Почему второй блок catch опускается? (Java SE 8 OCA — исключения) ⇐ JAVA

[Гость]

Я пытался решить этот пример вопроса для экзамена OCA Java SE 8, но ошибся.

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public static void main(String... a) {
System.out.print("a");
try {
System.out.print("b");
throw new IllegalArgumentException("1");
} catch (IllegalArgumentException e) {
System.out.println("c");
throw new RuntimeException("1");
} catch (RuntimeException e) {
System.out.println("d");
throw new RuntimeException("2");
} finally {
System.out.print("e");
throw new RuntimeException("3");
}
}

Here is the image of the try-catch sequence in question
I believe the answer should be:

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abcdeException in thread "main" java.lang.RuntimeException: 3

but the answer is actually:

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abceException in thread "main" java.lang.RuntimeException: 3

My thought process is that first, the main method prints "a", then prints "b". Then, an

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IllegalArgumentException

is thrown, and caught in the first catch block. Then, inside the first catch block, "c" is printed, and a

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RuntimeException

is thrown. I really do believe, then, that the second catch block catches the

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RuntimeException

and prints "d", while throwing another

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RuntimeException

, where the thread then "floats" over to the finally block, printing "e" and then throwing the

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RuntimeException("3")

instead. I really can't see why the second catch block is simply ignored.
Therefore,

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abcdeException in thread "main" java.lang.RuntimeException: 3

should be printed in my opinion.
Am I missing anything in terms of reasoning? How should my sequence of thoughts be instead?


Источник: https://stackoverflow.com/questions/781 ... exceptions