Vertex2 confusion with haskell openGL bindings

Question

I'm using Haskell openGL bindings to try to make a particle generator. I want to store information about a particle in a record where I can then pass in fields and update them when appropriate.

So a position in the record is stored as:

data Particle = Particle { pos :: Vertex2 GLfloat }

and set like this:

Particle { pos = Vertex2 1 (0::GLfloat) }

Then I pass in the particle and try to retrieve the values like so:

drawParticle :: Particle -> IO ()
drawParticle part = 
    (pos part) >>= \(Vertex2 x y) ->
    print x

The error I get:

Couldn't match type `Vertex2' with `IO'
Expected type: IO GLfloat
  Actual type: Vertex2 GLfloat
In the return type of a call of `pos'
In the first argument of `(>>=)', namely `(pos part)'
In the expression: (pos part) >>= \ (Vertex2 x y) -> print x

I'm mildly confused about the data type Vertex2 and why it has to be declared with a single GLfloat instead of two. How would I extract numbers from the data type Vertex2 GLfloat? (that is how would I extract Vertex2 1 (0::GLfloat) to be x = 1.0, y = 0.0?

Answer 1

To answer your questions:

• It would be possible to define Vertex2 to take two types, to allow X to have one type and Y another, e.g. data Vertex2 xtype ytype = Vertex2 xtype ytype. However, it's generally a bad idea to have two different types for X and Y so instead it's defined as: data Vertex2 sametype = Vertex2 sametype sametype to save problems.

• Since you have explicitly typed the Vertex2 in your declaration for Particle, you don't need to type the expression you list. Just Particle { pos = Vertex2 1 0 } should be enough. (Or: Particle (Vertex2 1 0)).

• You get the compile error because you don't need monadic bind. part has type Particle, not IO Particle, so you don't need bind to get the value out. You can write either:

  drawParticle part = let Vertex2 x y = pos part in print x

or:

drawParticle part = do
    let Vertex2 x y = pos part
    print x

(Note the two different forms of let, depending on whether it's in a do block; this confused me when I started out.)