How to translate C# 'readonly' keyword into D

[simendsjo]

On Monday, 4 February 2013 at 09:02:31 UTC, o3o wrote:
> I'm a C# programmer, when I apply IoC pattern  I use "readonly" 
> keyword 
> (http://msdn.microsoft.com/en-us/library/acdd6hb7%28v=vs.71%29.aspx) 
> in this manner:
>
> :// C# code
> :interface IFoo {
> :  void Fun();
> :}
> :
> :class Foo: IFoo {
> :  void Fun() {...}
> :}
> :class Bar {
> :  private readonly IFoo foo;
> :  // inject IFoo into Bar
> :  Bar(IFoo foo) {
> :    // assert(foo != null);
> :    this.foo = foo;
> :  }
> :  void Gun() {
> :    // foo = new Foo(); //// ERROR: foo is readonly!
> :    foo.Fun();
> :  }
> :}
>
> Can someone help me to translate "readonly IFoo foo;" so that 
> the dmd compiler raises an error when I write "foo = new 
> Foo();" ?
>
> I try:
> :// D code
> :interface IFoo {
> :   void fun();
> :}
> :
> :class Foo: IFoo {
> :   void fun() {
> :      writeln("fun...");
> :   }
> :}
> :
> :class Bar {
> :   private const IFoo service;
> :   this(const IFoo service) {
> :      this.service = service;
> :   }
> :
> :   void gun() {
> :      service.fun();
> :   }
> :}
> :unittest {
> :  const(IFoo) s = new Foo;
> :  auto bar = new Bar(s);
> :  bar.gun();
> :}
> but the compiler complains:
> Error: function main.IFoo.fun () is not callable using argument 
> types () const

In D, everything accessible from const is also const. You 
specifically state that you will not modify your IFoo instance. 
But then you call foo(). foo() in turn isn't marked as const, so 
this can modify IFoo, and thus break const.
So.. Every method you call through a const instance must also be 
const, otherwise you have the ability to change something that 
should be a constant.