About Question enthuware.ocajp.i.v7.2.1271 :

[shining_dragon]

why isn't the program will print ascii number equivalent to 'a'?

[admin]

Because that is how println method of PrintStream is coded. Please see this: http://docs.oracle.com/javase/7/docs/ap ... rint(char)

[nikolaj.mattrup]

Can someone please tell what would this print if c was incremented, thx!

[admin]

Please try it out and then ask if you do not understand the output.

[caminata]

Why ist the part '((int) (f + d)) of the if-statement valid?

When trying to add float and double value (e.g. float k = f + d;) it does not compile. Why does it work in the if-statement then?

[admin]

f+d returns a double. You can't assign a double to a float without a cast so float k = f + d; doesn't compile. But in case of the if statement, you are not assigning f+d to any variable. Further, there is an explicit cast.

[caminata]

Thanks for your explanations. I understand now.

[gparLondon]

What else shall we know about the characters for the exam? Shall we remember all the code for characters as well as ints? like int 0 is different from '0'?

Should we expect codes like this in the exam?

Code: Select all

public static void main(String args[])
	{
		char c=48;
		switch(c)
		{
		case 'a':System.out.println("a"); break;
		case '1': System.out.println("One"); break;
		case '0':System.out.println("Zero"); break;
		case 48:System.out.println("Fourty Eight"); break;
		}
		//System.out.println(++c);
	}

Regards,
GPAR

[admin]

You only need to know that '0' is not same as zero.
You don't need to know the int values for any character but again, you should know that a char 'a' is not same as int 0xa.
-Paul.

[gparLondon]

Thanks, for the quick reply.

Regards,
GPAR

[pbonito]

I don't understand the rule about == operator. I know that you can't apply it to variables referring to objects not in the same hierarchy, but what about primitives?
You can't compare a byte, char, .. to boolean ->ok
You can compare an int to flaot ->ok since int can be converted to float
but what about comparing int and short to char?

Thanks

[admin]

What happened when you tried it out?

[pbonito]

It works.
short i = 3;
char a= 'a';
System.out.println(i==a); -> print false
but a = i gives a compile error. So I don't understand which is the rule.

[admin]

a = i does not compile because char type has a different range than short type. So unless the compiler knows for sure that the value you are putting in a variable will fit in that type, it will not allow the assignment. It will work if i is final.

This should be covered in any regular Java book. Which book are you following?

[pbonito]

yes, I know that and this not surprise me. I don't understand why a==i is permitted since I know that == can't be applied to different type. == works if one of the two is promoted to the other type, but as you stated char can't be promoted to short, and short can't be promoted to char.

I'm following Bates & Sierra.

[admin]

In your previous post you asked about = and not ==.

but a = i gives a compile error. So I don't understand which is the rule.

My statement about why compiler won't allow a = i was for that. Not for a ==i.


a == i works fine because, for primitives, == operator checks the value and it is certainly possible that a and i contain the same value.

[aar2416]

how can i be promoted to float

[admin]

Java allows int to be promoted to float. For example, the following is valid:

Code: Select all

       float f = 0.0f;
       int i = 10;
       f = i;

[aar2416]

Yes but that is during the time of assignment...here comparison is being done...so will int be converted to float

[admin]

Yes, you should try System.out.println(f == i);

[aar2416]

ohhk got it

[JuergGogo]

Code: Select all

public class TestClass  {        
    public static void main( String[] args )  {        
        char c = 'a';
        System.out.println( ++c );     // 1 - output: b
        System.out.println( c+=1 );   // 2 - output: c
        System.out.println( c + 1 );   // 3 - output: 100      
    }
}

1+2: char remains char, then printed like a String, since SOP uses toString().
3: c is implicitly widened to int, so output is (int)char.