000094001c04 */   R0 = BlockIdx.x
/*0010*/        S2R R2, SR_TID.X;                      /* 0x2c00000084009c04 */   R2 = ThreadIdx.x
/*0018*/        IMAD R0, R0, c[0x0][0x8], R2;          /* 0x2004400020001ca3 */   R0 = R0 * c + R2
/*0020*/        ISETP.LT.AND P0, PT, R0, RZ, PT;       /* 0x188e0000fc01dc23 */   If statement
/*0028*/    @P0 BRA.U 0x58;                            /* 0x40000000a00081e7 */   Branch 1 - Jump to 0x58
/*0030*/   @!P0 IADD R2, R0, 0x2;                      /* 0x4800c0000800a003 */   Branch 2 - R2 = R0 + 2
/*0038*/   @!P0 ISCADD R0, R0, c[0x0][0x24], 0x2;      /* 0x4000400090002043 */   Branch 2 - Calculate gmem address
/*0040*/   @!P0 I2F.F32.S32 R2, R2;                    /* 0x180000000920a204 */   Branch 2 - R2 = R2 after int to float cast
/*0048*/   @!P0 ST [R0], R2;                           /* 0x900000000000a085 */   Branch 2 - gmem store
/*0050*/   @!P0 BRA.U 0x78;                            /* 0x400000008000a1e7 */   Branch 2 - Jump to 0x78 (exit)
/*0058*/    @P0 IADD R2, R0, 0x1;                      /* 0x4800c00004008003 */   Branch 1 - R2 = R0 + 1
/*0060*/    @P0 ISCADD R0, R0, c[0x0][0x20], 0x2;      /* 0x4000400080000043 */   Branch 1 - Calculate gmem address
/*0068*/    @P0 I2F.F32.S32 R2, R2;                    /* 0x1800000009208204 */   Branch 1 - R2 = R2 after int to float cast
/*0070*/    @P0 ST [R0], R2;                           /* 0x9000000000008085 */   Branch 1 - gmem store
/*0078*/        EXIT ;                                 /* 0x8000000000001de7 */

从上面的反汇编中，我希望你的分支发散测试的结果是相同的。

您是在 Debug模式还是 Release模式下编译？

关于linux - CUDA - 关于 “branch” 和 “divergent branch” 的 Visual Profiler 结果的混淆 (2)，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/19334589/

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