linux - CUDA - 关于“分支”和“发散分支”的 Visual Profiler 结果的混淆 (2)

Question

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，我从分析器中得到如下结果：

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，我从分析器中得到如下结果：

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 上使用具有 2.0 的 CUDA 功能和 NVIDIA Visual Profiler v4.2 的“GeForce GTX 570”。根据文件：

“branch” - “执行内核的线程所采用的分支数。如果 warp 中至少有一个线程采用该分支，则此计数器将加一。”

“发散分支”-“经线中发散分支的数量。如果经线中的至少一个步幅通过数据相关的条件分支发散（即遵循不同的执行路径），则此计数器将加一。”

但我真的对结果感到困惑。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“分歧分支”？

@JackOLantern：我在发布模式下编译。我按照你的方法拆开了。“gpu_test_divergency_4”的结果与你的完全相同，但“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，就像你说的那样，转换指令（在这种情况下为 I2F）不会添加额外的分支。

但是我看不到这些反汇编代码和 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/）中了解到，发散分支是根据 SM 上的实际线程（warp）运行情况计算的。所以我想我们不能仅仅根据这些反汇编代码推断出每次实际运行的分支分歧。我对吗？

score 1 · Accepted Answer

跟进 - 使用投票内在函数检查线程分歧

我认为检查经纱中线程分歧的最佳方法是使用投票内在函数，尤其是 the__ballot和__popc内在函数。__ballotShane Cook、CUDA Programming、Morgan Kaufmann的__popc书中提供了一个很好的解释。

的原型__ballot如下

unsigned int __ballot(int predicate);

如果 predicate 不为零，则__ballot返回一个设置了N第 th 位的值，其中Nis threadIdx.x。

另一方面，__popc返回使用32-bit 参数设置的位数。

因此，通过联合使用__ballot,__popc和atomicAdd，可以检查扭曲是否发散。

为此，我设置了以下代码

#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 1.2 卡上运行代码，因此在上面的示例中，我使用__ballot_non_atom的是非固有等效于__ballot，因为__ballot仅适用于计算能力 >= 2.0。换句话说，如果你有一张计算能力>= 2.0的卡，请取消注释__ballot内核函数中使用的指令。

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。

以前的答案

2.0我在发布模式下编译了您的代码以获得计算能力，并且我曾经-keep保留中间文件和cuobjdump实用程序来生成您的两个内核的反汇编，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 */

从上述反汇编中，我希望您的分支差异测试的结果是相同的。

你是在调试还是发布模式下编译？

linux - CUDA - 关于“分支”和“发散分支”的 Visual Profiler 结果的混淆 (2)

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