No need opUnary

[j]

On Sunday, 18 December 2022 at 16:21:05 UTC, Salih Dincer wrote:
> Don't you think it's interesting that it doesn't need unary 
> operator overloading?
>
> ```d
> import std.stdio;
>
> struct S
> {
>   int value;
>
>   alias opCall this;
>   this(int i) {
>     value = i;
>   }
>
>   alias opAssign = opCall;
>   @property opCall(int x) {
>     return value = x;
>   }
>
>   @property opCall() inout {
>     return value;
>   }
>
>   @property opOpAssign(string op)(int x) {
>     write(":"); // came here before
>     mixin("return value"~op~"=x;");
>   }
>   // no need: opUnary(string op)();
> }
>
> void main()
> {
>
>   S a = S(10),
>     b = S(-1);
>
>   writeln(a + b); // 9
>   writeln(++a + b); // :10
>
>   a += 10; // :
>
>   assert(a == 21);
>   writeln("\n--");
>
>   writeln(-b); // 1
> }
> ```
> SDB at 79



Why are you using @property everywhere?