java - Why is throwing a checked exception type allowed in this case?

Question

I noticed by accident that this throw statement (extracted from some more complex code) compiles:

void foo() {
    try {

    } catch (Throwable t) {
        throw t;
    }
}

For a brief but happy moment I thought that checked exceptions had finally decided to just die already, but it still gets uppity at this:

void foo() {
    try {

    } catch (Throwable t) {
        Throwable t1 = t;
        throw t1;
    }
}

The try block doesn't have to be empty; it seems it can have code so long as that code doesn't throw a checked exception. That seems reasonable, but my question is, what rule in the language specification describes this behavior? As far as I can see, §14.18 The throw Statement explicitly forbids it, because the type of the t expression is a checked exception, and it's not caught or declared to be thrown. (?)

Answer 1

This is because of a change that was included in Project Coin, introduced in Java 7, to allow for general exception handling with rethrowing of the original exception. Here is an example that works in Java 7 but not Java 6:

public static demoRethrow() throws IOException {
    try {
        throw new IOException("Error");
    }
    catch(Exception exception) {
        /*
         * Do some handling and then rethrow.
         */
        throw exception;
    }
}

You can read the entire article explaining the changes here.