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LambdaTest2 funktioniert nicht #164

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opened 2015-03-05 17:56:21 +00:00 by pl · 4 comments
pl commented 2015-03-05 17:56:21 +00:00
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class LambdaTest<A,B>{

Fun1<Fun1<A, Fun2<A, LambdaTest<A,B>, B>>, B> op = (m) -> (f) -> f.apply(this,m);

}

Es kommt zu
[
(Fun1< Fun1< TPH A1399499405, Fun2< TPH A1399499405, LambdaTest< TPH A1399499405, TPH B1399499405 >, TPH B1399499405 > >, TPH B1399499405 > <. Fun1< Fun1< TPH A1399499405, Fun2< TPH A1399499405, LambdaTest< TPH A1399499405, TPH B1399499405 >, TPH B1399499405 > >, TPH B1399499405 >),

(TPH ABJ <. Fun1< Fun1< TPH A1399499405, Fun2< TPH A1399499405, LambdaTest< TPH A1399499405, TPH B1399499405 >, TPH B1399499405 > >, TPH B1399499405 >),

(Fun1< ? extends TPH ABN, ? super TPH C > <. TPH ABJ),
(TPH C <. TPH T232863545),
(TPH B <. Fun2< TPH R32863545, TPH T132863545, TPH T232863545 >),
(TPH R32863545 <. TPH ABL),
(TPH ABL <. TPH ABM),
(Fun1< ? extends TPH ABM, ? super TPH B > <. TPH ABK),
(TPH ABK <. TPH ABN),
(LambdaTest< TPH A32863545, TPH B32863545 > <. TPH T132863545)]

[
(TPH C <. TPH T232863545),
(TPH R32863545 <. TPH ABL),
(TPH ABL <. TPH ABM),
(Fun1< ? extends TPH ABM, ? super Fun2< TPH R32863545, LambdaTest< TPH A32863545, TPH B32863545 >, TPH T232863545 > > <. TPH ABN),

(TPH ABJ = Fun1< Fun1< TPH A1399499405, Fun2< TPH A1399499405, LambdaTest< TPH A1399499405, ? super TPH C >, ? super TPH C > >, ? super TPH C >),

(Fun1< TPH A1399499405, Fun2< TPH A1399499405, LambdaTest< TPH A1399499405, ? super TPH C >, ? super TPH C > > = ? extends TPH ABN),

(TPH B1399499405 = ? super TPH C),

(TPH B = Fun2< TPH R32863545, LambdaTest< TPH A32863545, TPH B32863545 >, TPH T232863545 >),

(TPH ABK = Fun1< ? extends TPH ABM, ? super Fun2< TPH R32863545, LambdaTest< TPH A32863545, TPH B32863545 >, TPH T232863545 > >),

(TPH T132863545 = LambdaTest< TPH A32863545, TPH B32863545 >)]

class LambdaTest<A,B>{ Fun1<Fun1<A, Fun2<A, LambdaTest<A,B>, B>>, B> op = (m) -> (f) -> f.apply(this,m); } Es kommt zu [ (Fun1< Fun1< TPH A1399499405, Fun2< TPH A1399499405, LambdaTest< TPH A1399499405, TPH B1399499405 >, TPH B1399499405 > >, TPH B1399499405 > <. Fun1< Fun1< TPH A1399499405, Fun2< TPH A1399499405, LambdaTest< TPH A1399499405, TPH B1399499405 >, TPH B1399499405 > >, TPH B1399499405 >), (TPH ABJ <. Fun1< Fun1< TPH A1399499405, Fun2< TPH A1399499405, LambdaTest< TPH A1399499405, TPH B1399499405 >, TPH B1399499405 > >, TPH B1399499405 >), (Fun1< ? extends TPH ABN, ? super TPH C > <. TPH ABJ), (TPH C <. TPH T232863545), (TPH B <. Fun2< TPH R32863545, TPH T132863545, TPH T232863545 >), (TPH R32863545 <. TPH ABL), (TPH ABL <. TPH ABM), (Fun1< ? extends TPH ABM, ? super TPH B > <. TPH ABK), (TPH ABK <. TPH ABN), (LambdaTest< TPH A32863545, TPH B32863545 > <. TPH T132863545)] [ (TPH C <. TPH T232863545), (TPH R32863545 <. TPH ABL), (TPH ABL <. TPH ABM), (Fun1< ? extends TPH ABM, ? super Fun2< TPH R32863545, LambdaTest< TPH A32863545, TPH B32863545 >, TPH T232863545 > > <. TPH ABN), (TPH ABJ = Fun1< Fun1< TPH A1399499405, Fun2< TPH A1399499405, LambdaTest< TPH A1399499405, ? super TPH C >, ? super TPH C > >, ? super TPH C >), (Fun1< TPH A1399499405, Fun2< TPH A1399499405, LambdaTest< TPH A1399499405, ? super TPH C >, ? super TPH C > > = ? extends TPH ABN), (TPH B1399499405 = ? super TPH C), (TPH B = Fun2< TPH R32863545, LambdaTest< TPH A32863545, TPH B32863545 >, TPH T232863545 >), (TPH ABK = Fun1< ? extends TPH ABM, ? super Fun2< TPH R32863545, LambdaTest< TPH A32863545, TPH B32863545 >, TPH T232863545 > >), (TPH T132863545 = LambdaTest< TPH A32863545, TPH B32863545 >)]
pl commented 2015-03-06 16:05:21 +00:00
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2.Schritt von TUnify ergibt

