Explore tweets tagged as #binaryoperator
Why does Stream.reduce take BinaryOperator<T> rather than BiFunction<T, T, T>? https://t.co/6BrCuqAxFD
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@alioksuzdev Mantıklı, polymorphism kullanarak client kodundaki if-else yapısını elimine etmişsin. Ben de daha değişik bir yöntem kullandım, sırf deneek adına. Burada basit olsun diye IntBinaryOperator kullandım. BinaryOperator<T> şeklinde de tipten bağımsız yapılabilir.
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Java 11 Stream API - The best way to produce map with distinct values from a map (and use the right key using BinaryOperator) https://t.co/qGYiQ5Va0a
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What is the reason to create Stream.reduce(T identity, BinaryOperator<T> accumulator) while we have Stream.reduce(BinaryOperator<T> accumulator)? https://t.co/93zzRumeLy
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La interface funcional Reduce requiere un BinaryOperator, requiere el tipo de dato del stream y dos argumentos, el primero (sub) es el resultado de las operaciones y el segundo (elem) es el valor actual a ejecutar la operación. #Java
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Next up it's #UnknownToTheUnknown with #DAWK & #BinaryOperator on https://t.co/6hLtrMg0Aq + 106.8FM #RinseFM
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Explain the use of the BinaryOperator<T> interface. When would you choose BinaryOperator over BiFunction?
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<OperatedExpression> ::= <UnaryExpression> (<BinaryOperator> <UnaryExpression>)*
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そうかJavaって BinaryOperator extends BiFunction だから、メソッドの引数型に BinaryOperator を宣言してしまうと、BiFunction<T, T, T> をそのまま渡せないのか……
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BinaryOperator<String> sp = (l, r) -> l + r; String s = strings.stream().reduce("", sp); あ、出来た。
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JSONでプログラミング、考えたけど {iterator:’i’, for: {body: [ {return: “i”} ], conditional: binaryOperator: {left: 10, right: ‘i’, operator: ‘<’} }}みたいな
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ResultConsumerの組み合わせはBinaryOperatorで行われてて、今のところ既存→新規の順に組み合わせるものしか定義されていない 将来的に別の組み合わせ方を追加しやすいようにはなってる
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