CODEWITHSUNDEEP
operator-overloadingd
Abstract type
Abstract type

Reputation: 1921

Does D automatically rewrite opBinary to opOpAssign?

for example, assuming that T implements the right operator overloads:

T t1, t2, t3;
t3 = t1 + t2; // t3.opAssign(t1.opBinary!"+"(t2)) for sure
t3 = t3 + t2; // rewritten to t3.opOpAssign!"+"(t2) ?

Is the last operation optimized by D ?

Upvotes: 3

Views: 151

Answers (2)

Kozzi11
Kozzi11

Reputation: 2413

No, it is not. It is not possible, because opBinary and opOpAssign could have different semantic:

struct S
{
    int val = 5;

    S opBinary(string op)(S rhs) if (op == "+")
    {
        return S(val + rhs.val);
    }

    void opOpAssign(string op)(S rhs) if (op == "+")
    {
        val = val - rhs.val;
    }
}

void main()
{
    import std.stdio;

    S s1, s2, s3;
    writeln(s3); // S(5)

    s3 = s3 + s2;
    writeln(s3); // S(10)

    s3.val = 5;
    writeln(s3); // S(5)

    s3 += s2;
    writeln(s3); // S(0)
}

Upvotes: 3

Vladimir Panteleev
Vladimir Panteleev

Reputation: 25187

What is opOpBinary? Did you mean opOpAssign?

And no, it doesn't. For example, appending (~=) and concatenating (~) arrays are different operations (the former preallocates extra space at the end and the latter will always reallocate).

Upvotes: 2

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