• simendsjo (13/13) Sep 24 2011 Sorry about the possible double post. I cannot see my previous..
• Andrej Mitrovic (9/14) Sep 24 2011 I've reported the error message just recently because it's very
• Jonathan M Davis (4/24) Sep 24 2011 Property functions are still functions. A member function must be const ...
• Tobias Pankrath (2/5) Sep 25 2011 Since we have transitive const, why can't the compiler deduce which
• Jonathan M Davis (15/20) Sep 25 2011 You mean, why can't it just figure out whether a particular function can...
• Steven Schveighoffer (6/19) Sep 26 2011 If this is the error message it is a bug.

simendsjo <simendsjo gmail.com> writes:

Sorry about the possible double post. I cannot see my previous..

struct S {
      property int B() {
         return 1;
     }
}

void main() {
     S s1;
     auto a1 = s1.B; // ok
     const(S) s2;
     auto a2 = s2.B; // Error: function t.S.B () is not callable using 
argument types ()
}

Andrej Mitrovic <andrej.mitrovich gmail.com> writes:

On 9/24/11, simendsjo <simendsjo gmail.com> wrote:
 struct S {
       property int B() {
          return 1;
      }
 }

I've reported the error message just recently because it's very
uninformative. But the solution is to add const to your property
function:

struct S {
      property int B() const {
         return 1;
     }
}

Jonathan M Davis <jmdavisProg gmx.com> writes:

On Saturday, September 24, 2011 20:19:43 Andrej Mitrovic wrote:
 On 9/24/11, simendsjo <simendsjo gmail.com> wrote:
 struct S {
 
       property int B() {
      
          return 1;
      
      }
 
 }

 
 I've reported the error message just recently because it's very
 uninformative. But the solution is to add const to your property
 function:
 
 struct S {
       property int B() const {
          return 1;
      }
 }

Property functions are still functions. A member function must be const for it 
to be callable on a const object. But that error message is horrible.

- Jonathan M Davis

Tobias Pankrath <tobias pankrath.net> writes:

 
 Property functions are still functions. A member function must be const
 for it to be callable on a const object.

Since we have transitive const, why can't the compiler deduce which 
methods can be called on const instances? 

Jonathan M Davis <jmdavisProg gmx.com> writes:

On Sunday, September 25, 2011 12:10:42 Tobias Pankrath wrote:
 Property functions are still functions. A member function must be const
 for it to be callable on a const object.

 
 Since we have transitive const, why can't the compiler deduce which
 methods can be called on const instances?

You mean, why can't it just figure out whether a particular function can be 
const or not? Well technically, it probably could, but that complicates the 
compiler, and there are cases where you want to enforce that a particular 
function be const. If it were always inferred, you couldn't do that.

D _does_ now have inferrence for pure, nothrow, and  safe for templated 
functions, because it has to have that in order to be able to reasonably use 
pure, nothrow, or  safe with templated functions (since whether a templated 
function can be pure, nothrow, or  safe often depends on the template's 
arguments). But even with that added to the language, the compiler doesn't 
infer any of those attributes for non-templated functions. The programmer 
should be able to figure that out and decide whether they want them to be const 
or pure or whatever in those cases. The inference is only done when it's done, 
because it's needed..

- Jonathan M Davis

"Steven Schveighoffer" <schveiguy yahoo.com> writes:

On Sat, 24 Sep 2011 14:10:48 -0400, simendsjo <simendsjo gmail.com> wrote:

 Sorry about the possible double post. I cannot see my previous..

 struct S {
       property int B() {
          return 1;
      }
 }

 void main() {
      S s1;
      auto a1 = s1.B; // ok
      const(S) s2;
      auto a2 = s2.B; // Error: function t.S.B () is not callable using  
 argument types ()
 }

If this is the error message it is a bug.

It should say:

function t.S.B () is not callable using argument types () const

I believe this is the message you'd get if it were a member function.

-Steve