I don't know why f or F is placed after float values in Java or other languages? for instance,
float fVariable = 12.3f;
any features other than indicating that this is a float value?
By default 12.3 is double literal. To tell compiler to treat it as float explicitly -> use f or F.
See tutorial page on the primitive types.
Seeing as there are only so many ways to represent a number in your program, the designers of Java had to cherry pick and assign each form to the most common use case. For those forms selected as default, the suffix that denotes the exact type is optional.
For Integer literals (int, long), the default is int. For obvious
reasons.
For Floating point literals (float, double) the default is double.
Because using double potentially allows safer arithmetic on the
stored values.
So, when you type 12 in your program, thats an int literal, as opposed to 12L, which is a long.
And when you type 12.3, thats a double literal, as opposed to 12.3F, which is a float.
So where is this relevant? Primarily in handling downcasts, or narrowing conversions. Whenever you downcast a long to an int, or a double to a float, the possibility for data loss exists. So, the compiler will force you to indicate that you really want to perform the narrowing conversion, by signaling a compile error for something like this:
float f = 12.3;
Because 12.3 represents a double, you have to explicitly cast it to a float (basically signing off on the narrowing conversion). Otherwise, you could indicate that the number is really a float, by using the correct suffix;
float f = 12.3f;
So too summarize, having to specify a suffix for longs and floats is a compromise the language designers chose, in order to balance the need to specify what exactly a number is, with the flexibility of converting numbers from one storage type to another.
float and double can only provide approximate representation values for some values. e.g. 12.3 or 0.1
The difference is that float is not as accurate (as it has less precision, because its smaller)
e.g.
System.out.println("0.1f == 0.1 is " + (0.1f == 0.1));
System.out.println("0.1f is actually " + new BigDecimal(0.1f));
System.out.println("0.1 is actually " + new BigDecimal(0.1));
prints
0.1f == 0.1 is false
0.1f is actually 0.100000001490116119384765625
0.1 is actually 0.1000000000000000055511151231257827021181583404541015625
So 0.1 is the closest representation in double and 0.1f is the closest representation in float
float fVariable = 12.3; is fine. but when you use only float value(without any identifier) in any expression that time you need to tell compiler that value is float hence we use suffix "f" after value. example
float fl =13f/5f;
here 13 and 5 are float values.
In java we have many different basic variable types so in order to make it general , it has some default features. Like if we give an input 16.02 it automatically takes it as a double input. So if you want to specify it as float we mention that 'f' after the number or we can simply use:
float f = (float) 16.02;
or
float f = 16.02f;
In the same way we have to mention 16l if we want the number to be saved as a long type else it will automatically select the default type ie int type.
During compilation, all floating point numbers (numbers with decimal point) default to double.
Therefore, if you don't want your number to double and just want it as float, you have to explicitly tell the compiler by adding a f or F at end of the literal constant.
If you do not use f it will be interpreted as double, which is the default in Java.
You can also write it like this##
float f=(float) 32.5956; float f=32.5956f;
Float is single-precision 32-bit IEEE 754 floating point and Double is double-precision 64-bit IEEE 754 floating point. When you use a value with decimal points and if you don`t specify is as 0.23f (specifically float) java identifies it as a double.
For decimal values, double data type is generally the default choice taken by java.
Check This
[10 years after the original post]
why f or F is placed after float values
With the f, the initialization occurs with the closest float value. Without the f, it might differ.
Many code values like 12.3 are not exactly representable as a double or a float. Instead a nearby value is used.
One rounding
Code 12.3 converted to the closest float: 12.30000019073486328125
float fVariable1 = 12.3f;
Two roundings
Code 12.3 converted to the closest double 12.300000000000000710542735760100185871124267578125 and then to the nearest float: 12.30000019073486328125.
float fVariable2 = 12.3;
Sometimes that 2-step approach makes for a different value due to double rounding.
Related
Today, I defined two float variable f1 and f2. Then I perform an addition, "+", arithmetic operation and assign to float variable f.
float f1 = 0.5048076923076923F;
float f2 = 0.5048076923076923F;
float f = f1 + f2;
This is the output:
According to this picture, All floating point values (float and double) in an arithmetic operation (+, −, *, /) are converted to double type:
picture source: http://www.mathcs.emory.edu/~cheung/Courses/170/Syllabus/04/mixed.html
I found an identical question but it hasn't explain why. Why doesn't eclipse have any issue tips? Is it the reason why the value of "f1 + f2" is a float type? And, why will Java auto convert the double to float type if like the above picture saying?
You seem to be saying that there is a conversion to float in the following.
float f1 = 0.5048076923076923F;
float f2 = 0.5048076923076923F;
float f = f1 + f2;
In fact, there is no conversion. The values are all float.
