• XavierAP (5/5) Mar 09 2017 The same way as T.foo() is lowered to foo(T) if no such member is
• Adam D. Ruppe (4/6) Mar 09 2017 No.

XavierAP <n3minis-git yahoo.es> writes:

The same way as T.foo() is lowered to foo(T) if no such member is 
defined inside the type. It would allow me to extend 3rd party 
types with operator notation without wrapping them.

After trying and reading the specification, looks like nuts, but 
just wanted to confirm. Thx

Adam D. Ruppe <destructionator gmail.com> writes:

On Thursday, 9 March 2017 at 23:50:04 UTC, XavierAP wrote:
 The same way as T.foo() is lowered to foo(T) if no such member 
 is defined inside the type.

No.

But wrapping in a struct with alias this lets you extend them 
pretty easily too.