Re: how to define infix function

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On Saturday, 2 June 2018 at 23:17:48 UTC, Simen Kjærås wrote:
On Saturday, 2 June 2018 at 22:09:49 UTC, Neia Neutuladh wrote:
On Saturday, 2 June 2018 at 21:44:39 UTC, greatsam4sure wrote:
Sorry for the typo
is it possible to define infix function in D

3.min(5)// 3: where min is a function, works in D
3 min 5 // does not work.

thanks in advance

This is a horrible abuse of D's operator overloading 
discovered by FeepingCreature in the distant past.

You have to delimit your custom infix operator with slashes; 
you can't make `3 min 5` work, but you can make `3 /min/ 5` 
work.

Observe:

struct Min
{
    MinIntermediate!T opBinaryRight(string op, T)(T value) if 
(op == "/")
    {
        return MinIntermediate!T(value);
    }
}
struct MinIntermediate(T)
{
    T value;
    T opBinary(string op, T)(T value2) if (op == "/")
    {
        if (value < value2) return value;
        return value2;
    }
}
Min min;
void main()
{
    writeln(1 /min/ 2);
}

And of course, this can be generalized:

struct Operator(alias fn, string operator = "/")
{
    static auto opBinaryRight(string op : operator, T...)(T 
value1)
    {
        struct Result
        {
            auto opBinary(string op : operator, U...)(U value2)
                if (__traits(compiles, fn(value1, value2)))
            {
                return fn(value1, value2);
            }
        }

        Result result;
        return result;
    }
}

unittest
{
    import std.algorithm.comparison;

    alias min = Operator!(std.algorithm.comparison.min);

    assert(1 /min/ 3 == 1);
}

Note also the use of static opBinaryRight, allowing one to 
eschew the 'min' variable.

All of this said, I would suggest not using this in prod - it's 
a neat trick that shows off some of D's power, but I don't see 
a case where this would be easier to understand than a 
straightforward function call.

--
  Simen

That is interesting. I don't know yet whether it's good or bad, 
but certainly, it's interesting.