为什么这个程序可以用 Java 7 而不是 Java 8 编译? [英] Why does this program compile with Java 7 but not Java 8?

问题描述

考虑这个程序:

public class xx<T> {

    <T> Iterable<T> createIterable(Class<T> cls) {
        return null;
    }

    Iterable<? extends Number> createNumberIterable(boolean floatingPoint) {
        return this.createIterable(floatingPoint ? Integer.class : Float.class);
    }
}

在 Java 7 下编译:

Under Java 7 it compiles:

$ java -version
java version "1.7.0_45"
Java(TM) SE Runtime Environment (build 1.7.0_45-b18)
Java HotSpot(TM) 64-Bit Server VM (build 24.45-b08, mixed mode)
$ javac xx.java
$

在 Java 8 下没有:

Under Java 8 it does not:

$ java -version
java version "1.8.0_40"
Java(TM) SE Runtime Environment (build 1.8.0_40-b25)
Java HotSpot(TM) 64-Bit Server VM (build 25.40-b25, mixed mode)
$ javac xx.java
xx.java:8: error: method createIterable in class xx<T#2> cannot be applied to given types;
        return this.createIterable(floatingPoint ? Integer.class : Float.class);
                   ^
  required: Class<T#1>
  found: floatingPo[...]class
  reason: inferred type does not conform to equality constraint(s)
    inferred: Float
    equality constraints(s): Float,Integer
  where T#1,T#2 are type-variables:
    T#1 extends Object declared in method <T#1>createIterable(Class<T#1>)
    T#2 extends Object declared in class xx
1 error
$

这是真的:

1. 这是 Java 7 中的错误，在 Java 8 中得到修复(编译器过于宽容)；或
2. 这是 Java 8 中引入的一个新错误

如果答案是 #1，你能用普通语言解释为什么 JLS 不允许这样做，使用明显的解释吗?

If the answer is #1, can you explain in normal language the reason why the JLS doesn't allow this, using the obvious interpretation?

(注意:请不要解释如何解决问题，这不是问题)

(Note: please don't explain how to workaround the problem, that's not the question)

推荐答案

旧行为不是错误，新行为也不是错误.条件表达式类型的规则变得更加复杂，这在很多情况下都有帮助，而对您的情况并没有真正的伤害.

Neither was the old behavior a bug, nor is the new behavior a bug. The rules for the type of the conditional expression just got more complex, which helps in many cases and doesn't really hurt in yours.

编译器不允许，因为Integer.class 和Float.class 的类型不可比.没有类型 T 会使 Class 成为 Class 和 Class<的超类型/代码>.

The compiler does not allow it because the types of Integer.class and Float.class are incomparable. There is no type T which would make Class<T> a supertype of both Class<Integer> and Class<Float>.

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