If I hardcode a number in my code using the scientific notation (e.g. 1e9) what will the type of that number be (int, long, float, double..)?
When the significand or the exponent are floating point numbers it can't be an integer obviously, but what in the above case?
The e makes it a floating-point literal. From the JLS (§3.10.2. Floating-Point Literals):
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 (§4.2.3).
Therefore, 1e9 of type double, as is 1e9d. On the other hand, 1e9f is of type float.
These will usually be of type double.
If you put an f (of F) behind it, it's a float.
double d = 1e9;
float f = 1e9f;
It will be considered as a floating point literal, and without a trailing f or F it will be a double.
Integer literals can't use scientific notation.
It'll be 'double' unless you use 'f' or 'F' then it'll be float literal.
It is better to use Double. Because a it has decimal floating point is a computer arithmetic system closely related to scientific notation.
for example:
0.000 000 007 51
in scientific notation it is 7.51×10-9.
when you use it as int, 7.51 would be 7 while when you use int it will be 7.51.
Related
This question already has answers here:
Declaring floats, why default type double?
(3 answers)
Closed 6 years ago.
Ok I got a couple values here:
'F', 0xf, 070, 70L, 77e-1f, 7.7f, 7.77e1, 77.7
I know that 'F' is a char and that 70L has the type long. Also, I know that 0xf is hex(right?) and that 070 is octal. But what are those other numbers? And why the hell is 77.7 double and not float as well?
What you're looking for is this
About 77.7:
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.
0xf is not of type hex. There is no type hex. It's simply an int wirtten in hex. Just like 070 is an int written in octal.
The literal 77e-1fis clearly a float since it ends with f.
The e is exponent i.e. 77e-1f is in fact 77 * 10^(-1) or 7.7. The literal 7.77e1 is a double for the same reason 77.7 is a double, it's just that 7.77e1 is equal to 7.77 * 10 ^ 1 = 77.7.
In Java numerical literals that contain a dot are of type double by default. If you want a float, you neef to append the f.
The numbers containing the e are in exponential or scientific notation, but I am not going to explain it here because it has nothing to do with programming.
The only two which you didn't explain yourself:
7.7f
That f means float - it indicates that this should not be a double literal, but well, a float one! You see, by default any "floating point" literals are automatically of type double, you need the "f" in order to enforce the "smaller" float type.
And
77e-1f
is using scientific notation, see here for example for further explanations.
77e-1 means this: 77 is number-multiplier, e (or E) means base 10, and -1 means exponent.
So the result is 77 by (10 powered by -1), or 77 by .1, i. e. 7.7
The name of this notation is scientific or logarithmic and is good for very large or very small numbers, e. g. 1 million = 1.000.000 may be expressed as 1e6
And why the hell is 77.7 double and not float as well?
Float numbers has to be denoted with f on the end but double numbers don't require d on the end.
77e-1f is a float number 77 to the power of 10^-1, so it equals 7.7.
The same with 7.77e1 == 77.7
I was reading a book about Java and I found the following points unclear, please help me:
For integer literal expressed in any base other than base 10 (0b, 0, 0x) can we use the L suffix that stands for Long?
For floating point can we use any other base other than decimal? If yes can we specify float or double using F or D for other bases other than 10?
If yes with other bases than 10 could we use scientific notation or only decimal point is allowed?
1) Yes, it's possible for hexadecimal, octal and binary too. See jls-3.10.1
2) Yes, you can use hexadecimal notation, but you are restricted to binary exponents and specifying the exponent is required. See jls-3.10.2
Examples:
0xFF.Ap0d
0xFF.1p0f
0xFF.Ap1d
0xFF.Ap-1f
0xFF.Ap-1
0x.1p16
If you print these literals using System.out.println, you get:
255.625
255.0625
511.25
127.8125
63.90625
4096.0
The meaning of the binary exponent is as follows:
The value in front of the p or P is multipled by 2^z where z is the integer after the p (or P). The integer is in decimal format. E.g. 0xFF.1Ap0101d stands for 255.1015625 * 2^101.
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 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.
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
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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;