context-free grammar

[bearophile]

Simen kjaeraas:

> Well, obviously not. The grammar has one and only one meaning for that
> example - that of an a* called b, being set to c. This can be inferred
> with no other context.

This little program:

struct Foo {
    int x;
    Foo opBinary(string op:"*")(Foo other) {
        Foo result = Foo(x * other.x);
        return result;
    }
    void opAssign(Foo other) {
        x = other.x;
    }
}
void main() {
    Foo a, b, c;
    a * b = c;
}

Gives:
test.d(10): Error: a is used as a type
test.d(10): Error: cannot implicitly convert expression (c) of type Foo to _error_*
test.d(10): Error: declaration test.main.b is already defined


While this one gives no errors:

struct Foo {
    int x;
    Foo opBinary(string op:"*")(Foo other) {
        Foo result = Foo(x * other.x);
        return result;
    }
    void opAssign(Foo other) {
        x = other.x;
    }
}
void main() {
    Foo a, b, c;
    (a * b) = c;
}

Bye,
bearophile