[
(TPH R1860513229 <. TPH ABJ),
(TPH B <. Fun2< TPH R1860513229, TPH T11860513229, TPH T21860513229 >),
(LambdaTest< GTV A, GTV B > <. TPH T11860513229),
(TPH C <. TPH T21860513229),
(Fun1< ? extends TPH ABL, ? super TPH C > <. TPH ABH),
(TPH ABI <. TPH ABL),
(Fun1< ? extends TPH ABK, ? super TPH B > <. TPH ABI),
(TPH ABJ <. TPH ABK),
(TPH ABH <. Fun1< Fun1< GTV A,
Fun2< GTV A, LambdaTest< GTV A, GTV B >, GTV B > >,
GTV B >)]

Daraus kommt es dann über TPH ABH zu:

(Fun1< ? extends TPH ABL, ? super TPH C > =.
(TPH ABH <. Fun1< Fun1< GTV A,
Fun2< GTV A, LambdaTest< GTV A, GTV B >, GTV B > >,
GTV B >)]

Das ergibt:

? extends TPH ABL =. Fun1< GTV A,
Fun2< GTV A, LambdaTest< GTV A, GTV B >, GTV B > >

und

? super TPH C =. GTV B

Beides nicht lösbar.

Evlt. muss bei TYPE in der Lambda-Regel FunN<? ext Retty, ? super Argtyp>
auch nich FunN<rettyp, Argtyp> eingefügt werden, da

Rettyp keine Instanz von ? ext Rettyp und
Argtype keine Instanz ? super Argtyp ist.

2.Schritt von TUnify ergibt [ (TPH R1860513229 <. TPH ABJ), (TPH B <. Fun2< TPH R1860513229, TPH T11860513229, TPH T21860513229 >), (LambdaTest< GTV A, GTV B > <. TPH T11860513229), (TPH C <. TPH T21860513229), (Fun1< ? extends TPH ABL, ? super TPH C > <. TPH ABH), (TPH ABI <. TPH ABL), (Fun1< ? extends TPH ABK, ? super TPH B > <. TPH ABI), (TPH ABJ <. TPH ABK), (TPH ABH <. Fun1< Fun1< GTV A, Fun2< GTV A, LambdaTest< GTV A, GTV B >, GTV B > >, GTV B >)] Daraus kommt es dann über TPH ABH zu: (Fun1< ? extends TPH ABL, ? super TPH C > =. (TPH ABH <. Fun1< Fun1< GTV A, Fun2< GTV A, LambdaTest< GTV A, GTV B >, GTV B > >, GTV B >)] Das ergibt: ? extends TPH ABL =. Fun1< GTV A, Fun2< GTV A, LambdaTest< GTV A, GTV B >, GTV B > > und ? super TPH C =. GTV B Beides nicht lösbar. Evlt. muss bei TYPE in der Lambda-Regel FunN<? ext Retty, ? super Argtyp> auch nich FunN<rettyp, Argtyp> eingefügt werden, da Rettyp keine Instanz von ? ext Rettyp und Argtype keine Instanz ? super Argtyp ist.
pl commented 2015-03-13 21:01:57 +00:00
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Ich denke hier ist an der falschen Stelle in TYPEExpr das ODER-Constraint eingefügt worden.

Es sollte bei FunN<? Ext Ty, ? Super TY'> oder FunN<Ty, TY'> nicht bei den Argumenten eingefügt werden.

Bitte nochmals Rücksprache!

Ich denke hier ist an der falschen Stelle in TYPEExpr das ODER-Constraint eingefügt worden. Es sollte bei FunN<? Ext Ty, ? Super TY'> oder FunN<Ty, TY'> nicht bei den Argumenten eingefügt werden. Bitte nochmals Rücksprache!
pl commented 2015-03-20 01:03:05 +00:00
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Folgendes wird implementiert in
TypeExpr in Lambdaexpression:

FunN<? ext rTy, ? super argTy> (x) -> g(x)

argTy der Typ von x ist und
wenn g(x): rTy oder g(x): ? ext rTy

bei g(x): ? super Ty => Fehler
bei argTyp:? super Ty => Fehler

Folgendes wird implementiert in TypeExpr in Lambdaexpression: FunN<? ext rTy, ? super argTy> (x) -> g(x) argTy der Typ von x ist und wenn g(x): rTy oder g(x): ? ext rTy bei g(x): ? super Ty => Fehler bei argTyp:? super Ty => Fehler
stan commented 2017-10-10 15:22:17 +00:00
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OLD BUG

OLD BUG
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