Why doesn't Lombok.sneakyThrows throw a ClassCastException?

Question

I am trying to understand how the @SneakyThrows annotation in Lombok actually works under the hood.

From this SO answer, I can gather it uses the method Lombok.sneakyThrows() under the hood.

The code for Lombok.sneakyThrows() is as follows:

public static RuntimeException sneakyThrow(Throwable t) {
    if (t == null) throw new NullPointerException("t");
    return Lombok.<RuntimeException>sneakyThrow0(t);
}

@SuppressWarnings("unchecked")
private static <T extends Throwable> T sneakyThrow0(Throwable t) throws T {
    throw (T)t;
}

From what I can make out is that all it is doing is casting it to a java.lang.RuntimeException so the compiler does not complain.

But shouldn't that technically give a ClassCastException?

For example:

public static void main(String args[]) {
    Throwable i = new InterruptedException();
    RuntimeException rr = (RuntimeException) i;
}

The above code results in a ClassCastException.

Then why does the same thing not happen in case of Lombok?

your basic assumption regarding a cast to RuntimeException is wrong. Please see the documentation. — juwil
– juwil, Commented Nov 1, 2021 at 14:54
See gamlor.info/wordpress/2010/02/… which details this hack. — sp00m
– sp00m, Commented Nov 1, 2021 at 15:07

Sweeper · Accepted Answer · 2021-11-01 14:55:59Z

3

Notice this is not a cast to RuntimeException, but a cast to a generic parameter T, constrained to Throwable. Because of Type Erasure, this cast cannot be checked. The cast is a completely unchecked cast.

The runtime doesn't know what T is, except that it extends Throwable. So it can't check whether t actually is a T, and throw an exception if it is not. All it can do is check if t is Throwable, but t's type is already Throwable, so there's no need to do that either.

Normally, you would see a warning about this unchecked cast, and that is what @SuppressWarnings("unchecked") is suppressing.

answered Nov 1, 2021 at 14:55

Sweeper

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2 Comments

ng.newbie Over a year ago

I don't understand when you say "The runtime doesn't know what T is": how is this possible ? The runtime has access to all types. The programmer specified the generic type when writing the code, it should have everything.

Sweeper Over a year ago

@ng.newbie the compiler certainly have access to everything you wrote, but the JVM doesn’t know about generics, because the designers did not design it that way. Look up type erasure in Java, and check out the “type erasure” link in the answer.

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