The Java compiler seems inconsistent if there is some code that clearly can not throw an exception, and you write surrounding code that declares that the code can throw that
Just a try !!
For second case there might be a possibility that some other class overrides this method in the future and a code is written inside that which can throw this exception.
Throws
does not handle exception, It indicates the throwing exception upwards from where method will be called. In other words it will just pass exception to the caller.
While try...catch
block handles the exception, and that is why Java compiler will check if there is any exception to handle which is being caught in catch
block or not.
Those are two different thing, One is Throwing and another is Handling the exception and compiler will make his nose tilted on second one only...:p
From the JavaDoc :
Exception handlers can do more than just print error messages or halt the program. They can do error recovery, prompt the user to make a decision, or propagate the error up to a higher-level handler using chained exceptions.
So, by providing try...catch
implementation, you are requesting compiler to do some more thing than just printing exception.
Another specific reason :
public void testException() throws FileNotFoundException {
File file = new File("test.txt");
System.out.println(file.exists());
Scanner scanner = new Scanner(file);
}
If you will observe the compiled code of above example by javap -c Test.class
you will find an Exception table will be created.
public static void testException();
Code:
0: new #2 // class java/io/File
3: dup
4: ldc #3 // String test.txt
6: invokespecial #4 // Method java/io/File."<init>":(Ljava/lang/String;)V
9: astore_0
10: getstatic #5 // Field java/lang/System.out:Ljava/io/PrintStream;
13: aload_0
14: invokevirtual #6 // Method java/io/File.exists:()Z
17: invokevirtual #7 // Method java/io/PrintStream.println:(Z)V
20: new #8 // class java/util/Scanner
23: dup
24: aload_0
25: invokespecial #9 // Method java/util/Scanner."<init>":(Ljava/io/File;)V
28: astore_1
29: goto 37
32: astore_1
33: aload_1
34: invokevirtual #11 // Method java/io/FileNotFoundException.printStackTrace:()V
37: return
Exception table:
from to target type
20 29 32 Class java/io/FileNotFoundException
So, When compiler won't find any code that is not throwing exception in try block, an compile time error will be there.
and Exception table will not be generated in case of throws
.
Since you wanted other views -
Consider two implementations of the same signature in two different subclasses.
For example, (a made up example),
public class StudentLoader {
public abstract Student readStudentData() throws SQLException, IOException;
public static void main(String args[]) {
StudentLoader loader = getStudentLoader (); //may return any subclass instance
try {
Student s = loader.readStudentData();
}
catch(IOException e) {
//do something
}
catch(SQLException e) {
//do something
}
}
}
public class StudentFileReader extends StudentLoader {
public Student readStudentData() throws IOException {
//read from a file
}
}
public class StudentDBReader extends StudentLoader {
public Student readStudentData() throws SQLException {
//read from DB
}
}
why compiler do allow to write exceptions in throws section even if it can not be thrown?
Even though the subclass implementation StudentDBReader
is not throwing IOException
, the parent class StudentLoader
still has to say throws IOException
because other implementations of StudentLoader
may throw it. So even though the exception may not be thrown by the method, you have a way of indicating to the caller using StudentLoader
reference (that points to either of the two subclasses instance) that the caller must handle these exceptions.
In your snippet 1 showing method g()
, there is no scope of inheritance. The code is right there in the try block. If any statement in try is throwing a checked Exception, you will have to handle it. In case of throws clause, scope for inheritance has to be allowed. Compiler cannot decide which version of readStudentData( )
will be called at runtime.
I would expect that compiler should give an error in case of static methods, if Exception mentioned in throws clause is not thrown, since static methods do not participate in inheritance. I am not sure why throws clause in static methods can contain Exceptions which are never thrown in implementation. It will not be overridden anyway, so why not throw error here? There might be something I am missing.
public class UnreachableCatchBlock {
public static void main(String[] args) {
UnreachableCatchBlock ucb = new UnreachableCatchBlock();
System.out.println(ucb.getClass().getName() + " started.");
// ucb.method1();
ucb.method2();
}
private void method1() {
try {
System.out.println(getClass().getName() + ".method1() started.");
} catch (FileNotFoundException e) {//any checked exception
}
}
private void method2() {
System.out.println(getClass().getName() + ".method2() started.");
}
}
Above code compiles and runs as explained in @Raedwald answer.
I don't suggest this as an answer, just a place to post the code that I attempted to include in comments.
The compiler allows this because the throws
clause of the method is part of the signature of the method, rather than part of its implementation. It is possible that the implementation might change at some point, while keeping the signature the same. An old implementation might have thrown a checked exception, but the new one might not. Or the designer of the signature might have wanted to give the implementer the flexibility to throw a checked exception when that is not always necessary.