使用的方法引用具有返回類型Integer。String但在下面的示例中，不兼容是允許的。如何修復方法with聲明以獲得方法引用類型安全而無需手動轉換？import java.util.function.Function;public class MinimalExample { static public class Builder<T> { final Class<T> clazz; Builder(Class<T> clazz) { this.clazz = clazz; } static <T> Builder<T> of(Class<T> clazz) { return new Builder<T>(clazz); } <R> Builder<T> with(Function<T, R> getter, R returnValue) { return null; //TODO } } static public interface MyInterface { Integer getLength(); } public static void main(String[] args) {// missing compiletimecheck is inaceptable: Builder.of(MyInterface.class).with(MyInterface::getLength, "I am NOT an Integer");// compile time error OK: Builder.of(MyInterface.class).with((Function<MyInterface, Integer> )MyInterface::getLength, "I am NOT an Integer");// The method with(Function<MinimalExample.MyInterface,R>, R) in the type MinimalExample.Builder<MinimalExample.MyInterface> is not applicable for the arguments (Function<MinimalExample.MyInterface,Integer>, String) }}
4 回答
子衿沉夜
TA貢獻1828條經驗 獲得超3個贊
在第一個示例中，和MyInterface::getLength分別幫助"I am NOT an Integer"解析通用參數T和。RMyInterfaceSerializable & Comparable<? extends Serializable & Comparable<?>>
// it compiles since String is a Serializable
Function<MyInterface, Serializable> function = MyInterface::getLength;
Builder.of(MyInterface.class).with(function, "I am NOT an Integer");
MyInterface::getLength并不總是 a ，Function<MyInterface, Integer>除非您明確這么說，這會導致編譯時錯誤，如第二個示例所示。
// it doesn't compile since String isn't an Integer
Function<MyInterface, Integer> function = MyInterface::getLength;
Builder.of(MyInterface.class).with(function, "I am NOT an Integer");
翻過高山走不出你
TA貢獻1875條經驗 獲得超3個贊
類型推斷在這里發揮了作用。R考慮方法簽名中的泛型：
<R> Builder<T> with(Function<T, R> getter, R returnValue)
在所列情況中：
Builder.of(MyInterface.class).with(MyInterface::getLength, "I am NOT an Integer");
的類型R被成功推斷為
Serializable, Comparable<? extends Serializable & Comparable<?>>
并且 aString確實通過這種類型暗示，因此編譯成功。
要顯式指定 的類型R并找出不兼容性，只需將代碼行更改為：
Builder.of(MyInterface.class).<Integer>with(MyInterface::getLength, "not valid");
紫衣仙女
TA貢獻1839條經驗 獲得超15個贊
這是因為你的泛型類型參數R可以被推斷為 Object，即以下編譯：
Builder.of(MyInterface.class).with((Function<MyInterface, Object>) MyInterface::getLength, "I am NOT an Integer");
夢里花落0921
TA貢獻1772條經驗 獲得超6個贊
這個答案基于其他答案，這些答案解釋了為什么它不能按預期工作。
解決方案
下面的代碼通過將雙函數“with”拆分為兩個連貫函數“with”和“returning”來解決該問題：
class Builder<T> {
...
class BuilderMethod<R> {
final Function<T, R> getter;
BuilderMethod(Function<T, R> getter) {
this.getter = getter;
}
Builder<T> returning(R returnValue) {
return Builder.this.with(getter, returnValue);
}
}
<R> BuilderMethod<R> with(Function<T, R> getter) {
return new BuilderMethod<>(getter);
}
...
}
MyInterface z = Builder.of(MyInterface.class).with(MyInterface::getLength).returning(1L).with(MyInterface::getNullLength).returning(null).build();
System.out.println("length:" + z.getLength());
// YIPPIE COMPILATION ERRROR:
// The method returning(Long) in the type BuilderExample.Builder<BuilderExample.MyInterface>.BuilderMethod<Long> is not applicable for the arguments (String)
MyInterface zz = Builder.of(MyInterface.class).with(MyInterface::getLength).returning("NOT A NUMBER").build();
System.out.println("length:" + zz.getLength());
（有點陌生）
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