There's a general expectation in D that for all T, T[] is an array of T. Well, it's not always true... See,

    const(int)[] a;

does not declare a to be an array of "const(int)" at all. It declares
a to be an array of "const int" (without the parentheses). The
difference is that (under the peculiar syntax we have right now with
D2.008), "const(int)" is mutable (!), wheras "const int" is const - as
demonstrated by the following code.

    const(int) x;
    const int y;
    x = 1; // OK
    y = 2; // Error

Yet we write "const(int)[]", not "(const int)[]". Huh?

So, not only is my expectation that "const(X)" should mean the same thing as "const X" confounded, but now it turns out that my expectation that "T[]" means "array of T" is also confounded.

You can just imagine how easy this is going to be to explain to newbies. :-)

On Sun, 2 Dec 2007 21:10:25 +0000, Janice Caron wrote:

> There's a general expectation in D that for all T, T[] is an array of T. Well, it's not always true...

> ... my expectation that "const(X)" should mean the same thing as "const X" ...

"Walter says that's not going to change. You just have to deal with it." - Janice Caron


-- 
Derek Parnell
Melbourne, Australia
skype: derek.j.parnell

Please, use more descriptive subjects.

Thank you.

-- 
Leandro Lucarella (luca) | Blog colectivo: http://www.mazziblog.com.ar/blog/
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On Sun, 02 Dec 2007 21:10:25 +0000, Janice Caron wrote:

> There's a general expectation in D that for all T, T[] is an array of T. Well, it's not always true... See,
> 
>     const(int)[] a;
> 
> does not declare a to be an array of "const(int)" at all. It declares a
> to be an array of "const int" (without the parentheses). The difference
> is that (under the peculiar syntax we have right now with D2.008),
> "const(int)" is mutable (!), wheras "const int" is const - as
> demonstrated by the following code.
> 
>     const(int) x;
>     const int y;
>     x = 1; // OK
>     y = 2; // Error
> 
> Yet we write "const(int)[]", not "(const int)[]". Huh?
> 
> So, not only is my expectation that "const(X)" should mean the same thing as "const X" confounded, but now it turns out that my expectation that "T[]" means "array of T" is also confounded.
> 
> You can just imagine how easy this is going to be to explain to newbies. :-)

I haven't tried the new const yet (using 2.008),
but I did some simple test and I found the results beeing reasonable.

const(int) is a restriction of the const to the int data of the array.
So we get less const protection;
Therefore array.ptr and array.length are mutable (allowing assignment).

void main()
{
	const(int) x;
	const int y;
	x = 1; // OK
	y = 2; // Error

	const(int)[] xx;
	const int[] yy;

	xx = null; //OK
	xx[0] = 1; //Error

	yy = null; //Error
	yy[0] = 1; //Error
}

Janice Caron wrote:
> There's a general expectation in D that for all T, T[] is an array of
> T. Well, it's not always true... See,
> 
>     const(int)[] a;
> 
> does not declare a to be an array of "const(int)" at all.

Yes, it does.

> It declares
> a to be an array of "const int" (without the parentheses).

Why do you think it does?

> The
> difference is that (under the peculiar syntax we have right now with
> D2.008), "const(int)" is mutable (!), wheras "const int" is const - as
> demonstrated by the following code.
> 
>     const(int) x;
>     const int y;
>     x = 1; // OK
>     y = 2; // Error
> 
> Yet we write "const(int)[]", not "(const int)[]". Huh?

There is the type, and the storage class. Const with parens sets the type, const without parens sets the storage class. That's why we don't write (const int).

> So, not only is my expectation that "const(X)" should mean the same
> thing as "const X" confounded, but now it turns out that my
> expectation that "T[]" means "array of T" is also confounded.

Why? The expectation is that const(int)[] is an array of ints that cannot be modified through the array reference, and it is.

Walter Bright wrote:
> Janice Caron wrote:
>> There's a general expectation in D that for all T, T[] is an array of
>> T. Well, it's not always true... See,
>>
>>     const(int)[] a;
>>
>> does not declare a to be an array of "const(int)" at all.
> 
> Yes, it does.
> 
>> It declares
>> a to be an array of "const int" (without the parentheses).
> 
> Why do you think it does?
> 
>> The
>> difference is that (under the peculiar syntax we have right now with
>> D2.008), "const(int)" is mutable (!), wheras "const int" is const - as
>> demonstrated by the following code.
>>
>>     const(int) x;
>>     const int y;
>>     x = 1; // OK
>>     y = 2; // Error
>>
>> Yet we write "const(int)[]", not "(const int)[]". Huh?
> 
> There is the type, and the storage class. Const with parens sets the type, const without parens sets the storage class. That's why we don't write (const int).
> 
>> So, not only is my expectation that "const(X)" should mean the same
>> thing as "const X" confounded, but now it turns out that my
>> expectation that "T[]" means "array of T" is also confounded.
> 
> Why? The expectation is that const(int)[] is an array of ints that cannot be modified through the array reference, and it is.

I think what Janice is saying is that if const(int)[] is an array of integers which cannot be modified, then why can the value "const(int) x" be modified?  There is an apparent lack of consistency with the meaning of "const(int)" between array and standalone declarations.


Sean