Re: immuable method address and export method address.

[deadalnix]

On Sunday, 5 May 2013 at 19:34:34 UTC, Igor Stepanov wrote:
Hello, I've two simple questions:
1. I have a structure:

struct Foo
{
  public int bar() const {return 0;}
  public int bar() immutable {return 1;}
}

How can I get immutable bar address?

When I've wrote next code I've got a const bar address. Is it a 
bug?

void main()
{
    immutable Foo boo;
    int delegate() immutable dg = &boo.bar;
}


Yes. First, the immutable keyword before dg don't do what you 
think in D. Right now this is an hole in the type 
system/spec/implementation (pick one). But boo being immutable it 
should return the immutable delegate anyway. So you have two bugs 
in one. Congrat !

Hopefully, I've been able to discuss this with Andrei at DConf, 
who agreed on the problem and we were able to discuss solutions.

Next question:
If I write туче code, it'll be builded successfully

void main()
{
    foreach(cur; __traits(getOverloads, Foo, "bar"))
    {
        void* p = &cur;
        writeln(cur); //prints a function pointer.
    }
}

If I change protection of bar to export I've got a error:

struct Foo
{
  export int bar() const {return 0;}
  export int bar() immutable {return 1;}
}

void main()
{
    foreach(cur; __traits(getOverloads, Foo, "bar"))
    {
        void* p = &cur; //Error: need 'this' to access member 
bar
        writeln(cur);
    }
}

How should it be?

The method aren't static, so you should need this all the time. I 
can understand both behavior, but it is clearly inconsistent. We 
should pick one and stick to it.