declaration float question!??

[huanyu zhao]

14 which three are valid declaraction of a float?
A. float foo=-1;
B. float foo=1.0;
C. float foo=42e1;
D. float foo=2.02f;
E. float foo=3.03d;
F. float foo=0x0123;
why A F is right!??? especially why A is rihgt??? B C is wrong???

[Corey McGlone]

By default, integral values are interpreted as ints and decimal values are interpreted as doubles.
Therefore, these two are considered int values:

float foo=-1; 
float foo=0x0123;

And ints can be automatically cast (through a widening conversion) to a float.
However, these are interpreted as doubles:

float foo=1.0; 
float foo=42e1; 

Since a double can't be explicitly cast to a float (due to loss of precision), this will cause a compiler error.
Corey
Of course, if you would have simply compiled these, the compiler would have given you the same answer. (and much faster)
[ June 24, 2002: Message edited by: Corey McGlone ]

[Chung Huang]

B C E are invalid
A D F are valid
A is valid because -1 is an int which is 32-bit size that is same size as float, so automatic cast is allowed.
B is invalid because 1.0 is considered as double? So, if I am to declare a float string, I must make sure the number fits into the range and I must add F at the end? Otherwise I'll get a double instead of float?
C is invalid because it is double? Why is it double? Does Java interpret all decimal number literals as double unless specified by appending F?
D is valid since it has a F appended. So if the F is dropped, would it be invalid as well because it is interpreted as double?
E is invalid since it has a D appened.
F is valid for same reason as A. However, I thought int and float both has same bit size, therefor the cast is a simple conversion not a widening conversion. Am I wrong?
Thanks a bunch for the question and and answers.

[Tybon Wu]

The thing to remember is that all integral types can be implicitly converted to float. The following would compile and run.

byte b = 0;
short s = 0;
int i = 0;
long l = 0;
char c = 0;
float f;
f = b;
f = s;
f = i;
f = l;
f = c;

[Corey McGlone]

Originally posted by Chung Huang:
...However, I thought int and float both has same bit size, therefor the cast is a simple conversion not a widening conversion...

True, they are the same size, but it is still considered a widening conversion to go from an int to a float. Therefore, this is legal:

int i = 10;
float f = i;

While this is not:

float f = 10.0f;
int i = f;

The reason for this is that a float has a much wider range than an int (due to exponential notation). Therefore, any value that can fit in an int can fit in a float, but not vice versa. Take a look at this example:

float f = Integer.MAX_VALUE + 10;
int i = f;

Obviously, the resulting value fits in a float, but won't fit in the int.
Of course, with this, you may lose precision. Take a look at this thread for more details about loss of precision.
I hope that helps,
Corey

[huanyu zhao]

float foo=-1;
yes it`s a widening conversion,but should it
float foo=-1f; why there is no suffix f?
and then
another question i have quite puzzle form a lot of books:
byte-[short]
---int---long---float----double
[char]
is it a widen order? why?
long is 64 bits and float is only 32 bits? why is widening conversion???
[ June 24, 2002: Message edited by: huanyu zhao ]

[Corey McGlone]

Originally posted by huanyu zhao:
float foo=-1;
yes it`s a widening conversion,but should it
float foo=-1f; why there is no suffix f?

Huh? Why would you want to put a 'f' on the end of an integral value. It wouldn't serve much purpose. The use of the trailing 'f' is to tell the compiler to refer to a decimal value as a float, not a double.

byte-[short]
---int---long---float----double
[char]
is it a widen order? why?
long is 64 bits and float is only 32 bits? why is widening conversion???

It's all about the range of the data types. The range of a float is much wider than that of a long, even though a long is twice as large. The reason for this is the fact that a float uses exponential notation. As there is no value that can be contained in a long that doesn't fall within the range of a float, it is considered a widening conversion to go from a long to a float.
Again, take a look at the thread I referred to earlier. That thread goes into pretty great detail about this matter.
Corey

[Paul Villangca]

Hi Huanyu,
float f1 = -1; // no error here
float f2 = -1.1; // error!
In this example, -1 is an int, while -1.1 is a double. Since a double is wider than a float, there's a compiler error. To solve this,
float f2 = -1.1F; // no more error!
Note that appending the 'F' on anything less than a double will be redundant.
float f3 = -1F; // not needed
That's because the int value -1 is automatically promoted to float in the assignment.

[Alan Chong]

To make thing clear, it depends on the range instead of the number of bits. So
float f = 100000000L; // float <-- long, OK.
A float's range can cover the entire range of a long, though with some lost of precision.