## digitalmars.D - You know, it's just occurred to me...

- Janice Caron (17/17) Dec 02 2007 There's a general expectation in D that for all T, T[] is an array of
- Derek Parnell (7/10) Dec 02 2007 "Walter says that's not going to change. You just have to deal with it."...
- Leandro Lucarella (10/10) Dec 02 2007 Please, use more descriptive subjects.
- mandel (22/46) Dec 02 2007 I haven't tried the new const yet (using 2.008),
- Walter Bright (8/30) Dec 02 2007 Why do you think it does?
- Sean Kelly (6/43) Dec 03 2007 I think what Janice is saying is that if const(int)[] is an array of

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. :-)

Dec 02 2007

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

Dec 02 2007

Please, use more descriptive subjects. Thank you. -- Leandro Lucarella (luca) | Blog colectivo: http://www.mazziblog.com.ar/blog/ ---------------------------------------------------------------------------- GPG Key: 5F5A8D05 (F8CD F9A7 BF00 5431 4145 104C 949E BFB6 5F5A 8D05) ---------------------------------------------------------------------------- Do not get mad with others Because they know more than you It is not their fault

Dec 02 2007

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 }

Dec 02 2007

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.

Dec 02 2007

Walter Bright wrote:Janice Caron wrote: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. SeanThere'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.

Dec 03 2007