day7
asked on
insanceof question
Why can't I get this to compile?
public class InstanceofTest {
public static void main(String args[]){
Double a = new Double(0.0d);
if(a instanceof Integer){
System.out.println("true")
}
else{
System.out.println("false"
}
}
}
here's what I get:
F:\>javac InstanceofTest.java
InstanceofTest.java:4: inconvertible types
found : java.lang.Double
required: java.lang.Integer
if(a instanceof Integer){
^
1 error
this kind of error doesn't make any sense to me being I thought instanceof was precisely to compare objects and simply return true or false
public class InstanceofTest {
public static void main(String args[]){
Double a = new Double(0.0d);
if(a instanceof Integer){
System.out.println("true")
}
else{
System.out.println("false"
}
}
}
here's what I get:
F:\>javac InstanceofTest.java
InstanceofTest.java:4: inconvertible types
found : java.lang.Double
required: java.lang.Integer
if(a instanceof Integer){
^
1 error
this kind of error doesn't make any sense to me being I thought instanceof was precisely to compare objects and simply return true or false
Would compile with
Object a = new Double(0.0d);
Object a = new Double(0.0d);
ASKER
ok, but doesn't that defeat the purpose of instanceof? why does it matter that a couldn't be an Integer? why doesn't it just return false?
What are the rules, then? so I know how to use it correctly...I haven't seen that explained in any of the tutorials, etc. on instanceof
What are the rules, then? so I know how to use it correctly...I haven't seen that explained in any of the tutorials, etc. on instanceof
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SOLUTION
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ASKER
ok, I think I got the idea,
thanks
thanks
>> What are the rules, then?
The compiler tries to catch as many errors as possible during compilcation time.
As said before, Becuase Double apparent type can never hold an Integer instance (due to java single inheritence), the compiler
can tell that and will ommit an error (as this line does not make any sense, if you want it to fail all the time then do if (false) instead).
If Double was an interface or a parent of Integer then the compiler will pass it as it can't tell if its wrong during compilation time.
The compiler tries to catch as many errors as possible during compilcation time.
As said before, Becuase Double apparent type can never hold an Integer instance (due to java single inheritence), the compiler
can tell that and will ommit an error (as this line does not make any sense, if you want it to fail all the time then do if (false) instead).
If Double was an interface or a parent of Integer then the compiler will pass it as it can't tell if its wrong during compilation time.
:-)
To make it compile try changing the declaration to
Number a = new Double(0.0d);