cuda kernel for循环太长？

Question

经过长时间的手动调优和半成品的调试，我终于找到了导致内核崩溃的原因。是不是你不能在内核中创建一个太长的for循环？下面是最小的完整和可验证的代码:错误是:未指明的启动失败

#include "cuda_runtime.h"
#include "device_launch_parameters.h"
#include <iostream>
#include <stdio.h>
using namespace std;
typedef unsigned char uchar;

struct NodePointer {
    int id;
    uchar dist;
    int idintree;
    NodePointer() :id(0), dist(0), idintree(-1){}
};
struct TreeNode {
    NodePointer father;
    NodePointer children[4];
    int id;
    int childrenNum;
    int idintree;
    TreeNode() :id(0), childrenNum(0), idintree(-1){}
};

__global__ void
kernel4(int NumTN, TreeNode* tempthistree, int size)
{
    int index = blockIdx.x*blockDim.x + threadIdx.x;
    if (index < NumTN)
    {
        for (int i = 0; i < size; i++)
        {
            TreeNode node1 = tempthistree[i];
            printf("              node %d in tree %d, its id in tree is %d, its child num is %d\n", i, index, node1.idintree, node1.childrenNum);

        }
    }
}


int main()
{
    int n1 = 33417;
    TreeNode * testtree;
    cudaMallocManaged(&testtree, n1*sizeof(TreeNode));
    for (int i = 0; i < n1; i++)
    {
        TreeNode c;
        c.idintree = i;
        c.id = i;
        c.father.id = i - 1;
        c.father.dist = 1;
        c.childrenNum = i % 4;
        int aaa = i % 4;
        for (int j = 0; j < 4; j++)
        {
            c.children[j].dist = j;
            c.children[j].id = 1;
            c.children[j].idintree = 10;
        }
        testtree[i] = c;
    }


    kernel4 << <1, 1 >> >(4000, testtree, n1);
    cudaDeviceSynchronize();
    cudaError_t err = cudaGetLastError();
    if (err != cudaSuccess)
    {
        printf("Kernel3 error :%s\n", cudaGetErrorString(err));
        system("pause");
    }
}