opMul

[Bill Baxter]

Denton Cockburn wrote:
> On Tue, 04 Mar 2008 05:56:57 +0900, Bill Baxter wrote:
> 
>> Denton Cockburn wrote:
> 
>>> so what I want is this:
>>>
>>> struct Foo
>>>  {
>>>  	int bar;
>>>  	void doStuff(const Foo b)
>>>  	{
>>>  		bar += b.bar; /* I want an error here */
>>>  	}
>> You're not changing b, so you're not going to get an error there.
> 
> You're right, that was a mistake.
> 
> I misphrased my whole query (I misread the compiler error message).
> The concern is still there though, here's a sample of code that produces
> the message.
> 
> struct Foo
> {
> 	int x;
> 	
> 	Foo opMul(const Foo b)
> 	{
> 		Foo f;
> 		f.x = x * b.x;
> 		
> 		return f;
> 	}
> }
> 
> void main()
> {
> 	Foo f;
> 	Foo y;
> 	f.x = 6;
> 	y.x = 7;
> 	
> 	const(Foo) t = f;
> 	Foo p = t * y;
> }
> 
> produces this error:
> test.d(24): function test.Foo.opMul (const(Foo)) does not match parameter types (Foo)
> test.d(24): Error: t.opMul can only be called on a mutable object, not const(Foo)
> 
> 
> So my question SHOULD have been:
> 
> Why can't opMul et al be called on a constant object (reference)?
> 
> Sorry for the previous confusion.

Ok *that's* because saying the parameter is const doesn't mean the 
'this' pointer is const.

To make the hidden 'this' parameter const too you need to use

  const Foo opMul(const Foo b) {...}

That const out front may look like it's modifying the return value, but 
its not.  That would be

  const(Foo) opMul(const Foo b) {...}


--bb