exceptions and type casting confusion

[Jorge Bendahan]

Hi, i have this problem, when dealing with this Array of Objects.I need to be able to tell whether this object is a type String or is a type Boolean, etc,
and throwing an exception when the content of the first column it's not of the type that it says on the second column, the tricky part is that every time it runs, different objects have different datatypes..
Example:

 Class myClass{
 
Object[5][1] arr; 
 
arr[0][0]= pedro , arr[0][1] = String
arr[1][0]= 123   , arr[1][1] = int
arr[2][0]= c     , arr[2][1] = char 
arr[3][0]= bill  , arr[3][1] = String
 
..and so on.....
 

//some other Stuff ....
/////////////////
 

if(verifyForError(arr)==true){
do this
}
else{
show a message...
}
 

private boolean verifyForError(arr){
 
boolean flag = false;
 
try{
for(int i = 0; i<arr.length; i++){

if(arr[i][1].toString()=="boolean"){
boolean b = new Boolean(arr[i][0].toString());
}
 
else if(arr[i][1].toString()=="String"){
String b = new String(arr[i][0].toString());
}
 
else if(arr[i][1].toString()=="int"){
Integer b = new Integer(arr[i][0].toString());
}

}
 

}
catch(Exception e){
e.printStackTrace();
flag = true;
}
 
}
   

[Jorge Bendahan]

I forgot, this is what i've done, but always catches an error, even when it supposed not to.

[Jorge Bendahan]

I've got it, here it is :

long l = new Long(arr[i][1].toString().trim()).longValue();

[bart zagers]

Some remarks:

if(arr[i][1].toString()=="boolean")

It is usually not a good idea to compare Strings with the "==" operator. What you normally want is comparing using the equals method.
if(arr[i][1].equals("boolean"))

String b = new String(arr[i][0].toString());

When you know you have a String object, it is not needed to create a new one. Just cast the object to String
String b = (String)arr[i][0];

long l = new Long(arr[i][1].toString().trim()).longValue();

There is no need to create a Long object (or any Number object) here as Long has a useful static method to parse a String
long l = Long.parseLong(arr[i][1].toString().trim());

Btw, Is there a reason you have an array of Objects? All your elements are Strings, so you could just as well have an array of Strings.
[ May 20, 2008: Message edited by: bart zagers ]

[Jorge Bendahan]

the objects in the array aren't strings, i'm using the toString method just because the constructors of that clases(Integer, Boolean, etc) require a String parameters. But thanks for those observations.

[bart zagers]

If they aren't Strings, what are they?

[Jorge Bendahan]

they are Objects, after the datatype comparison they are casted to the correspondant datatype.

[bart zagers]

Ok, you have an array of Objects, but the actual elements of the array will not be instances of Object, but something else.
For example if you have your array
Object[][] arr
and you do
arr[0][0] = new Long(5);
arr[0][0] will contain an object of type Long.
So again, what will be the type(s) of the elements in the array?

[Jorge Bendahan]

Suppose i'm calling this method once, and this time the array will be:

arr[0][0]= pedro , arr[0][1] = String
arr[1][0]= 123 , arr[1][1] = int
arr[2][0]= c , arr[2][1] = char
arr[3][0]= bill , arr[3][1] = String

Now i'm calling the method for the second time and the parameter passed is:

arr[0][0]= -235 , arr[0][1] = short
arr[1][0]= 1.2 , arr[1][1] = float
arr[2][0]= asd123 , arr[2][1] = String
arr[3][0]= x , arr[3][1] = char

[Jorge Bendahan]

As you can see the elements of the array may vary in each call, you may wonder what is this for, well i'm doing a research in uml models and ocl parsing, so different models or constraints have different elements with different datatypes and values.

[Jorge Bendahan]

BTW this is for academic non commercial purposes(i'm student, as you may noticed for the poor knowledge of the java api ), so i'm not "taking advantage of your knowledge", actually i'm doing this research pretty much solo(just me).

[bart zagers]

I guessed that, that's why I asked some addtional questions.
It is just that the code you showed seems not very efficient and not very elegant, although it might work.
Depending on the content of the array, there are several improvements and simplifications possible.
For example take your last example and the line
arr[1][0]= 123 , arr[1][1] = int
what is now the type of arr[1][0] (you can check it for example by printing arr[1][0].getClass().getName() )?
It is most likely either a String, which case your conversion code could be simplified to

if(arr[i][1].toString()=="int"){
   Integer b = Integer.valueOf((String)arr[i][0]);
}

or the object already is an Integer object, in which case you don't even need the second column and the code becomes something like

if(arr[i][0] instanceof Integer){
   Integer b = (Integer)arr[i][0];
}

[Jorge Bendahan]

     
    if(arr[i][1].toString()=="int"){
       Integer b = Integer.valueOf((String)arr[i][0]);
    }
 

or the object already is an Integer object, in which case you don't even need the second column and the code becomes something like
 
    code:
 
    if(arr[i][0] instanceof Integer){
       Integer b = (Integer)arr[i][0];
    }   

This cant work, here is an example: arr[i][0]= 12, arr[i][1]="short"
if you use the instanceof method with Integer i will return true.

But thanks for the responses anyway, this issue has been solved, now i have some new ones