While loop fails in CUDA kernel

Question

I am using GPU to do some calculation for processing words. Initially, I used one block (with 500 threads) to process one word. To process 100 words, I have to loop the kernel function 100 times in my main function.

for (int i=0; i<100; i++)
    kernel <<< 1, 500 >>> (length_of_word);

My kernel function looks like this:

__global__ void kernel (int *dev_length)
{
   int length = *dev_length;
   while (length > 4)
   {   //do something;
          length -=4;
   }
}

Now I want to process all 100 words at the same time.

Each block will still have 500 threads, and processes one word (per block).

dev_totalwordarray: store all characters of the words (one after another)

dev_length_array: store the length of each word.

dev_accu_length: stores the accumulative length of the word (total char of all previous words)

dev_salt_ is an array of of size 500, storing unsigned integers.

Hence, in my main function I have

   kernel2 <<< 100, 500 >>> (dev_totalwordarray, dev_length_array, dev_accu_length, dev_salt_);

to populate the cpu array:

    for (int i=0; i<wordnumber; i++)
    {
        int length=0;
        while (word_list_ptr_array[i][length]!=0)
        {
            length++;
        }

        actualwordlength2[i] = length;
    }

to copy from cpu -> gpu:

    int* dev_array_of_word_length;
    HANDLE_ERROR( cudaMalloc( (void**)&dev_array_of_word_length, 100 * sizeof(int) ) );
    HANDLE_ERROR( cudaMemcpy( dev_array_of_word_length, actualwordlength2, 100 * sizeof(int),

My function kernel now looks like this:

__global__ void kernel2 (char* dev_totalwordarray, int *dev_length_array, int* dev_accu_length, unsigned int* dev_salt_)
{

  tid = threadIdx.x + blockIdx.x * blockDim.x;
  unsigned int hash[N];

  int length = dev_length_array[blockIdx.x];

   while (tid < 50000)
   {
        const char* itr = &(dev_totalwordarray[dev_accu_length[blockIdx.x]]);
        hash[tid] = dev_salt_[threadIdx.x];
        unsigned int loop = 0;

        while (length > 4)
        {   const unsigned int& i1 = *(reinterpret_cast<const unsigned int*>(itr)); itr += sizeof(unsigned int);
            const unsigned int& i2 = *(reinterpret_cast<const unsigned int*>(itr)); itr += sizeof(unsigned int);
            hash[tid] ^= (hash[tid] <<  7) ^  i1 * (hash[tid] >> 3) ^ (~((hash[tid] << 11) + (i2 ^ (hash[tid] >> 5))));
            length -=4;
        }
        tid += blockDim.x * gridDim.x;
   }
}

However, kernel2 doesn't seem to work at all.

It seems while (length > 4) causes this.

Does anyone know why? Thanks.

the code of do //something doesn't use the variable length.. — user1807890

CygnusX1 CygnusX1 · Accepted Answer · 2012-11-08T08:06:49

I am not sure if the while is the culprit, but I see few things in your code that worry me:

Your kernel produces no output. The optimizer will most likely detect this and convert it to an empty kernel
In almost no situation you want arrays allocated per-thread. That will consume a lot of memory. Your hash[N] table will be allocated per-thread and discarded at the end of the kernel. If N is big (and then multiplied by the total amount of threads) you may run out of GPU memory. Not to mention, that accessing the hash will be almost as slow as accessing global memory.
All threads in a block will have the same itr value. Is it intended?
Every thread initializes only a single field within its own copy of hash table.
I see hash[tid] where tid is a global index. Be aware that even if hash was made global, you may hit concurrency problems. Not all blocks within a grid will run at the same time. While one block will initialize a portion of hash, another block might not even start!

While loop fails in CUDA kernel

1 Answers