pass a 2D array from a C++ class to a CUDA function

Question

I am a Java guy jumping into CUDA and the syntax is tripping me. I'm trying to create a matrix in the .cpp file then pass that off to the .cu file to be processed. I see examples where the CUDA function expects the 2D array to come in looking like

void handleMatrix(float* A){
    // do stuff
}

when I create the matrix I'm used to doing it in code that looks like this:

    int main()
{

   const int row=8;
   const int column=8;
   int rnum;
   srand(time(0));
   rnum = (rand() % 100) + 1;  

  float table[row][column];  
    for(int r=0; r<row; r++){ 
      for(int c=0; c<column;c++){       
        table[row][column] = (rand()%100) + 1.f;    
      }
      cout << "\n";
    }
   
   handleMatrix(table);
   return 0;
}

When I compile the code I'm getting the error cannot convert ‘float ()[8]’ to ‘float*’ for argument ‘1’ to ‘void handleMatrix(float*)’*

Is there a different way I should be declaring the matrix or creating it?

Thanks in advance for the help.

Eugene Smith · Accepted Answer · 2010-11-13 04:09:18Z

1

You can do

handleMatrix(table[0]);

or, equivalently,

handleMatrix(&table[0][0]);

That's if 'handleMatrix' is host code. If it's device code, you can't allocate the buffer like that. You'll want to assemble an array in local memory, fill the entries, allocate another array in the device memory using cudaMalloc() or cudaMallocPitch(), and then copy from local to device using cudaMemcpy() or cudaMemcpy2D().

edited Nov 13, 2010 at 4:09

answered Nov 13, 2010 at 4:02

Eugene Smith

9,4186 gold badges39 silver badges43 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Heatsink · Accepted Answer · 2010-11-13 00:53:32Z

0

You want a pointer to an array. The syntax for declaring a pointer to an array is

void handleMatrix(float (*A)[8][8]) {
  // do stuff
}

That is, when you dereference A, you get a reference to an 8 by 8 array of floats.

Even for 1D arrays, there is a distinction between pointer to array (float (*anArray)[100]) and pointer to element (float *anArray). C++ will transparently convert the former to the latter, which means that for 1D arrays (but not arrays of higher dimension) you can usually ignore the difference.

answered Nov 13, 2010 at 0:53

Heatsink

5042 silver badges3 bronze badges

Comments

aschepler · Accepted Answer · 2010-11-13 00:54:46Z

0

In your example, table[0] converts to a valid float* pointer to 64 consecutive float numbers. But it looks highly suspicious that handleMatrix takes a pointer meant to be an array and doesn't take any information about the dimensions of that array.

answered Nov 13, 2010 at 0:54

aschepler

72.5k11 gold badges114 silver badges174 bronze badges

Comments

AndersK · Accepted Answer · 2010-11-13 01:00:38Z

0

the handleMatrix() function, is this a function from you or is part of a library? If the latter you may need to create the 2-d array as a long row x col 1-d array. If the former you need to change the function to accept a 2-d array e.g. handleMatrix(float**m) and pass the dimensions of the matrix to the function;

preferably though you should use vector<> when programming in C++ then the dimensions are known by the callee.

e.g.

  #include <vector>
  typedef std::vector<std::vector<float > > matrix;

  void handleMatrix( matrix& m ) {..}

answered Nov 13, 2010 at 1:00

AndersK

36.1k6 gold badges63 silver badges86 bronze badges

1 Comment

Heatsink Over a year ago

This is fine in host-only code, however, an std::vector can't be passed to a CUDA kernel.

Collectives™ on Stack Overflow

pass a 2D array from a C++ class to a CUDA function

4 Answers 4

Comments

Comments

Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

Comments

Comments

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related