In particular, the literal 0.5048076923076923F is a float. The trailing F makes it a float. If you want a double literal, leave off the F or replace it with D.
When your teacher says "all floating point values are converted to double" he is wrong. The JLS says (in effect) that a numeric primitive operand will be converted to double when the other operand is a double. If both operands are float, then the operation will performed using single-precision floating point arithmetic.
The JLS reference is JLS 5.6.2: Binary Numeric Promotion.
It has been pointed out that there may be additional conversions happening at the hardware level. For example, the JLS says this:
Within an expression that is not FP-strict, some leeway is granted for an implementation to use an extended exponent range to represent intermediate results; the net effect, roughly speaking, is that a calculation might produce "the correct answer" in situations where exclusive use of the float value set or double value set might result in overflow or underflow.
However:
This is only allowed if the expression is not strictfp.
These are not the "conversions" that the JLS talks about in JLS 5.6.2.
This still contradicts what the OP's teacher is saying. He (or she) states that the all floating point calculations are done using double. The JLS states that under certain circumstances, a hardware platform may use extended precision arithmetic (possibly with more precision than 64 bit floating point), and in other circumstances it must not do this.
Your teacher is incorrect, kindly refer him to section 5.6.2 of the Java Language Specification, which states:
Widening primitive conversion (§5.1.2) is applied to convert either or both operands as specified by the following rules:
If either operand is of type double, the other is converted to double.
Otherwise, if either operand is of type float, the other is converted to float.
Otherwise, if either operand is of type long, the other is converted to long.
Otherwise, both operands are converted to type int.
As you can see, floats are only converted to double if the other operand is a double, not in all cases as your syllabus incorrectly states.
That is, the sum of two floats is a float, but the sum of a float and a double is a double.
PS: And no, Java will never convert a double to a float unless you explicitly ask it to, as in:
float f = (float) Math.PI;
Hi for floating points values JVM uses instructions like this:
'+' : fadd : pops two floats, adds them, and pushes the float result
'-' : fsub : pops two floats, subtracts them and pushes the flot result
similarly for there are other instructions set for other operators like * and / etc.
So as per JVM implementation point of view float + float will result to float.
But float + double will result to double.
For more information, you can read chapter 14. Floating-Point Arithmetic
of the book "Inside Java Virtual Machine"
What the other answers are missing is the hardware level. Floating point processors, in general, convert all operations to double (or quad), and then return the result in the requested precision. The Java specification may not recognize the conversion, but if it uses the floating point hardware then the conversion happens.
I have a primitive float and I need as a primitive double. Simply casting the float to double gives me weird extra precision. For example:
float temp = 14009.35F;
System.out.println(Float.toString(temp)); // Prints 14009.35
System.out.println(Double.toString((double)temp)); // Prints 14009.349609375
However, if instead of casting, I output the float as a string, and parse the string as a double, I get what I want:
System.out.println(Double.toString(Double.parseDouble(Float.toString(temp))));
// Prints 14009.35
Is there a better way than to go to String and back?
It's not that you're actually getting extra precision - it's that the float didn't accurately represent the number you were aiming for originally. The double is representing the original float accurately; toString is showing the "extra" data which was already present.
For example (and these numbers aren't right, I'm just making things up) suppose you had:
float f = 0.1F;
double d = f;
Then the value of f might be exactly 0.100000234523. d will have exactly the same value, but when you convert it to a string it will "trust" that it's accurate to a higher precision, so won't round off as early, and you'll see the "extra digits" which were already there, but hidden from you.
When you convert to a string and back, you're ending up with a double value which is closer to the string value than the original float was - but that's only good if you really believe that the string value is what you really wanted.
Are you sure that float/double are the appropriate types to use here instead of BigDecimal? If you're trying to use numbers which have precise decimal values (e.g. money), then BigDecimal is a more appropriate type IMO.
I find converting to the binary representation easier to grasp this problem.
float f = 0.27f;
double d2 = (double) f;
double d3 = 0.27d;
System.out.println(Integer.toBinaryString(Float.floatToRawIntBits(f)));
System.out.println(Long.toBinaryString(Double.doubleToRawLongBits(d2)));
System.out.println(Long.toBinaryString(Double.doubleToRawLongBits(d3)));
You can see the float is expanded to the double by adding 0s to the end, but that the double representation of 0.27 is 'more accurate', hence the problem.
