Hi all,
I'm sure, I didn't find something obvious, but:

Given this:

´´´
void main()
{
	auto s = S();
	s.operator;
	assert(s.myOp(42));
	assert(42 in s);

	auto sptr = new S();
	sptr.operator;
	assert(sptr.myOp(42));
	//assert(42 in sptr);  //<-- does not compile
}

struct S
{
	void operator() const
	{
		assert(true);
	}

	bool opBinaryRight(string op)(size_t input) const if(op == "in")
	{
		return true;
	}

	bool myOp(size_t input)
	{
		return input in this;
	}
}
´´´

The last line in the main does not compile with the message
source/app.d(9,9): Error: incompatible types for `(42) in (sptr)`: `int` and `S*`

This behaves differently, w.r.t. to an arbitrary method, like "operator". Why? Is there any workaround?

On 5/11/18 8:53 AM, Alex wrote:

> 
> This behaves differently, w.r.t. to an arbitrary method, like "operator". Why? Is there any workaround?

operators don't follow pointers.

Imagine if you had a struct that overloads "+" and then you wanted to use pointer arithmetic, but instead it called ptr.opBinary.

The workaround is to dereference the pointer. e.g. 42 in *sptr;

-Steve

On Friday, 11 May 2018 at 15:24:08 UTC, Steven Schveighoffer wrote:
> On 5/11/18 8:53 AM, Alex wrote:
>
>> 
>> This behaves differently, w.r.t. to an arbitrary method, like "operator". Why? Is there any workaround?
>
> operators don't follow pointers.
>
> Imagine if you had a struct that overloads "+" and then you wanted to use pointer arithmetic, but instead it called ptr.opBinary.

Ah!

> The workaround is to dereference the pointer. e.g. 42 in *sptr;
>
> -Steve

Thanks a lot!