Easiest is to do the multiply yourself:

float * f = new float[ size*3 ];

float0fromnth = f[ n*3 ];
float1fromnth = f[ n*3+1 ];
float2fromnth = f[ n*3+2 ];

If you really want the array like that, that should work fine, as the "last" sizer of the array can be dynamic, but the others cannot.

float * f = new float[size][3]; // works
float * f = new float[3][size]; // doesn''t work


Here''s some code:

float (*f)[3];

void func( int size )
{
f = new float[size][3];
}

Remember that declarations read from in to out, first right, and then left (paying attention to parens).

Thus:

float *f[3]; // f is array of 3 elements of pointer to float
float (*f)[3]; // f is pointer to array of 3 elements of float

* typedef propaganda *

typedef float vector3[3];vector3 * v = new vector3[size];

Oh, btw, a more C++-ish way (yeah I know new is kinda C++ already) :

vector<vector<float> > v;v.resize( size, vector( 3 ) );// and if you have some boost or other lib with stackvector<stack<3,float> > v;v.resize( size );