A few related questions here.
As per the title, why is it a requirement if we are specifying the variable type as long or float, double? Doesn't the compiler evaluate the variable's type at compile time?
Java considers all integral literals as int - is this to lessen the blow of inadvertent memory waste? And all floating-point literals as double - to ensure highest precision?
When you have a constant there are subtle differences between value which look the same, but are not. Additionally, since autoboxing was introduce, you get a very different result as less.
Consider what you get if you multiply 0.1 by 0.1 as a float or as a double and convert to a float.
float a = (float) (0.1 * 0.1);
float b = 0.1f * 0.1f;
System.out.println("a= "+new BigDecimal(a));
System.out.println("b= "+new BigDecimal(b));
System.out.println("a == b is " + (a == b));
prints
a= 0.00999999977648258209228515625
b= 0.010000000707805156707763671875
a == b is false
Now compare what you get if you use either float or int to perform a calculation.
float a = 33333333f - 11111111f;
float b = 33333333 - 11111111;
System.out.println("a= "+new BigDecimal(a));
System.out.println("b= "+new BigDecimal(b));
System.out.println("a == b is " + (a == b));
prints
a= 22222220
b= 22222222
a == b is false
Compare int and long
long a = 33333333 * 11111111; // overflows
long b = 33333333L * 11111111L;
System.out.println("a= "+new BigDecimal(a));
System.out.println("b= "+new BigDecimal(b));
System.out.println("a == b is " + (a == b));
prints
a= -1846840301
b= 370370362962963
a == b is false
compare double with long
double a = 333333333333333333L / 333333333L;
double b = 333333333333333333D / 333333333D;
System.out.println("a= "+new BigDecimal(a));
System.out.println("b= "+new BigDecimal(b));
System.out.println("a == b is " + (a == b));
prints
a= 1000000001
b= 1000000000.99999988079071044921875
a == b is false
In summary its possible to construct a situation where using int, long, double or float will produce a different result compared with using another type.