Functional programming function confusion

Question

I'm learning functional programming and am using Ocaml, but I'm having a bit of a problem with functions. Anyway, I have a tuple and I want to return its first value. (Very simple I know, sorry) ``` let bach (x,y):(float*float) = (x,y);; val bach : float * float -> float * float = ``` All well and good up here. ``` let john (x,y):(float*float) = y;; val john : 'a * (float * float) -> float * float = ``` Now this is what confuses me. Why is there a `'a` there? I know that it stands for a variable with an unknown type, but I'm confused as to how changing the return value adds that there. I am a self professed n00b in functional programming, please don't eat me :)

Accepted Answer

You were bitten by a subtle syntax mistake that is really non-obvious for beginners:

```
 let foo x : t = bar
```

is not the same as

```
 let foo (x : t) = bar
```

it is on the contrary equivalent to

```
 let foo x = (bar : t)
```

constraining the return type of the function.

.

So you have written

```
let john (x, y) = (y : float * float)
```

The input type is a pair whose second element, `y`, has type `float * float`. But `x` can be of any type, so the function is polymorphic in its type, which it represents as a type variable `'a`. The type of the whole function, `'a * (float * float) -> float * float`, indicates that for any type `'a`, you may pass a tuple of an `'a` and a `(float * float)`, and it will return a `(float * float)`.

This is a particular case of the `snd` function:

```
let snd (x, y) = y
```

which has type `'a * 'b -> 'b`: for any `'a` and `'b`, you take a pair `('a * 'b)` and return a value of type `'b`.

Functional programming function confusion

Solution

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