Blaszard - 6 months ago 7

Swift Question

I want to define a new operator and multiply each element of the array

`[Int]`

`Int`

`[3, 2, 10] * 10`

However, because

`Int`

`protocol Numeric {}`

extension Int: Numeric {}

extension Double: Numeric {}

extension Float: Numeric {}

And then, I tried defining the operator, like:

`func *<T: Numeric> (left: [T], right: T) -> [T] {`

var newArray: [T] = []

for i in left {

newArray.append(i * right)

}

return newArray

}

However, this spits out an error:

`Cannot convert value of type 'T' to expected argument type '[_]'`

I'm not sure what the type

`[_]`

`T`

`T`

`Numeric`

So I defined another operator, like:

`func *<T: Numeric> (left: T, right: T) -> T {`

return left * right

}

However, while this has been compiled without problems, the runtime error occurred with a lot of a lot of

`static * infix <A where ...> (A, A) -> A`

I'm not sure why this operator was executed so many times, but now I wonder if it is possible in the first place to define a custom

`*`

`Int`

`*`

So is it still possible to define

`[Int] * Int`

Answer

You have to require the multiplication operation in the `Numeric`

protocol:

```
protocol Numeric {
func *(lhs: Self, rhs: Self) -> Self
}
```

Otherwise the multiplication in `newArray.append(i * right)`

is
not defined.

Your

```
func *<T: Numeric> (left: T, right: T) -> T {
return left * right
}
```

(which calls itself recursively, resulting in a stack overflow) is then not needed.

The implementation of your new operator itself can be simplified (as already described in a now deleted answer) to

```
func *<T: Numeric> (left: [T], right: T) -> [T] {
return left.map { $0 * right }
}
```

Source (Stackoverflow)

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