Float literals.

[Shivani Chandna]

Java 1.4//
Why am I getting error on line 3 and not one line 1 and 2?

float f1 =45.64e-1f; // No Compile time error //1
        float f2 = 4.1e-3f; // No Compile time error //2
float f3 =45.l4e-1f; // Compile time error //3
                             // possible loss of precision found   
                             // : double required: float 

Thanks,
--

Shivani,
[ September 25, 2005: Message edited by: Shivani Chandna ]

[Barry Gaunt]

Please check your question.

[Barry Gaunt]

On the line 3 you have an invalid digit. You have the letter "l" (el) and not a "1" (one)

[Shivani Chandna]

Sad - :-/ what an error ! :roll:
I looked up - on EditPlus both l and 1 were looking alike .

[Jayant Raj]

Can you please explain the type conversions taking place here?
According to my understanding,

 
float f1 =45.64e-1f; // No Compile time error //1

will throw a compile time error. If we dont append 45.64e with f, it will implicitly be a double.Thus the right hand expression will be a double which cannot be assigned to a float.

Please clarify.

[A Kumar]

Hi,

  
 
float f1 =45.64e-1f; // No Compile time error //1
 

There is no compilation error...As you said

45.64e is double and hence the whole expression would be implicitly a double expression...

After the evaluation it is converted back to float..

here 45.64e is within the range of float and hence there is no compilation error.

But if the evaluated value exceeds the range of float it needs to be

explicitly casted else a compilation error..



An interesting thing in your example...

  
 
float f1 =45.64e-1.0f; 
 
 

If the literal is defined as 1.0f then error occurs ..Why???



Regards
[ September 26, 2005: Message edited by: A Kumar ]

[Joji Doi]

>>If the literal is defined as 1.0f then error occurs ..Why???
Look at the following as a float literal rather than literals.

float f1 =45.64e-1f;

The above statement is NOT 45.64e - 1f as minus arithmetic.
45.64e-1 is double literal. The exponent cannot be floating point.

Thus the following is equivalent:
float f1 =(float)45.64e-1;

Cheers!

[A Kumar]

Hi,

That means my explanation above is not in tune with the above code...


I overlooked that


Regards

[Jenny Thompson]

Hi ,
The error is nothing bcoz of logical thing i belive..
when i compiled ur code i got the following error message..


Test.java:5: ';' expected
float f3 =45.l4e-1;
^
1 error

instead of numerical value one(1)in ur code lowercase of L is present..
it caused the problem..if u change it to one it s compiling nicely...
no error occurs.

[Akhilesh Trivedi]

Guys, before you answer, don't just read the question alone, do also read the opinions and discussion that follows under it. I guess the author has already discovered the fault.

Originally posted by Shivani Chandna:
Sad - :-/ what an error ! :roll:
I looked up - on EditPlus both l and 1 were looking alike .

[ September 27, 2005: Message edited by: Akhil Trivedi ]