ocaml help needed: cant get return type of float

Hi, I am rather new to functional programming. I am going to try to
use it for a component of a commercial application I'm working on. I
have so far decided to focus on ocaml because it seems to be catching
fire right now, but I may end up using bigaloo scheme if I keep
running into issues.

What I'm doing to learn the language is work through the classic text,
"Structure and Interpretation of Computer Programs", converting the
scheme examples to ocaml as I go.

I have run into a problem right away. I am trying to write a really
simple Absolute function for floats.

Here is the function that I think should work:
let abs x = if x >= 0.0 then x else -x;;

Here is the error:
Characters 37-38:
let abs x = if x >= 0.0 then x else -x;;
^
This expression has type float but is here used with type int

If I do this it compiles (but gives the wrong answer):
let abs x = if x >= 0.0 then x else x;;

or if I do this it works correctly:
let abs x = if x >= 0.0 then x else 0.0 -. x;;

Now I have tried to specify that x is a float and the return type is a
float but I maybe haven't got the syntax right or something?

could someone show me what I'm doing wrong here and perhaps a correct
version of the function? I hit this again in one of the functions in a
later exercise, so I really need to figure this out before I can move
on (or just go over to scheme).

thanks for any assistance!

Shanon Fernald schrob:
[...]
> I have run into a problem right away. I am trying to write a really
> simple Absolute function for floats.

> Here is the function that I think should work:
> let abs x = if x >= 0.0 then x else -x;;

> Here is the error:
> Characters 37-38:
> let abs x = if x >= 0.0 then x else -x;;
> ^
> This expression has type float but is here used with type int

$ ocaml
# (~-);;
- : int -> int = <fun>

In other words, unary - is a function that expects and returns an int.
x is a float, so this is a type error.

[...]
> or if I do this it works correctly:
> let abs x = if x >= 0.0 then x else 0.0 -. x;;

Yes, because the type of -. is float -> float -> float.

[...]
> could someone show me what I'm doing wrong here and perhaps a correct
> version of the function? I hit this again in one of the functions in a
> later exercise, so I really need to figure this out before I can move
> on (or just go over to scheme).

let abs x = if x >= 0.0 then x else -.x ;;

All you need is "-.", the floating point version of "-".

HTH, Lukas
--
exit sub{&{$_[0]}}->(sub{$_[1]&&print("$_[1] bottle",$_[1]!=1&&s=>
" of beer on the wall,$/$_[1] bottle",$_[1]!=1&&s=>" of beer,
take one down and pass it around,$/",$_[1]-1,' bottle',$_[1]!=2&&s=>
" of beer on the wall.$/$/")&&$_[0]->($_[0],$_[1]-1)},99)