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Why do you need to place an 'f' at the end of the decimal number when declaring a float variable?

I have seen this question asked before. However, I am not satisfied with the answers that are given. Typical responses are that Java sees the number as a double, since that is the default in JAVA, and gives a mismatch error on compiling.
This behavior is apparently totally ignoring my use of a float declaration.
My question is: If I am declaring a variable type as float, why does Java think it is a double? I am specifically declaring:
float x = 3.14;
I have told Java that I want this variable to be a float type by using the float keyword. Why is this required?
float x = 3.14f;
If I want a double type variable, I use the double keyword specifically declaring the variable as a double.
double x = 3.14;
Why does Java not recognize this when declaring a variable and insists the literal needs to be cast from a double to a float by adding the 'f' at the end? Shouldn't Java recognize the use of the keyword float and properly assign the type to the literal?

Size matters
32-bit float → float x = 3.14 ; ← 64-bit double
The default for a fractional numeric literal in Java is the 64-bit primitive double type, as shown in the Answer by Mark Rotteveel. 3.14 is a double, as is 3.14d and 3.14D.
The primitive float type is smaller, 32-bit.
If Java were to infer that what you said is a double but what you meant is a float, Java would have to reduce the 64-bit value to a 32-bit value, discarding half the data, 32-bits. This could involve data loss for larger numbers. And this would definitely mean a loss of capacity.
So the designers of Java decided to be conservative in this matter. They decided to not be so presumptive as to throw away half your data, nor to reduce your specified capacity by half. They decided to take you at your word — if you said 64-bit double, they give you a double.
If you then go on to say you want to cram that 64-bit number into a 32-bit container, they flag the error, as 64-bits does not fit into 32-bits. Square peg, round hole.
round hole, 32-bit float → float x = 3.14 ; ← 64-bit double, square peg
Java does support the converse, a widening conversion. You can put a 32-bit float into a 64-bit double. And you can put a 32-bit int into a 64-bit long. These operations are safe, as the extra 32-bits can be filled safely with zeros. So Java is willing and able to support these operations.
Be explicit
I suggest you make a habit of being explicit. Do not rely on the default. If you mean a 32-bit float, append the f or F to your literal. If you mean a 64-bit double, append the d or D to your literal.
Doing so not only quiets the compiler, it makes your intentions clear to the humans reading your code.
float pi = 3.14F ;
double pi = 3.14d ;
BigDecimal
And, by the way, if you care about accuracy rather than performance, you’ll be using BigDecimal rather than float/double. In which case this issue is moot.
BigDecimal pi = new BigDecimal( "3.14" ) ;

The reason is that the Java Language Specification specifies that a floating point literal without the f or F suffix, is a double. From Floating-Point Literals in the Java 15 Language Specification:
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.
Java doesn't try to infer the type of the literal based on the type of the variable you're assigning to. A double-literal is simply always a double. In other words, semantically, it behaves the same as if your code did:
double temp = 3.14;
float x = temp;
Which will also fail to compile, because assigning a double to a float can result in loss of information, if the magnitude of the value is too large to fit in a float. This is known as narrowing primitive conversion, and requires an explicit cast to tell the compiler you're aware of the risk and you accept it.
In some cases, it may seem as if Java will perform inference of type from the variable (e.g. when assigning an int or long literal to a double or float variable), but that is because the entire range of int and long can be stored in a float or double (possibly with loss of precision). This is known as widening primitive conversion.

Why will Java convert the double to float type in this situation?

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.

Is scientific notation interpreted as int or float?

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.

Why is the letter f used at the end of a float no.?

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

why f is placed after float values?

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.