Thread overview
Understand signature of opOpAssign in std/complex.d
3 days ago
René Heldmaier
3 days ago
Adam D. Ruppe
3 days ago
René Heldmaier
1 day ago
Q. Schroll
1 day ago
Adam D. Ruppe
3 days ago
Hi,
i'm currently implementing a dual number datatype. Dual numbers are very similar to complex numbers, but instead i (i^2 = -1) you have epsilon (epsilon^2 = 0).
This sounds strange but it is really useful for something called "automatic derivation". I will probably explain it in more detail when i'm ready to publish it as a dub package...

Because of the similarity to complex numbers i looked at how they are implemented in std/complex.d, to learn from how it is done there.

I came across a function signature which i don't understand:

ref Complex opOpAssign(string op, C)(const C z)
        if ((op == "+" || op == "-") && is(C R == Complex!R))

The part i don't understand is "is(C R == Complex!R)".
What does that mean?
3 days ago
On Saturday, 9 November 2019 at 12:30:46 UTC, René Heldmaier wrote:
> The part i don't understand is "is(C R == Complex!R)".
> What does that mean?

That's checking the if the template argument C is a case of Complex!R, while at the same time declaring a symbol R to hold the inner type.

So let's say we passed it a Complex!int. In that case, C is Complex!int, and then the new name, R, is declared to be `int`.

(though the name R in here is only valid inside that one expression because it is in the constraint... if you need to use it inside the function body, you will need to do it again:

void foo(C)(C c) if(is(C R == Complex!R)) {
        static if(is(C R == Complex!R)) // gotta redeclare in this scope
          pragma(msg, R); // will print the type passed now
}


Can read more here:

https://dlang.org/spec/expression.html#IsExpression


I just like to think of it as a variable declaration pattern. A normal variable is declared

int a = 0;

where it goes `type name OPERATOR initializer`

so the

is(C R == Y)

is kinda similar: the existing type is C, the new name is R, then the == Y is the initializer.

still magic syntax but it helps remember the order at least.


And then the Y is like writing out the variable with placeholders.
3 days ago
On Saturday, 9 November 2019 at 13:26:12 UTC, Adam D. Ruppe wrote:
> That's checking the if the template argument C is a case of Complex!R, while at the same time declaring a symbol R to hold the inner type.

I got it. Thank you very much ;)
1 day ago
On Saturday, 9 November 2019 at 13:26:12 UTC, Adam D. Ruppe wrote:
> On Saturday, 9 November 2019 at 12:30:46 UTC, René Heldmaier wrote:
>> The part i don't understand is "is(C R == Complex!R)".
>> What does that mean?
>
> That's checking the if the template argument C is a case of Complex!R, while at the same time declaring a symbol R to hold the inner type.

What's the difference of
    is(C R == Complex!R)
to
    is(C == Complex!R, R)
then?

1 day ago
On Monday, 11 November 2019 at 16:59:57 UTC, Q. Schroll wrote:
> What's the difference of
>     is(C R == Complex!R)
> to
>     is(C == Complex!R, R)
> then?

Nothing, they do the same thing.

(I'm of the opinion that the first one should actually be illegal - I thought it was until I checked for this thread - since it is weird. The general rule is placeholders need to be defined after the pattern, not before. The `C R == X` thing usually defines R to be an alias of something that already exists - maybe C itself, maybe a specialization of X like __parameters - but meh it did work here so weird and as far as I can tell they are identical.)