context-free grammar

[SiegeLord]

bearophile Wrote:

> 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

Yeah, and this:

struct Foo {
    
}

void main() {
    Foo a, b, c;
    (a * b) = c;
}

Gives an error. I don't see any problem here:

a * b; // always a pointer declaration
(a * b); // always a binary expression

-SiegeLord