Multiplizieren Sie rechteckige Matrizen in CUDA

In dieser Hausaufgabe muss ich den Code vervollständigen, um zwei Rechteckmatrizen mit CUDA C zu multiplizieren. Nachdem ich den Code vervollständigt hatte, übermittelte ich und die Lösung war für den Datensatz korrekt, als die Matrizen quadratisch waren, während das Ergebnis nicht mit dem erwarteten übereinstimmte Wert, wenn die Matrizen nicht quadratisch waren.

Hier ist der Code, nachdem ich die fehlenden Teile hinzugefügt habe:

#include    <wb.h>

#define wbCheck(stmt) do {                             \
    cudaError_t err = stmt;                            \
    if (err != cudaSuccess) {                          \
        wbLog(ERROR, "Failed to run stmt ", #stmt);    \
        return -1;                                     \
    }                                                  \
} while(0)

// Compute C = A * B
__global__ void matrixMultiply(float * A, float * B, float * C,
               int numARows, int numAColumns,
               int numBRows, int numBColumns,
               int numCRows, int numCColumns) {
   //@@ Insert code to implement matrix multiplication here
   int Row = blockIdx.y * blockDim.y + threadIdx.y;
   int Col = blockIdx.x * blockDim.x + threadIdx.x;
   if (numAColumns != numBRows) return ;
   if ((Row < numARows) && (Col < numBColumns)){
       float Cvalue = 0;
       for (int k = 0 ; k < numAColumns ; ++k )
       Cvalue += A[Row*numAColumns + k] * B[k * numBRows + Col];
       C[Row*numAColumns + Col] = Cvalue;
     }

    }



int main(int argc, char ** argv) {
   wbArg_t args;
   float * hostA; // The A matrix
   float * hostB; // The B matrix
   float * hostC; // The output C matrix
   float * deviceA;
   float * deviceB;
   float * deviceC;
   int numARows; // number of rows in the matrix A
   int numAColumns; // number of columns in the matrix A
   int numBRows; // number of rows in the matrix B
   int numBColumns; // number of columns in the matrix B
   int numCRows; // number of rows in the matrix C (you have to set this)
   int numCColumns; // number of columns in the matrix C (you have to set this)

   args = wbArg_read(argc, argv);

   wbTime_start(Generic, "Importing data and creating memory on host");
   hostA = (float *) wbImport(wbArg_getInputFile(args, 0), &numARows, &numAColumns);
   hostB = (float *) wbImport(wbArg_getInputFile(args, 1), &numBRows, &numBColumns);
   //@@ Set numCRows and numCColumns  
   numCRows = 0;
   numCColumns = 0;
   numCRows = numARows;
   numCColumns = numBColumns;  
   //@@ Allocate the hostC matrix
   hostC = (float*) malloc(sizeof(float)*numCRows*numCColumns);  
   wbTime_stop(Generic, "Importing data and creating memory on host");

   wbLog(TRACE, "The dimensions of A are ", numARows, " x ", numAColumns);
   wbLog(TRACE, "The dimensions of B are ", numBRows, " x ", numBColumns);

   wbTime_start(GPU, "Allocating GPU memory.");
   //@@ Allocate GPU memory here
   cudaMalloc((void**)&deviceA ,sizeof(float)*numARows*numAColumns );
   cudaMalloc((void**)&deviceB , sizeof(float)*numBRows*numBColumns);
   cudaMalloc((void**)&deviceC , sizeof(float)*numCRows*numCColumns);  

   wbTime_stop(GPU, "Allocating GPU memory.");

   wbTime_start(GPU, "Copying input memory to the GPU.");
   //@@ Copy memory to the GPU here

   cudaMemcpy(deviceA, hostA, sizeof(float)*numARows*numAColumns, cudaMemcpyHostToDevice);
   cudaMemcpy(deviceB, hostB, sizeof(float)*numBRows*numBColumns, cudaMemcpyHostToDevice);
   wbTime_stop(GPU, "Copying input memory to the GPU.");

   //@@ Initialize the grid and block dimensions here

   dim3 DimGrid(numARows / 8 , numBColumns / 8, 1);
   dim3 DimBlock(8 , 8, 1);

   wbTime_start(Compute, "Performing CUDA computation");

   //@@ Launch the GPU Kernel here
   matrixMultiply<<<DimGrid , DimBlock>>>(deviceA , deviceB , deviceC , numARows , numAColumns, numBRows ,numBColumns , numCRows , numCColumns);  

   cudaThreadSynchronize();
   wbTime_stop(Compute, "Performing CUDA computation");

   wbTime_start(Copy, "Copying output memory to the CPU");
   //@@ Copy the GPU memory back to the CPU here
   cudaMemcpy(hostC, deviceC, sizeof(float)*numCRows*numCColumns , cudaMemcpyDeviceToHost);  

   wbTime_stop(Copy, "Copying output memory to the CPU");

   wbTime_start(GPU, "Freeing GPU Memory");
   //@@ Free the GPU memory here

   cudaFree(deviceA);
   cudaFree(deviceB);
   cudaFree(deviceC);
   wbTime_stop(GPU, "Freeing GPU Memory");

   wbSolution(args, hostC, numCRows, numCColumns);

   free(hostA);
   free(hostB);
   free(hostC);

   return 0;
}

Ich hoffe, Sie können mir helfen, das falsche Teil zu finden.

Antworten auf die Frage(5)

Ihre Antwort auf die Frage