On Monday, 12 February 2018 at 02:05:16 UTC, aliak wrote:
From spec: Cast expression: "cast ( Type ) UnaryExpression" converts UnaryExpresssion to Type.


And https://dlang.org/spec/operatoroverloading.html#cast makes no mention of the return type of opCast. One could think that the return type of opCast would be the return type. But it seems it must be the same as the template parameter of opCast else you get a compile error that seems like it can be much better.

---

import std.stdio;

struct B(T) {
    T t;
}

struct A(T) {
    T t;
    auto opCast(U)() {
        return B!U(cast(U)t);
    }
}

void main() {
    auto a = A!int(3);
    auto b = cast(float)a; // error
}

Error: cannot implicitly convert expression a.opCast() of type B!float to float

Is this deliberate?

The use case I have is making an optional type that you can cast to a different type:

auto opCast(U)() const {
    static if (isOptional!U)
    {
        alias V = OptionalTarget!U;
return empty ? no!V : some!V(cast(V)front); // it's a range so "front" is the raw value
    }
    else
    {
            return empty ? no!U : some!U(cast(U)front);
    }
}

It would allow for scenarios like:

Optional!int a = 3;
auto b = cast(float)a;
// b == some!float


Cheers
- Ali

I think the best way to do this is to implement `map` for your optional type.

Optional!U map(U, alias f)()
{
    return empty? no!U : some!U(f(t));
}

Optional!int a = 3;
auto b = a.map!(v => cast(float)v);
assert(is(typeof(b) == Optional!float));

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