C++ 对于具有CUDA的嵌套循环

我对一些嵌套循环有问题，我必须将它们从C/C++转换成CUDA。基本上，我有4个for嵌套循环，它们共享相同的数组并进行位移位操作

#define N 65536

// ----------------------------------------------------------------------------------

int a1,a2,a3,a4, i1,i2,i3,i4;

int Bit4CBitmapLookUp[16] = {0, 1, 3, 3, 7, 7, 7, 7, 15, 15, 15, 15, 15, 15, 15, 15};

int _cBitmapLookupTable[N];

int s = 0;  // index into the cBitmapLookupTable

for (i1 = 0; i1 < 16; i1++)
{
    // first customer
    a1 = Bit4CBitmapLookUp[i1] << 12;

    for (i2 = 0; i2 < 16; i2++)
    {
        // second customer
        a2 = Bit4CBitmapLookUp[i2] << 8;

        for (i3 = 0; i3 < 16; i3++)
        {
            // third customer
            a3 = Bit4CBitmapLookUp[i3] << 4;

            for (i4 = 0;i4 < 16;i4++)
            {
                // fourth customer
                a4 = Bit4CBitmapLookUp[i4];

                // now actually set the sBitmapLookupTable value
                _cBitmapLookupTable[s] = a1 | a2 | a3 | a4;

                s++;

            } // for i4
        } // for i3
    } // for i2
} // for i1

#定义N 65536
// ----------------------------------------------------------------------------------
int a1、a2、a3、a4、i1、i2、i3、i4；
int Bit4CBitmapLookUp[16]={0,1,3,3,7,7,7,7,15,15,15,15,15,15}；
intcbitmaplookuptable[N]；
int s=0；//索引到cBitmapLookupTable
对于（i1=0；i1<16；i1++）
{
//第一位顾客
a1=Bit4CBitmapLookUp[i1]因为leftaroundabout已经指出初始化有问题。我建议您按照如下方式重写程序

int i1 = blockDim.x * blockIdx.x + threadIdx.x;
int i2 = blockDim.y * blockIdx.y + threadIdx.y;
int i3;
int i4;

while(i1 < N && i2 < N){
  a1 = ..;
  a2 = ..;
  for(i3 = i4 = 0; i3 < N, i4 < N; i3++, i4++){
    // third customer
    a3 = Bit4CBitmapLookUp_device[i3] << 4;

    // fourth customer
    a4 = Bit4CBitmapLookUp_device[i4];

    // now actually set the sBitmapLookupTable value
    _cBitmapLookupTable[s] = a1 | a2 | a3 | a4;
    s ++;
  }
  s += blockDim.x*gridDim.x*blockDim.y*gridDim.y;
  i1 += blockDim.x*gridDim.x;
  i2 += blockDim.y*gridDim.y;
}

int i1=blockDim.x*blockIdx.x+threadIdx.x；
int i2=blockDim.y*blockIdx.y+threadIdx.y；
int i3；
int i4；
而（i1a3=Bit4CBitmapLookUp\u设备[i3]提示：您正在将变量
i1
..
i4
初始化为从未使用过的值。请参见此->但是，索引i3和i4如何？我是否应该将它们声明为正常整数？抱歉，我认为它们显然是正常整数。更新了我的答案。
int i1 = blockDim.x * blockIdx.x + threadIdx.x;
int i2 = blockDim.y * blockIdx.y + threadIdx.y;
int i3;
int i4;

while(i1 < N && i2 < N){
  a1 = ..;
  a2 = ..;
  for(i3 = i4 = 0; i3 < N, i4 < N; i3++, i4++){
    // third customer
    a3 = Bit4CBitmapLookUp_device[i3] << 4;

    // fourth customer
    a4 = Bit4CBitmapLookUp_device[i4];

    // now actually set the sBitmapLookupTable value
    _cBitmapLookupTable[s] = a1 | a2 | a3 | a4;
    s ++;
  }
  s += blockDim.x*gridDim.x*blockDim.y*gridDim.y;
  i1 += blockDim.x*gridDim.x;
  i2 += blockDim.y*gridDim.y;
}