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STL Vector problem

Started by stodge Mar 21, 2003 at 3:45 PM 11 replies 300+ views

Original Post

stodge

Mar 21, 2003 at 3:45 PM

I'm trying to do the following:

quote:
std::vector<float[3]> HBvertexArray;

I'm getting an error:

quote:
g:\sdk\stlport\stlport\stl\_construct.h(85) : error C2538: new : cannot specify initializer for arrays

Does this mean I can't do the above? Thanks [edited by - stodge on March 21, 2003 4:45:40 PM] [edit: fixed <'s & >'s, use the html & g t ; and & l t ; to display angle-brackets (without spaces)] [edited by - Magmai Kai Holmlor on March 21, 2003 9:25:54 PM]

Feydrex

Mar 21, 2003 at 3:55 PM

Are you giving the vector a type, since it is a template? Something like:

  std::vector<int> iVec;

stodge

Mar 21, 2003 at 3:56 PM

Argh! I used the quote tags instead of source. The forum script wiped out some of the code:

   typedef std::vector<float[3]> HBvertexArray;

civguy

Mar 21, 2003 at 4:19 PM

Seems like you can't do it then. But maybe you could do this:

    
template <class T, int size>
class Array {
public:
  T& operator[](int i) {
    return array[i];
  }

  const T& operator[](int i) const {
    return array[i];
  }
private:
  T array[size];
};

std::vector<Array<float, 3> > HBvertexArray;

//and access it normally, like:


HBvertexArray[3][0] = 2;

[edited by - civguy on March 21, 2003 5:20:10 PM]

stodge

Mar 21, 2003 at 4:24 PM

Thanks civguy. I think I''ll try doing this:

  		typedef struct
		{
			HBfloat x;
			HBfloat y;
			HBfloat z;
		} HBvertex;

typedef std::vector<HBvertex>	HBvertexArray;

Shannon Barber

Mar 21, 2003 at 8:32 PM

That will work. You can use this sneaky code so that you can use either .x/y/z or [1,2,3] to access the vertex coordinates:


struct point3D
{
float x,y,z;
float& operator[](size_t index)
   {
   //do bounds check in debug build,
   // size_t is unsigned, so index can never be negative
   assert(index < 3);
   return (&x)[index];
   }
};

point3D v;
v.y = 1.00;
v[1] == 1.00;

Er, wait. That's not a vertex yet, it's just a point until you associate some other information with the point (e.g. color, normal, texture mapping coor.).

[edited by - Magmai Kai Holmlor on March 21, 2003 9:36:25 PM]

Beer Hunter

Mar 21, 2003 at 10:00 PM

quote:
Original post by Magmai Kai Holmlor
...this sneaky code...

Why "sneaky"? Point3D is a POD type, and is layout-compatible with an array of three floats. It''s valid code.

stodge

Mar 21, 2003 at 10:12 PM

Thanks Magmai; I''ll try that code snippet out.

Shannon Barber

Mar 22, 2003 at 12:24 AM

It''s not obvious to a beginner, and as you indicate there are issues in the dark if the structure definition changes.

civguy

Mar 22, 2003 at 2:04 AM

quote:
Original post by Beer Hunter
Why "sneaky"? Point3D is a POD type, and is layout-compatible with an array of three floats. It's valid code.

Is it really? Is there a quarantee that there can't be padding between the members? Even if it worked in all compilers, I wouldn't call it valid code. More like kludgy but working code.

[edited by - civguy on March 22, 2003 3:06:30 AM]

YoshiN

Mar 22, 2003 at 2:11 AM

quote:
Original post by Magmai Kai Holmlor
That will work. You can use this sneaky code so that you can use either .x/y/z or [1,2,3] to access the vertex coordinates:

struct point3D
{
float x,y,z;
float& operator[](size_t index)
{
//do bounds check in debug build,
// size_t is unsigned, so index can never be negative
assert(index < 3);
return (&x)[index];
}
};

point3D v;
v.y = 1.00;
v[1] == 1.00;

Er, wait. That''s not a vertex yet, it''s just a point until you associate some other information with the point (e.g. color, normal, texture mapping coor.).

[edited by - Magmai Kai Holmlor on March 21, 2003 9:36:25 PM]

Are you sure that x, y and z will always be stored linearly and in that order?

Kylotan

Mar 22, 2003 at 8:15 AM

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