Cannot compile program that contains method that has pointer type receiver

[shouji-kazuo]

Steps to reproduce:

1. Enter program list as below:

struct Foobar {
  member int
}

fn (f *Foobar) do() { // method of pointer type receiver
  println('I am foobar')
}

fn main() {
  f := Foobar{}
  f.do() // I think f is type of "Foobar", but I assume that f can treat as pointer type with implicit conversion ( like Golang )
} 

2. Click Run button
3. The output was shown as below.
   You just found a bug. V can't compile this program, but it should. Please create a GitHub issue.

[medvednikov]

The correct syntax is

struct Foobar {
	member int
}

fn (f mut Foobar) do() {
	// method of pointer type receiver
	println('I am foobar')
}

fn main() {
	f := Foobar{}
	f.do()// I assume f is type of "Foobar", but can treat as pointer type with implicit conversion ( like Golang )
}

[medvednikov]

(f mut Foobar) instead of (f *Foobar)

I'll update the compiler so that (f *Foobar) doesn't compile.

[medvednikov]

I assume f is type of "Foobar", but can treat as pointer type with implicit conversion ( like Golang )

This assumption is correct :)

[shouji-kazuo]

OK, I understood.
ありがとうございました m(_ _)m

[medvednikov]

Fixed.

You should now get use (f mut Foobar) instead of (f *Foobar)

[shouji-kazuo]

I have confirmed .
thanks a lot!

[ntrel]

f is type of "Foobar", but can treat as pointer type with implicit conversion

I don't think letting pointers implicitly dereference is a good idea, and it seems to go against the simple philosophy of V:

mut i := 5
p := &i
i = p // confusing, should be an error
i = *p // OK

Edit: See #61