Question

我希望从CUDA中的输入数据中选择一些项目（不是全部）。我的输入数组d_in大小为53 * 53，（对不起，它很长）：

$abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz
z$abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxy
yz$abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwx
xyz$abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvw
wxyz$abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuv
vwxyz$abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstu
uvwxyz$abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrst
tuvwxyz$abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrs
stuvwxyz$abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqr
rstuvwxyz$abcdefghijklmnopqrstuvwxyzabcdefghijklmnopq
qrstuvwxyz$abcdefghijklmnopqrstuvwxyzabcdefghijklmnop
pqrstuvwxyz$abcdefghijklmnopqrstuvwxyzabcdefghijklmno
opqrstuvwxyz$abcdefghijklmnopqrstuvwxyzabcdefghijklmn
nopqrstuvwxyz$abcdefghijklmnopqrstuvwxyzabcdefghijklm
mnopqrstuvwxyz$abcdefghijklmnopqrstuvwxyzabcdefghijkl
lmnopqrstuvwxyz$abcdefghijklmnopqrstuvwxyzabcdefghijk
klmnopqrstuvwxyz$abcdefghijklmnopqrstuvwxyzabcdefghij
jklmnopqrstuvwxyz$abcdefghijklmnopqrstuvwxyzabcdefghi
ijklmnopqrstuvwxyz$abcdefghijklmnopqrstuvwxyzabcdefgh
hijklmnopqrstuvwxyz$abcdefghijklmnopqrstuvwxyzabcdefg
ghijklmnopqrstuvwxyz$abcdefghijklmnopqrstuvwxyzabcdef
fghijklmnopqrstuvwxyz$abcdefghijklmnopqrstuvwxyzabcde
efghijklmnopqrstuvwxyz$abcdefghijklmnopqrstuvwxyzabcd
defghijklmnopqrstuvwxyz$abcdefghijklmnopqrstuvwxyzabc
cdefghijklmnopqrstuvwxyz$abcdefghijklmnopqrstuvwxyzab
bcdefghijklmnopqrstuvwxyz$abcdefghijklmnopqrstuvwxyza
abcdefghijklmnopqrstuvwxyz$abcdefghijklmnopqrstuvwxyz
zabcdefghijklmnopqrstuvwxyz$abcdefghijklmnopqrstuvwxy
yzabcdefghijklmnopqrstuvwxyz$abcdefghijklmnopqrstuvwx
xyzabcdefghijklmnopqrstuvwxyz$abcdefghijklmnopqrstuvw
wxyzabcdefghijklmnopqrstuvwxyz$abcdefghijklmnopqrstuv
vwxyzabcdefghijklmnopqrstuvwxyz$abcdefghijklmnopqrstu
uvwxyzabcdefghijklmnopqrstuvwxyz$abcdefghijklmnopqrst
tuvwxyzabcdefghijklmnopqrstuvwxyz$abcdefghijklmnopqrs
stuvwxyzabcdefghijklmnopqrstuvwxyz$abcdefghijklmnopqr
rstuvwxyzabcdefghijklmnopqrstuvwxyz$abcdefghijklmnopq
qrstuvwxyzabcdefghijklmnopqrstuvwxyz$abcdefghijklmnop
pqrstuvwxyzabcdefghijklmnopqrstuvwxyz$abcdefghijklmno
opqrstuvwxyzabcdefghijklmnopqrstuvwxyz$abcdefghijklmn
nopqrstuvwxyzabcdefghijklmnopqrstuvwxyz$abcdefghijklm
mnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz$abcdefghijkl
lmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz$abcdefghijk
klmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz$abcdefghij
jklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz$abcdefghi
ijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz$abcdefgh
hijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz$abcdefg
ghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz$abcdef
fghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz$abcde
efghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz$abcd
defghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz$abc
cdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz$ab
bcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz$a
abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz$

我想从输入到输出d_out中选择每行的最后一项。这样，输出大小应为53.这是我的代码。用于处理preSort到d_in和temp到d_out的数据，为两个指针分配内存并启动内核。

//variables declared
const int ARRAY_BYTES_IN = CAPACITY * sizeof(char);
const int ARRAY_BYTES_ST = CAPACITY * CAPACITY * sizeof(char);
const int CAPACITY = 53;
char preSort[CAPACITY * CAPACITY];
char temp[CAPACITY];

void getLast(){
     //two pointers
     char* d_in;
     char* d_out;

     //allocate gpu memory     
     cudaMalloc(&d_in, ARRAY_BYTES_ST);
     cudaMalloc(&d_out, ARRAY_BYTES_IN);

     //transfer input into gpu
     cudaMemcpy(d_in, preSort, ARRAY_BYTES_ST, cudaMemcpyHostToDevice);

     int size = CAPACITY*CAPACITY;
     int blockSize = 1024;
     int numbBlock = (size + blockSize - 1) / blockSize;

     //Launch the kernel
     DoGetLast<<<numbBlock, blockSize>>>(d_out, d_in);

     //Copy back to the  host
     cudaMemcpy(temp, d_out, ARRAY_BYTES_IN, cudaMemcpyDeviceToHost);

     cudaFree(d_in);
     cudaFree(d_out);
}

GPU内核

__global__ void DoGetLast(char* d_out, char* d_in){

     int CAP = 53*53;
     int idx = blockDim.x * blockIdx.x + threadIdx.x;

     char f;

     //get the output trmo the input, It's a 1-D array actually, so pick 
     //only one character through every 53 characters from d_in

     if(idx % CAP == (CAP - 1)){
          f = d_in[idx];
          d_out[idx] = f;
     }
}

在main中，我只调用getLast()方法，并使用循环显示输出。我希望输出看起来像：

zyxwvutsrqponmlkjihgfedcbazyxwvutsrqponmlkjihgfedcba$

但是，我的输出只有一个字母输出，只有z。任何人都可以在我的代码中说出问题？并给予帮助？

Answer 1

你的内核有一些错误。您应该像这样修改内核。

__global__ void DoGetLast(char* d_out, char* d_in)
{
    int CAP = 53; // not 53*53
    int idx = blockDim.x * blockIdx.x + threadIdx.x;
    if (idx < CAP * CAP) // d_in boundary check
        if (idx % CAP == (CAP - 1)) {
            char f = d_in[idx];
            d_out[idx / CAP] = f; // not d_out[idx]
        }
}

如果您使用proper CUDA error checking method和cuda-memcheck工具检查了代码，您就会知道内核存在内存访问错误，您可能已经发现自己的代码出了什么问题。< / p>

（如果我使用cuda-memcheck运行您的代码，则会在Invalid __global__ write of size 1行显示d_out[idx] = f。）

另外，请注意您的代码的GPU利用率较低，因为在您启动的numBlock * blockSize个线程中，只有53个线程可以正常工作。其余的（超过52 * 53个线程）没有做任何事情。

Answer 2

在CUDA中，最好创建相当于输出大小或更小的线程。在您的情况下，您应该只创建53个线程;

 int size = CAPACITY;
 int blockSize = 64;
 int numbBlock = (size + blockSize - 1) / blockSize;

现在，在内核端，每个线程都会选择行的最后一项，如：

__global__ void DoGetLast(char* d_out, char* d_in){

     int idx = blockDim.x * blockIdx.x + threadIdx.x;
     int CAP = 53;

      if(idx >= CAP)
        return;

     d_out[idx] = d_in[idx*CAP];

}

如何在CUDA中处理不同大小的输入和输出

2 个答案: