Understand signature of opOpAssign in std/complex.d
Adam D. Ruppe
destructionator at gmail.com
Sat Nov 9 13:26:12 UTC 2019
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.
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