Q: how to do AA of delegates keyed by int?

Re: Q: how to do AA of delegates keyed by int? (Mac)

Feb 02, 2005

Feb 03, 2005

Re: how to do AA of delegates keyed by int?

Feb 01, 2005

Help? I'm having trouble figuring out how to get a delegate out of an AA. void delegate() hash [int]; int key = 5; hash[key] = delegate void() {printf ("whee!\n");}; Now how do I go about getting that value out? void delegate() fn = (key in hash); doesn't work and neither does: void delegate()* fn = (key in hash); which according to the docs I would think is how it is supposed to work seeing how it's supposed to return a pointer to the value type. But the compiler says: "cannot implicitly convert expression key in hash of type int to void delegate()*" I give up. What am I doing wrong? Thanks in advance for putting up with my lameness. ;-)

In article <ctns13$8fu$1@digitaldaemon.com>, Brian Chapman says... > >Help? I'm having trouble figuring out how to get a delegate out of an AA. > > void delegate() hash [int]; > > int key = 5; > hash[key] = delegate void() {printf ("whee!\n");}; > >Now how do I go about getting that value out? > > void delegate() fn = (key in hash); > >doesn't work and neither does: > > void delegate()* fn = (key in hash); > This one works for me (dmd 0.111, winxp) >which according to the docs I would think is how it is supposed to work seeing how it's supposed to return a pointer to the value type. But the compiler says: > >"cannot implicitly convert expression key in hash of type int to void delegate()*" > >I give up. What am I doing wrong? > >Thanks in advance for putting up with my lameness. ;-) > ---------------- Carlos Santander

Brian Chapman wrote: > Help? I'm having trouble figuring out how to get a delegate out of an AA. > > void delegate() hash [int]; > int key = 5; > hash[key] = delegate void() {printf ("whee!\n");}; > > Now how do I go about getting that value out? Have you tried the 'normal' method? void delegate() fn = hash[key]; > void delegate() fn = (key in hash); > > doesn't work Correct, as the InExpression should be of type pointer to delegate. > and neither does: > > void delegate()* fn = (key in hash); > > which according to the docs I would think is how it is supposed to work seeing how it's supposed to return a pointer to the value type. But the compiler says: > > "cannot implicitly convert expression key in hash of type int to void a > delegate()*" <snip> This should work. It looks as if the in operator has a bug or two with determining its return type. Stewart. -- My e-mail is valid but not my primary mailbox. Please keep replies on the 'group where everyone may benefit.

In article <cto6bg$kkb$1@digitaldaemon.com>, Stewart Gordon says... >> "cannot implicitly convert expression key in hash of type int to void a >> delegate()*" ><snip> > >This should work. It looks as if the in operator has a bug or two with determining its return type. > Actually, its not a bug. The 'in' operator is a 'boolean' operation ("does this key exist in the map?"), so it returns 'int' (true/false actually). Also, Stewart's suggestion of "hash[key]" is the correct approach, however it does have the side-effect of generating an empty value for 'key' (the AA grows as a result) if the key doesn't already exist in the map. The result is 'fn' gets assigned null, but future checks to "key in hash" will return true, not false. The side-effect free code is like this: > void delegate fn; > if(key in hash){ > fn = hash[key]; > } - EricAnderton at yahoo

pragma wrote: > Actually, its not a bug. The 'in' operator is a 'boolean' operation ("does this > key exist in the map?"), so it returns 'int' (true/false actually). The return type of "in" was changed in DMD 0.107, from bit to pointer. http://www.digitalmars.com/d/changelog.html#new0107 > InExpressions now, instead of returning a bit, return a pointer to the > associative array element if the key is present, null if it is not. This > obviates the need for many double lookups. http://www.digitalmars.com/d/arrays.html#associative > The InExpression yields a pointer to the value if the key is in the > associative array, or null if not: > > int* p; > p = ("hello" in b); > if (p != null) > ... Not that D cares about the difference between false and null anyway. --anders

In article <cto98f$njs$1@digitaldaemon.com>, =?ISO-8859-1?Q?Anders_F_Bj=F6rklund?= says... > >pragma wrote: > >> Actually, its not a bug. The 'in' operator is a 'boolean' operation ("does this >> key exist in the map?"), so it returns 'int' (true/false actually). > >The return type of "in" was changed in DMD 0.107, from bit to pointer. > >http://www.digitalmars.com/d/changelog.html#new0107 >> InExpressions now, instead of returning a bit, return a pointer to the associative array element if the key is present, null if it is not. This obviates the need for many double lookups. Well, crap. Sorry if I mislead anyone. :( I'll read the changelog a little more closely for now on! - EricAnderton at yahoo

pragma wrote: >>The return type of "in" was changed in DMD 0.107, from bit to pointer. > Well, crap. Sorry if I mislead anyone. :( > > I'll read the changelog a little more closely for now on! On the contrary, your "no side-effects" code still works - and seems to be the only one not affected by new "in" bugs ? "if (key in hash)" still works, just as "if (object)" does. --anders

"Brian Chapman" <nospam-for-brian@see-post-for-address.net> wrote in message news:ctns13$8fu$1@digitaldaemon.com... > I give up. What am I doing wrong? I think you might be using an older DMD version. Try upgrading to 0.111.