opMul

[Denton Cockburn]

On Mon, 03 Mar 2008 15:08:55 +0100, Simen Kjaeraas wrote:

> On Mon, 03 Mar 2008 04:10:36 +0100, Denton Cockburn <diboss at hotmail.com>  
> wrote:
>> A function does not have to be constant to allow a constant parameter.
>> It just has to promise not to change the parameter.
> 
> And that is exactly what const does.
> 
>> I don't understand why opMul would change the parameter, or more
>> importantly, why it can't be stated that it WONT change the parameter in
>> any cases.
> 
> Given
> 
> struct Foo
> {
> 	int bar;
> 	void doStuff(Foo b)
> 	{
> 		bar += b.bar;
> 	}
> }
> 
> would you expect
> 
> const Foo f,g;
> f.doStuff(g);
> 
> to work without problems? This is exactly the same, only doStuff is not  
> considered special by the compiler.
> 
> You might argue that the compiler should enforce opMul, opDiv, etc. to be  
> const, but that does carry a load of other troubles with it.
> 
>> I think this is related to what Janice spoke about.  The constant changes
>> that have taken place need to be properly incorporated into the language
>> and library.  There are a lot of places where constant doesn't work as
>> would be intuitively expected.
> 
> This is intuitive to me. (Well, apart from what I mentioned in the  
> response to Bill :p)

You are unfortunately completely missing my point.
In the example you give, you don't want a const parameter, because you
change the object.

I WANT a const parameter, because I have NO INTENTION of changing the
parameter.

so what I want is this:

struct Foo
 {
 	int bar;
 	void doStuff(const Foo b)
 	{
 		bar += b.bar; /* I want an error here */
 	}

	/* now more specifically, I want to do the same as above...but with */
	/* opMul */

	Foo opMul(const Foo b)
	{
		Foo res;
		res.bar = bar * b.bar;

		return res;
	}
 }

See, I make NO changes to the object in that opMul, so my question
is...why is that illegal (intuitively)?