111110100010100011110101110001
11111111010001010001111010111000100000000000000000000000000000
11111111010001010001111010111000010100011110101110000101001000
This is due the contract of Float.toString(float), which says in part:
How many digits must be printed for
the fractional part […]? There
must be at least one digit to
represent the fractional part, and
beyond that as many, but only as many,
more digits as are needed to uniquely
distinguish the argument value from
adjacent values of type float. That
is, suppose that x is the exact
mathematical value represented by the
decimal representation produced by
this method for a finite nonzero
argument f. Then f must be the float
value nearest to x; or, if two float
values are equally close to x, then f
must be one of them and the least
significant bit of the significand of
f must be 0.
I've encountered this issue today and could not use refactor to BigDecimal, because the project is really huge. However I found solution using
Float result = new Float(5623.23)
Double doubleResult = new FloatingDecimal(result.floatValue()).doubleValue()
And this works.
Note that calling result.doubleValue() returns 5623.22998046875
But calling doubleResult.doubleValue() returns correctly 5623.23
But I am not entirely sure if its a correct solution.
I found the following solution:
public static Double getFloatAsDouble(Float fValue) {
return Double.valueOf(fValue.toString());
}
If you use float and double instead of Float and Double use the following:
public static double getFloatAsDouble(float value) {
return Double.valueOf(Float.valueOf(value).toString()).doubleValue();
}
Use a BigDecimal instead of float/double. There are a lot of numbers which can't be represented as binary floating point (for example, 0.1). So you either must always round the result to a known precision or use BigDecimal.
See http://en.wikipedia.org/wiki/Floating_point for more information.
Floats, by nature, are imprecise and always have neat rounding "issues". If precision is important then you might consider refactoring your application to use Decimal or BigDecimal.
Yes, floats are computationally faster than decimals because of the on processor support. However, do you want fast or accurate?
A simple solution that works well, is to parse the double from the string representation of the float:
double val = Double.valueOf(String.valueOf(yourFloat));
Not super efficient, but it works!
For information this comes under Item 48 - Avoid float and double when exact values are required, of Effective Java 2nd edition by Joshua Bloch. This book is jam packed with good stuff and definitely worth a look.
Does this work?
float flt = 145.664454;
Double dbl = 0.0;
dbl += flt;
There is a way to convert Float value into Double without adding the extra precision
Float aFloat= new Float(0.11);
String s = aFloat.toString();
Double aDouble = Double.parseDouble(s);
This Approach will not add an extra precisions to your Float value while converting. The only Problem with this approach is memory usage of the JVM by creating an extra tamp String object.
When calling an toString() (aDouble.toString()) on Double will never add an extra precisions. The precisions will be added while type conversion.
I just wanted some information about this.
float n = 2.99944323200023f
What does the f at the end of the literal do? What is it called? Why is it used?
The f indicates it's a floating point literal, not a double literal (which it would implicitly be otherwise.) It hasn't got a particular technical name that I know of - I tend to call it the "letter suffix" if I need to refer to it specifically, though that's somewhat arbitrary!
For instance:
float f = 3.14f; //Compiles
float f = 3.14; //Doesn't compile, because you're trying to put a double literal in a float without a cast.
You could of course do:
float f = (float)3.14;
...which accomplishes near enough the same thing, but the F is a neater, more concise way of showing it.
Why was double chosen as the default rather than float? Well, these days the memory requirements of a double over a float aren't an issue in 99% of cases, and the extra accuracy they provide is beneficial in a lot of cases - so you could argue that's the sensible default.
Note that you can explicitly show a decimal literal as a double by putting a d at the end also:
double d = 3.14d;
...but because it's a double value anyway, this has no effect. Some people might argue for it advocating it's clearer what literal type you mean, but personally I think it just clutters code (unless perhaps you have a lot of float literals hanging around and you want to emphasise that this literal is indeed meant to be a double, and the omission of the f isn't just a bug.)
The default Java type which Java will be using for a float variable will be double. So, even if you declare any variable as float, what compiler has to actually do is to assign a double value to a float variable, which is not possible.So, to tell compiler to treat this value as a float, that 'f' is used.
In Java, when you type a decimal number as 3.6, its interpreted as a double. double is a 64-bit precision IEEE 754 floating point, while floatis a 32-bit precision IEEE 754 floating point. As a float is less precise than a double, the conversion cannot be performed implicitly.
If you want to create a float, you should end your number with f (i.e.: 3.6f).
For more explanation, see the primitive data types definition of the Java tutorial.
It's to distinguish between floating point and double precision numbers. The latter has no suffix.
You need to put the 'f' at the end, otherwise Java will assume its a double.
From the Oracle Java Tutorial, section Primitive Data Types under Floating-Point Literals
A floating-point literal is of type float if it ends with the letter F or f; otherwise its type is double and it can optionally end with the letter D or d.
