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FUNvoidN<...> implementieren #157

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opened 2015-03-03 15:38:34 +00:00 by pl · 2 comments
pl commented 2015-03-03 15:38:34 +00:00
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Als Return-Typ eines Lambda-Ausdrucks kann void derzeit nicht inferiert werden.

Als Return-Typ eines Lambda-Ausdrucks kann void derzeit nicht inferiert werden.
stan commented 2015-03-03 16:05:35 +00:00
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Lösungsmöglichkeit:

Einführen einer neuen Reihe von Interfaces: FunVoidN<N1,...Nn>{ void apply(N1 n1, ... , Nn nn);}

Lösungsmöglichkeit: Einführen einer neuen Reihe von Interfaces: FunVoidN<N1,...Nn>{ void apply(N1 n1, ... , Nn nn);}
stan commented 2016-11-15 23:15:52 +00:00
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FunVoidN wurde eingeführt. siehe commit 388685d3b2

FunVoidN wurde eingeführt. siehe commit 388685d3b2e54463fb2e1d1e780db224e34968e4
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