Ruben Safir ruben-at-mrbrklyn.com=
writes:

Why are we using the ADDRESS of a pointer to a float as in &d_in

Because the first parameter of cudaMalloc is what, in some =
programming languages, would be called an =E2=80=9Coutput parameter=E2=80=
=9D. The cudaMalloc function is going to overwrite the value o=
f the pointer with a new pointer to the allocated memory. For it to do that=
, we have to pass the address of the pointer.

In C++, we could do it as a =E2=80=9Creference parameter=E2=80=9D instea=
d, that type would be void*& and then you don=E2=80=99t ha=
ve to do &d_in but just d_in. (It still passe=
s the address behind the scenes.)

Compare cudoMalloc to the regular malloc in ode>stdlib.h =E2=80=93 it returns the pointer, so there=E2=
=80=99s no need for an =E2=80=9Coutput parameter=E2=80=9D.

ceCode cpp">  float* d_in =3D (>float*) malloc(ARRAY_BYTES);

CL

ceCode cpp">#include <stdio.h>

int main(int argc, n class=3D"dt">char ** argv)
{
  const int ARRAY_SIZE =
=3D 64;
  const int ARRAY_BYTES=
 =3D sizeof(float) * AR=
RAY_SIZE;
  //generate the input array on host
  float h_in[ARRAY_SIZE];
  float h_out[ARRAY_SIZE];
  for( int i =3D class=3D"dv">0; i < ARRAY_SIZE; i++)
  {
    h_in[i] =3D float(i);
  }
  //declare gpu memory pointers
  float *d_in *d_out;
  cudaMalloc(( void **) &d_in, ARRAY_BYTES );
  cudaMalloc(( void **) &d_out, ARRAY_BYTES );
}