It means that it's a single precision floating point literal rather than double precision. Otherwise, you'd have to write float n = (float)2.99944323200023; to cast the double to single.
When you write 1.0, it's ambiguous as to whether you intend the literal to be a float or double. By writing 1.0f, you're telling Java that you intend the literal to be a float, while using 1.0d specifies that it should be a double (which is also default if you do not specify that explicitely).
If f is not precised at the end, value is considered to be a double.
And a double leads to more bytes in memory than float.
For the new versions of C lang, the conversion from double input to a float variable (assigning operation) is done by the compiler without the 'F'
float fraction1 = 1337;
float fraction2 = 1337.0;
float fraction3 = 1337.0F;
printf("%f, %f, %f", fraction1, fraction2, fraction3);
output(C GNU): 1337.000000, 1337.000000, 1337.000000
I wrote this code:
float b = 3.6;
and I get this:
Error:Unresolved compilation problem:
Type mismatch: cannot convert from double to float
Why? Whats the definition of float?
In Java, when you type a decimal number as 3.6, its interpreted as a double. double is a 64-bit precision IEEE 754 floating point, while floatis a 32-bit precision IEEE 754 floating point. As a float is less precise than a double, the conversion cannot be performed implicitly.
If you want to create a float, you should end your number with f (i.e.: 3.6f).
For more explanation, see the primitive data types definition of the Java tutorial.
Make it
float b= 3.6f;
A floating-point literal is of type float if it is suffixed with an ASCII letter F or f; otherwise its type is double and it can optionally be suffixed with an ASCII letter D or d
Read More
The thing is that decimal numbers defaults to double. And since double doesn't fit into float you have to tell explicitely you intentionally define a float. So go with:
float b = 3.6f;
In JAVA, values like:
8.5
3.9
(and so on..)
Is assumed as double and not float.
You can also perform a cast in order to solve the problem:
float b = (float) 3.5;
Another solution:
float b = 3.5f;
In another Bruce Eckel exercise, the code I've written takes a method and changes value in another class. Here is my code:
class Big {
float b;
}
public class PassObject {
static void f(Letter y) {
y.c = 'z';
} //end f()
static void g(Big z) {
z.b = 2.2;
}
public static void main(String[] args ) {
Big t = new Big();
t.b = 5.6;
System.out.println("1: t.b : " + t.b);
g(x);
System.out.println("2: t.b: " + t.b);
} //end main
}//end class
It's throwing an error saying "Possible loss of precision."
PassObject.java:13: possible loss of precision
found: double
required : float z.b = 2.2
passobject.java:20: possible loss of precision
found : double
required : float t.b = 5.6
Can't doubles be floats as well?
Yes, but you have to specify that they are floats, otherwise they are treated as doubles:
z.b = 2.2f
The 'f' at the end of the number makes it a float instead of a double.
Java won't automatically narrow a double to a float.
No, floats can be automatically upcast to doubles, but doubles can never be floats without explicit casting because doubles have the larger range.
float range is 1.40129846432481707e-45 to 3.40282346638528860e+38
double range is 4.94065645841246544e-324d to 1.79769313486231570e+308d
By default, Java will treat a decimal (e.g. "4.3") as a double unless you otherwise specify a float by adding an f after the number (e.g. "4.3f").
You're having the same problem on both lines. First, the decimal literal is interpreted as a double by the compiler. It then attempts to assign it to b, which is of type float. Since a double is 64 bits and a float is only 32 bits (see Java's primitives documentation), Java gives you an error indicating that the float cannot fit inside the double. The solution is to add an f to your decimal literals.
If you were trying to do the opposite (i.e. assign a float to a double), that would be okay since you can fit a float's 32 bits within a double's 64.
Don't use float. There is almost never a good reason to use it and hasn't been for more than a decade. Just use double.
can't doubles be floats as well?
No. Each value or variable has exactly one type (double, float, int, long, etc...). The Java Language Specification states exactly what happens when you try to assign a value of one type to a variable of another type. Generally, assignments of a "smaller" value to a "larger" type are allowed and done implicitly, but assignments where information could be lost because the target type is too "small" to hold all values of the origin type are not allowed by the compiler, even if the concrete value does fit into the target type.
That's why the compiler complains that assigning a double value (which the literal implicitly is) to a float variable could lose information, and you have to placate it by either making the value a float, or by casting explicitly.
One area that often causes confusions is calculations, because these are implicitly "widened" to int for technical reasons. So if you multiply two shorts and try to assign the result to a short, the compiler will complain because the result of the calculation is an int.
float range is lower than double so a float can be easily represented in double, but the reverse is not possible because, let say we take a value which is out of float range then during convention we will lose the exact data