需要解释：java中的三元运算符

发布于 2024-12-05 06:35:32 字数 655 浏览 0 评论 0原文

有问题的行是 return pFile.exists() ？ true：空；。由于它不会引发任何编译错误，对此有何解释。它最终筹集了NPE。

import java.io.File;
public class Main {
  public static void main(String... args) {
    boolean accept = accept(new File(""));
    System.out.println("accept = " + accept);
  }
  public static boolean accept(File pFile) {
    System.out.println(pFile.exists()); // prints: false, so pFile is not null
    return pFile.exists() ? true : null; //this line should throw compilation error
  }
}

pFile 不为 null；如您所见，一个 File 被实例化。但显然该文件不存在。问题不在于pFile。我对运算符如何处理 null 感兴趣。

原文

The line in question is return pFile.exists() ? true : null;. As it does not raise any compilation error, what is the explanation for this. It ended up raising NPE.

import java.io.File;
public class Main {
  public static void main(String... args) {
    boolean accept = accept(new File(""));
    System.out.println("accept = " + accept);
  }
  public static boolean accept(File pFile) {
    System.out.println(pFile.exists()); // prints: false, so pFile is not null
    return pFile.exists() ? true : null; //this line should throw compilation error
  }
}

pFile is not null; a File is instantiated as you can see. But obviously the file is not there. The question is not about pFile. I am interested in how the operator is dealing with null.

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信仰 2024-12-12 06:35:32

您的代码等效于：

public static boolean accept(File pFile) {
    System.out.println(pFile.exists()); // prints: false, so pFile is not null
    Boolean tmp = pFile.exists() ? true : null;
    return (boolean) tmp;
}

换句话说，条件运算符的类型在这种情况下为boolean，然后将其拆箱以返回boolean。当null被拆箱时，您会得到异常。

来自第15.25节 Java语言规范的第15.25节

否则，第二和第三操作数分别为S1和S2类型。令T1为将拳击转换转换为S1而产生的类型，然后将T2作为将拳击转换为S2施加的类型。条件表达式的类型是将捕获转换（§5.1.10）应用于lub（T1，T2）的结果（§15.12.2.7）。

我相信在这里适用的情况，尽管我会授予的那样清楚。

You code is equivalent to:

public static boolean accept(File pFile) {
    System.out.println(pFile.exists()); // prints: false, so pFile is not null
    Boolean tmp = pFile.exists() ? true : null;
    return (boolean) tmp;
}

On other words, the type of the conditional operator is Boolean in this case, and then the value is being unboxed to return a boolean. When null is unboxed, you get an exception.

From section 15.25 of the Java Language Specification:

Otherwise, the second and third operands are of types S1 and S2 respectively. Let T1 be the type that results from applying boxing conversion to S1, and let T2 be the type that results from applying boxing conversion to S2. The type of the conditional expression is the result of applying capture conversion (§5.1.10) to lub(T1, T2) (§15.12.2.7).

I believe that's the case that's applicable here, although I'll grant it's not as clear as it might be.

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