c++ CUDA:结构的共享数据成员和该结构的引用成员具有不同的地址、值

zbq4xfa0  于 2023-05-20  发布在  其他
关注(0)|答案(1)|浏览(168)

好了,问题来了:
使用CUDA 1.1计算GPU,我试图为每个线程维护一组索引(可能数量不同,这里固定为4个),我将其作为结构变量的成员保留。
我的问题是,当访问成员数组时,获取对结构体的引用会导致错误的结果:我用0初始化成员数组值,当我使用原始结构变量读取数组值时,我得到正确的值(0),但当我使用对结构变量的引用读取它时,我得到垃圾(-8193)。即使使用class而不是struct,也会发生这种情况。
为什么tmp小于/不等于0?
C++不是我的主要语言,所以这可能是一个概念问题,也可能是在CUDA中工作的一个怪癖。

struct DataIdx {
    int numFeats;
    int* featIdx;
};
extern __shared__ int sharedData[];

__global__  void myFn(){
    int tidx = blockIdx.x * blockDim.x + threadIdx.x;
    
    DataIdx myIdx;  //instantiate the struct var in the context of the current thread
    myIdx.numFeats = 4;
    size_t idxArraySize = sizeof(int)*4;
    //get a reference to my array for this thread. Parallel Nsight debugger shows myIdx.featIdx address = 0x0000000000000000e0
    myIdx.featIdx = (int*)(&sharedData[tidx*idxArraySize]);  
    
    myIdx.featIdx[0] = 0x0;  //set first value to 0 
    int tmp = myIdx.featIdx[0];  // tmp is correctly eq to 0 in Nsight debugger -- As Expected!!
    tmp = 2*tmp;    antIdx.featIdx[0] = tmp; //ensure compiler doesn't elide out tmp
    
    DataIdx *tmpIdx = &myIdx;  //create a reference to my struct var
    tmp = tmpIdx.featIdx[0];   // expected 0, but tmp = -8193 in debugger !! why?  debugger shows address of tmpIdx.featIdx = __devicea__ address=8
    tmpIdx.featIdx[0] = 0x0;
    tmp = tmpIdx.featIdx[0]; // tmp = -1; cant even read what we just set
    
    //forcing the same reference as myIdx.featIdx, still gives a problem! debugger shows address of tmpIdx.featIdx = __devicea__ address=8
    tmpIdx->featIdx =  (int*)(&sharedData[tidx*idxArraySize]); 
    tmp = tmpIdx.featIdx[0]; //tmp = -8193!! why != 0?

    DataIdx tmpIdxAlias = myIdx;
    tmp = tmpIdx.featIdx[0]; //aliasing the original var gives correct results, tmp=0
    
    
     myIdx.featIdx[0] = 0x0;
     mySubfn(&myIdx); //this is a problem because it happens when passing the struct by reference to subfns
     mySubfn2(myIdx);
}
__device__ mySubfn(struct DataIdx *myIdx){
  int tmp = myIdx->featIdx[0]; //tmp == -8193!! should be 0
}
__device__ mySubfn2(struct DataIdx &myIdx){
  int tmp = myIdx.featIdx[0]; //tmp == -8193!! should be 0
}
dtcbnfnu

dtcbnfnu1#

我不得不修改你的代码来编译。排队

tmpIdx->featIdx[0] = 0x0

编译器无法理解指针指向共享内存。它不是存储到共享内存(R2G),而是存储到越界的全局地址0x10

DataIdx *tmpIdx = &myIdx;
0x000024c8  MOV32 R2, R31;  
0x000024cc  MOV32 R2, R2;  
    tmp = tmpIdx->featIdx[0];
    tmpIdx->featIdx[0] = 0x0;
0x000024d0  MOV32 R3, R31;  
0x000024d4  MOV32 R2, R2;  
0x000024d8  IADD32I R4, R2, 0x4;  
0x000024e0  R2A A1, R4;  
0x000024e8  LLD.U32 R4, local [A1+0x0];  
0x000024f0  IADD R4, R4, R31;  
0x000024f8  SHL R4, R4, R31;  
0x00002500  IADD R4, R4, R31;  
0x00002508  GST.U32 global14 [R4], R3;   // <<== GLOBAL STORE vs. R2G (register to global register file)
    tmp = tmpIdx->featIdx[0];

Nsight CUDA内存检查器捕获到全局内存的越界存储。

Memory Checker detected 1 access violations.
error = access violation on store (global memory)
blockIdx = {0,0,0}
threadIdx = {0,0,0}
address = 0x00000010
accessSize = 0

如果你编译compute_10,sm_10(实际上<= 1.3),你应该看到下面的警告,每一行编译器不能确定访问是共享内存:

kernel.cu(46): warning : Cannot tell what pointer points to, assuming global memory space

如果在启动后添加cudaDeviceSynchronize,您应该会看到由越界内存访问引起的错误代码cudaErrorUnknown
__shared__是一个可变内存限定符,而不是类型限定符,所以我知道你会如何告诉编译器featIdx将始终指向共享内存。在CC >= 2.0中,编译器应该将(int*)(&sharedData[tidx*idxArraySize])转换为泛型指针。

相关问题