首页以下Java方法将返回什么值?public static int calculate(int a, int b)try return a 了 catch (Arithmeticbxception e System.out.println("Exception caught!");return -1； B finallyí System.out.println("Finally block executed!"); public static void main(String[] args)1 int result = calculate(10, 0); System.out.println("Result: "+ wesult); Result:-1 Exception caught! Followed by Result:-1 Finally block executed Followed by Result:-1 Exception caught! Followed by Finally block executed! Followed by Result: -1

以下Java方法将返回什么值?public static int calculate(int a, int b)try return a 了 catch (Arithmeticbxception e System.out.println("Exception caught!");return -1； B finallyí System.out.println("Finally block executed!"); public static void main(String[] args)1 int result = calculate(10, 0); System.out.println("Result: "+ wesult); Result:-1 Exception caught! Followed by Result:-1 Finally block executed Followed by Result:-1 Exception caught! Followed by Finally block executed! Followed by Result: -1

时间: 2025-04-04 10:06:05 浏览: 19

从给定的问题来看，这里涉及到 Java 中异常处理的关键字 `try-catch-finally` 的工作机制。我们先逐步分析所提供的代码及其运行结果： ```java public static int calculate(int a, int b){ try{ return a / b; // 这里会发生 ArithmeticException 因为 b=0; }catch (ArithmeticException e){ System.out.println("Exception caught!"); return -1； // 捕获到异常之后，此处将返回-1. } finally { System.out.println("Finally block executed!"); // 不管有没有发生异常，finally块都会被执行。 } } ``` #### 具体流程解析： 1. 调用了方法 `calculate(10, 0)` 。由于除数b为0，这一步会产生一个算术异常 (`ArithmeticException`)。 2. 异常被捕获到了 catch 块内，所以首先打印出 `"Exception caught!"` 然后准备返回 `-1`. 3. 关键点在于 **无论是否抛出了异常或者已经决定了返回值，`finally` 块总是会被执行**。所以在实际返回之前，“Finally block executed!” 将被输出至控制台。 4. 最终整个函数返回的是在捕获异常后的指定数值 `-1`. ### 所以最终程序输出将是这样的顺序： ``` Exception caught! Finally block executed! Result: -1 ``` ---

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