I have come across with the following two codes. Why does it not throw an exception for floating point where as in other case it will throw a runtime exception.
class FloatingPoint
    {
      public static void main(String [] args)
       {
         float a=1000f;
         float b=a/0;
        System.out.println("b=" +b);
       }
    }
OUTPUT:b=Infinity.
If I try with int values then it will throw a runtime exception. Why is it like this?
When an exception is thrown using the throw keyword, the flow of execution of the program is stopped and the control is transferred to the nearest enclosing try-catch block that matches the type of exception thrown. If no such match is found, the default exception handler terminates the program.
It will cause integer overflow error. The value of each integer variable is stored in a block of memory of a fixed size.
We have used the parseInt method of the Integer class before. The method throws a NumberFormatException if the string it has been given cannot be parsed into an integer. The above program throws an error if the user input is not a valid number. The exception will cause the program execution to stop.
In that case the float value containing variable may contain same data but reason behind is to mathematical rules. The only two value are consider after the point & remaining are value get neglected. so the we get two different addresses of the same data containing variables.
The short answer
Integral types (JLS 4.2.1) are categorically different from floating point types (JLS 4.2.3). There may be similarities in behavior and operations, but there are also characteristically distinguishing differences such that confusing the two can lead to many pitfalls.
The difference in behavior upon division by zero is just one of these differences. Thus, the short answer is that Java behaves this way because the language says so.
The values of the integral types are integers in the following ranges:
byte: from -128 to 127, inclusive, i.e. [-27, 27-1]
short: from -32768 to 32767, inclusive,  i.e. [-215, 215-1]
int: from -2147483648 to 2147483647, inclusive, i.e. [-231, 231-1]
long: from -9223372036854775808 to 9223372036854775807, inclusive, i.e. [-263, 263-1]
char, from '\u0000' to '\uffff' inclusive, that is, from 0 to 65535, i.e. [0, 216-1]
The floating-point types are float and double, which are conceptually associated with the single-precision 32-bit and double-precision 64-bit format IEEE 754 values and operations.
Their values are ordered as follows, from smallest to greatest:
0.0 == -0.0),Additionally, there are special Not-a-Number (NaN) values, which are unordered. This means that if either (or both!) operand is NaN:
<, <=, >, and >= return false
== returns false
!= returns true
In particular, x != x is true if and only if x is NaN.
For e.g. double, the infinities and NaN can be referred to as:
Double.POSITIVE_INFINITYDouble.NEGATIVE_INFINITYDouble.NaN, testable with helper method boolean isNaN(double)
The situation is analogous with float and Float.
Numerical operations may only throw an Exception in these cases:
NullPointerException, if unboxing conversion of a null reference is requiredArithmeticException, if the right hand side is zero for integer divide/remainder operationsOutOfMemoryError, if boxing conversion is required and there is not sufficient memoryThey are ordered by importance, with regards to being common source for pitfalls. Generally speaking:
null
For integer operation:
ArithmeticException if the right hand side is zeroFor floating point operation:
NaN or 0, the result is NaN.NaN
The general rule for all floating point operation is as follows:
NaN.NaN as an operand produce NaN as a result.There are still many issues not covered by this already long answer, but readers are encouraged to browse related questions and referenced materials.
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