问题
Possible Duplicate:
Is 1/0 a legal Java expression?
Why does this code compile?
class Compiles {
public final static int A = 7/0;
public final static int B = 10*3;
public static void main(String[] args) {}
}
If I take a look in the compiled class file, I can see that B has been evaluated to 30, and that A still is 7/0.
As far as I understand the JSL an expression where you divide by zero is not a constant.
Ref: JLS 15.28
My above statement is due to this line:
A compile-time constant expression is an expression denoting a value of primitive type
Hence dividing by zero is not evaluated to a primitive value.
What I really dont understand is why the compiler allows this anyway? Just to be clear, my code above crashes runtime with a "java.lang.ExceptionInInitializerError"
As it seems to me the compiler threats any final static variable as a constant and evaluates it compile time. That means that the compiler already has tried to evaluate A, but since it was a division by zero it just let it go through. No compile time error. But this seems very very bizarre... The compiler knows it is a divide by zero and that it will crash runtime but nevertheless it doesn't flag a compile error!
Can anyone explain to me why?
回答1:
To throw an java.lang.ExceptionInInitializerError is the only correct behavior.
If your code did not compile, a perfectly valid Java program would have been rejected, and that would have been a bug.
The only correct alternative to putting 7/0 in the compiled code, would actually be to explicitly throw a ExceptionInInitializerError, but how much more useful is that?
The compiler knows it is a divide by zero and that it will crash runtime but nevertheless it does flag a compile error!
Actually, I wouldn't agree with that... would this program crash?
class Compiles {
public final static int A = 7/0;
public final static int B = 10*3;
public static void main(String[] args) {}
}
public class Test {
// Application entry point.
public static void main(String[] args) {
try {
new Compiles();
launchTheMissiles();
} catch (ExceptionInInitializerError e) {
doUsefulStuff();
}
}
}
回答2:
JLS 15.28 Constant Expression:
A compile-time constant expression is an expression denoting a value of primitive type or a String that does not complete abruptly and is composed using only the following:
...
Therefore 7/0 is not a compile-time constant, since its evaluation completes abruptly due to division by zero. So, it's treated as a regular run-time expression, and throws an exception in runtime.
来源：https://stackoverflow.com/questions/4980360/why-doesnt-a-java-constant-divided-by-zero-produce-compile-time-error