Java Exception StackTrace is determined when new

I wrote the following code that "new exceptions in advance and reuse them".

package exceptiontest;

public class ExceptionTest {
    
    //Generate an exception here.
    private static final RuntimeException _e = new RuntimeException("hello"); // line: 6

    public static void main(String[] args) {
        //It is here to actually throw.
        throw _e; // line: 10
    }
}

The result is ...

Exception in thread "main" java.lang.RuntimeException: hello
	at exceptiontest.ExceptionTest.<clinit>(ExceptionTest.java:6)

Why?

Java seems to determine the StackTrace when it creates an instance of Exception.

With python, StackTrace appears when raise ( throw) is done, so I wrote it with the same feeling.

For example, even if you rewrite the code as follows ...

package exceptiontest;

public class ExceptionTest {
    
    //Generate an exception here.
    private static final RuntimeException _e = new RuntimeException("hello"); // line: 6
    
    private static void f1() {
        f2();
    }
    
    private static void f2() {
        throw _e;
    }

    public static void main(String[] args) {
        // main() --> f1() --> f2() --> throw _e
        f1();
    }
}

You will get a StackTrace on line 6 that is always new to Exception.

Exception in thread "main" java.lang.RuntimeException: hello
	at exceptiontest.ExceptionTest.<clinit>(ExceptionTest.java:6)

<clinit> is a static initializer for a class. The variable _e is declared as a static element of the class, so the StackTrace will be determined when the ExceptionTest class is loaded.

Bonus (in Python)

It works as expected (?).

_e = Exception('hello')

def f1():
    raise _e

if __name__ == '__main__':
    f1()

Traceback (most recent call last):
  File "/Users/username/Documents/workspace/raisetest/main.py", line 7, in <module>
    f1()
  File "/Users/username/Documents/workspace/raisetest/main.py", line 4, in f1
    raise _e
Exception: hello