About Question enthuware.ocajp.i.v7.2.985 :

[nickeasyuptech]

For this question, why does 'float f = 0x0123;' work but not 'float f = 43e1;' ?

why is 43e1 a double but 0x0123 is not?

Thank you for your help!

[admin]

That is just a rule of the language. There is no particular reason for it except that the designers designed it that way. They might have had some reason but I haven't seen it.
-Paul.

[nickeasyuptech]

Thank you!

[javaman]

I thought 43e1 is scientific notation meaning 43 * 10 to the power of 1... Why is it a double and not 430?

[admin]

That is just how Java language was designed. A number with e notation is considered a double. Please check Java Language Specification for more details.

HTH,
Paul.

[coder007]

To note:
this line will compile, because 10 is of integer type, default type for integer literals.
float f1 = 10;

But this one won't compile, because 10.0 is of double type, which is default type for floating-point literals. And we try to cast double to float -> narrowing casting:
float f2 = 10.0;

[coder007]

For this question, why does 'float f = 0x0123;' work but not 'float f = 43e1;' ?
why is 43e1 a double but 0x0123 is not?

You should apply a general rule - real numbers are double by default, and integer numbers are int by default.
Here 0x0123 represents an integer number "291", that is why it compiles as well (implicit casting int to float).
How recognize real number in hexadecimal form? Look for decimal points in it:
0x123.456p0 - it is equal to "291.271", than it's double by default, and statement:
float f = 0x123.456p0;
will not compile, because of narrowing casting double to float. To say compiler that this is a float, just put f or F in the end (as with ordinary real numbers in decimal form), and than it will compile:
float f = 0x123.456p0F;

[nsbrasco]

The explanation for this problem states the following: "implicit narrowing is permitted only among byte, char, short, and int." However, it seems to me that ALL narrowing must be done explicitly, as the following two lines of code throw a type mismatch error.

int i = 1;
byte b = i;

Is the explanation just wrong, or am I not understanding it?

[admin]

The explanation is correct but incomplete. Implicit narrowing is permitted only only among byte, char, short, and int but the right hand side also has to be a compile time constant. This point is covered in other questions on this topic.

In your case, i is a variable and is not a compile time constant. It will work if you make it final.

HTH,
Paul.

[nsbrasco]

A very interesting distinction. Thank you!

[jamesmccreary]

The explanation states: "Although the values in the option 1 and 2 are compile time constants"

For reference, Option 1 and 2 are the following

1. float f1 = 1.0;
2. float f = 43e1;

How are these compile time constants? From my research, they are just variables. Compile time constants are denoted by the final keyword, which is absent from both options.

Can you confirm whether a compile time constant requires the keyword final? The JLS https://docs.oracle.com/javase/specs/jl ... jls-4.12.4 does mention any condition in which this question in particular regarding the floats applies.

[admin]

In an assignment statement, the right hand side of = is the value that is being assigned to the "variable" on the left hand side.
f1 is the variable.
1.0 is the value that f1 is being is assigned. It is this value that the statement is talking about.
1.0 is a compile time constant.

What you are talking about is a compile time constant "variable". For example, final float f1 = 1.0;. Here, f1 is a compile time constant because its value cannot change.

[davidshinabarger]

Why do these float values not need the f at the end of the value?

For example:
float f = 4f;
float f = -1f;
etc...

Won't java try and "correct" these values to be double values without the f?

Maybe someone can point me in the direction of a better article to understand why these declarations are valid.

[davidshinabarger]

Well, I found a really thorough article on this topic over on codingranch:

https://coderanch.com/t/238351/certific ... tion-float

[Arold Aroldson]

how do i know that 0x0123 can be float and 43e1 is not? Is there any table or rule to know?

[admin]

For a floating point number either a decimal point, an exponent, or a float type suffix (i.e. f or d ) are required. So, you can see that 0x123 cannot be a floating point number while 43e1 is.
You may want to check out section 3.10.2 of JLS for the details.