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Array allocation from memory pool

General and Gameplay Programming Programming

Started by Misery January 19, 2012 07:19 PM

6 comments, last by Misery 12 years, 3 months ago

Misery

354

Author

January 19, 2012 07:19 PM

Hi,

I would like to allocate memory from pool for a matrix. How to do this correctly?
Do I have to allocate the memory for a vector of pointers first? Or just the matrix entries?
I would like to allocate the whole matrix in one memory block.



 float **matrix=NULL;

 int Cols=someValue, Rows=someValue;

How to allocate this with memory pool like this:



 void *Allocate(int Size);

Thanks in advance,
regards

BeerNutts

4,410

January 19, 2012 07:59 PM

If you just want a 2d array of floats, then you should just be able to allocate a piece of memory like this:



float **matrix;

int Cols=someValue, Rows=someValue;



matrix = (float **)Allocate(sizeof(float)*Cols*Rows);

Then you have a 2d array you can access via matrix[SomeCol][SomeRow] or even matrix[SomeRow][SomeCol] (assuming SomeRow < Rows and SomeCol < Cols), but just stick to one.

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SiCrane

11,840

January 19, 2012 08:07 PM

Then you have a 2d array you can access via matrix[SomeCol][SomeRow] or even matrix[SomeRow][SomeCol] (assuming SomeRow < Rows and SomeCol < Cols), but just stick to one.

No, that doesn't perform the array arithmetic necessary to index matrix as a 2D array. You either need to perform the multiplication and addition manually or store an indirection array to the individual elements.

Misery

354

Author

January 19, 2012 10:45 PM

Should it be something like this:



int Rows,Cols //have some values

matrix=(float**)Allocate(Rows*sizeof(float*)+Rows*(Cols-1)*sizeof(float));

  for (int i=0;i<Rows;i++) matrix=matrix[i*Rows];

?? :]

fastcall22

10,918

January 20, 2012 05:52 AM

You're close:



void* mem = malloc( rows*sizeof(float*) + rows*cols*sizeof(float) );

float** mat = (float**)mem;

float* data = (float*)mem + rows;



for ( int row = 0; row < rows; ++row )

    mat[row] = data + row*cols;



mat[3][2] = 1.f;

Though you could do it this way:



float* mat = (float*)malloc( sizeof(float)*rows*cols );



mat[ 3*rows + 2 ] = 1.f;

Or this way:



class Matrix {

private:

    int rows, cols;

    float* data;



public:

   Matrix( int rows, int cols ) { resize( rows, cols ); }



   void resize( int newRows, int newCols ) { /* ... */ }  



   float& operator() (int row, int col) { return data[ row*cols + col ]; }

   const float& operator() (int row, int col) const { return data[ row*cols + col ]; }

}



void test() {

   Matrix mat( 4, 4 );



   mat(3, 2) = 1.f;

}

Misery

354

Author

January 20, 2012 06:37 AM

Thanks a lot :]
So allocating a float matrix of size NxN from a pool requires more memory than using new[] operator?
The difference is a vector of pointers.



  int totalsize=sizeof(float*)*Rows+sizeof(float)*Rows*Cols;

fastcall22

10,918

January 20, 2012 06:52 AM

So allocating a float matrix of size NxN from a pool requires more memory than using new[] operator?

No. The extra memory comes from setting up the row pointers, which are required if you want to use the matrix[row][col] notation. However, you can avoid the overhead by using a single dimension array and use this notation: matrix[ row*numRows + col ].

EDIT:
I just remembered you can do this:



const int N = 4;

/* 1: */ {

     float (*mat)[N] = new float[N][N];

     mat[3][2] = 1.f;



     delete[] mat;

}



/* 2: */ {

     float (*mat)[N] = (float(*)[N]) malloc( sizeof(float)*N*N );

     mat[3][2] = 1.f;



     free( mat );

}

The best of both worlds: No need for allocating and setting up row pointers, and the mat[row][col] notation.