Why is "float f4 = 42e7" not considered a valid declaration of a float in Java?

[Nilay Jhaveri]

Hi Friends,

I was reading the Java Sun Certification Book and one of the sample questions in the first chapter is:

"
Which 3 are valid declarations of a float? (Choose 3)
A. float f1 = -343;
B. float f2 = 3.14;
C. float f3 = 0x12345;
D. float f4 = 42e7;
E. float f5 = 2001.0D;
F. float f6 = 2.81F;
"

The Book says the answer is A, C, and F. It is clear to me why F is one of the answers since the "F" suffix is attached to the end of the number. It is also clear why A and C are correct based on the explanation at the end of the chapter which says: "A and C are integer literals (32 bits), and integers can be legally assigned to floats (also 32 bits.)"

But what I don't understand is why D is not one of the correct answers, based upon the same logic which makes A and C correct. Specifically, isn't D (42e7) and integer literal as well, making it able to be legally assigned to a float. Since 42e7 represents (240,000,000) , isn't it an int and not a double by default as a literal?

Thank you for your help.

Nilay

[Keith Lynn]

No 42e7 is a double.

[Nilay Jhaveri]

Can you explain why its a double given that it is an int and in the solution they said ints can be assigned to floats?

[Keith Lynn]

No, it is not an int. When you express a number in scientific notation, it is considered a double unless you add a f or F.

Note what happens when you try to compile this.

public class Test {
   
   public static void main(String[] args) {
      float f4 = 1e2;
   }
 
}

Test.java:4: possible loss of precision
found   : double
required: float
      float f4 = 1e2;
                 ^
1 error

[ June 11, 2006: Message edited by: Keith Lynn ]