[Haskell-cafe] Begginer question

[Jules Bean]

On 6 Jan 2005, at 01:37, Maurício wrote:
> import Complex;
>
> complex_root :: (Float, Float, Float) -> (Complex Float, Complex Float)
> complex_root (a,b,c) = (x1,x2) where {	
> 	delta = b * b - 4 * a * c :: Float;
> 	sqr_delta = if delta >= 0 then (sqrt delta) :+ 0 else 0 :+ (sqrt 
> delta) :: (Complex Float);
> 	x1 = (b + sqr_delta)/(2 * a);
> 	x2 = (b - sqr_delta)/(2 * a);
> }


> Couldn't match `Float' against `Complex Float'
>         Expected type: Float
>         Inferred type: Complex Float
>     In the second argument of `(+)', namely `sqr_delta'
>     In the definition of `x1': x1 = (b + sqr_delta)

>   Can you help me finding what is wrong? Shouldn't "b" be converted to 
> Complex Float and be summed to sqr_delta?
>

Haskell will not automatically convert b from Float to Complex Float. 
The arguments of (+) should have the same type.

One alternative is to use b :+ 0 instead of b. (and similarly for a).

Another approach is to define a 'cast' function like:

toComplex x = (fromRational.toRational) x :: Complex Float

and then you can use toComplex b instead of b :+ 0.  that's more 
characters to type, though...

Note that sqr_delta isn't going to be defined as you expect, either. 
Since delta has type Float, sqrt delta has type Float. (And sqrt -1 :: 
Float is Not A Number). If you want to do it by hand this way, then you 
want:

> sqr_delta = if delta >= 0 then (sqrt delta) :+ 0 else 0 :+ (sqrt 
> (-delta)) :: (Complex Float);

If delta was itself already Complex, then sqrt would do the right thing 
automatically.

Jules