On Mar 7, 11 15:16, %u wrote:
That's essentially the example that's been under discussion - though in this
case it's a ref instead of
a temporary for the lvalue. Regardless, it's context free because a * b is by
definition a variable
declaration, not a call to the multiplication operator. If you want it to use
the multiplication
operator, then use parens: (a * b) = b. It's context free, because it just
assumes one of the two and
it's _always_ that one, so there's no ambiguity. It is, _by definition_, a
variable declaration.
Oh, I see. So is multiplication being special-cased in the grammar, or is it
part of a more general rule
in the language?
Only statement of the form 'p*q=r' is "special-cased". Actually it's not
quite fair to say it is "special-cased", just because declarations
(which accepts 'p*q=r') are processed before expressions.
import std.stdio;
struct Foo {
ref Foo opBinary(string x) (int a) pure nothrow {
return this;
}
ref Foo opBinaryRight(string x) (int a) pure nothrow {
return this;
}
}
void main() {
Foo a;
int b;
a + 1 = a;
a * 1 = a;
1 + a = a;
1 * a = a;
a + b = a;
//a * b = a;
}
Also, opMul is on its way to deprecation. binaryOp should be used for
overloading the multiplication
operator.
Whoa! I did not know that; thanks.
