Nio - 1 year ago 81

Scala Question

From the following example, I think it is correct to say that

`String`

`String`

`""`

`scala> ("" + "Jane") + "Doe" == "" + ("Jane" + "Doe")`

res0: Boolean = true

From the various texts I have been reading on the subject lately, it seems that the correct use of the term

`String`

For example, here is a theoretical

`Monoid`

`trait Monoid[A] {`

def op(a1: A, a2: A): A

def zero: A

}

val stringMonoid = new Monoid[String] {

def op(a1: String, a2: String) = a1 + a2

val zero = ""

}

I know that we do not need

`trait Monoid[A]`

`stringMonoid`

My issue (and I am very possibly thinking about it way too much) is the purist definition. I know that the underlying

`java.lang.String`

`StringBuilder`

`""`

Question:-

Is

`String`

`""`

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Answer Source

I think that you already understand the concept right and indeed "think too much" about it (;

A monoid is a *triple* as they say in mathematics: a set (think of a type with its values), an associative binary operator on it and a neutral element. You define such triple — you define a monoid. So the answer to

Is

`String`

a monoid under the concatenation operation implicitly, just because we happen to know that using an empty string`""`

provides the identity?

is simply **yes**. You named the set, the associative binary operation and the neutral element — bingo! You got a monoid! About

already defines the associative operation, but I do not think that there is an explicit definition of the identity element

there's a bit of confusion. You can choose different associative operations on `String`

with their corresponding neutral elements and define various monoids. Concatenation is not like some kind of "orthodox" associative operation on `String`

to define monoid, just, probably, the most obvious.

If you define a table as something "with four legs and a flat horizontal surface on them", then anything that fits this definition *is* a table, regardless of the material it's made and other variable characteristics. Now, when do you need to "certify" it explicitly is a table? Only when you need to use its "table-properties", say if to sell it and advertise that you can put things on it and they won't fall off, because the surface is guaranteed to be *flat* and *horizontal*.

Sorry, if the example is kind of stupid, I'm not very good in such analogies. I hope it is still helpful.

Now about instantiating the mentioned "theoretical `Monoid`

type". Such types are usually called a type class. Neither existence of such type itself (which can be defined in various ways), nor its instance are necessary to call the triple `(String, (_ ++ _), "")`

a monoid and *reason* about it as a monoid (i.e. use general monoid properties). What it is actually used for is ad-hoc polymorphism. In Scala it is done with implicits. One can, for example, define a *polymorphic* function

```
def fold[M](seq: Seq[M])(implicit m: Monoid[M]): M = seq match {
case Seq.empty => m.zero
case (h +: t) => m.op(h, fold(t))
}
```

Then if your `stringMonoid`

value is declared as `implicit val`

, whenever you use `fold[String]`

and `stringMonoid`

is in scope, it will use its `zero`

and `op`

inside. Same `fold`

definition will work for other instances of `Monoid[...]`

.

Another topic is what happens when you have several instances of `Monoid[String]`

. Read Where does Scala look for implicits?.

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