Why does the use of integer variables throw an exception?_问答_开发者

I have come across with the following two codes. Wh开发者_开发知识库y 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?

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.

On integral and floating point values

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:

negative infinity,
negative finite nonzero values,
positive and negative zero (i.e. 0.0 == -0.0),
positive finite nonzero values, and
positive infinity.

Additionally, there are special Not-a-Number (NaN) values, which are unordered. This means that if either (or both!) operand is NaN:

numerical comparison operators <, <=, >, and >= return false
numerical equality operator == returns false
numerical inequality operator != 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_INFINITY
Double.NEGATIVE_INFINITY
Double.NaN, testable with helper method boolean isNaN(double)

The situation is analogous with float and Float.

On when exceptions may be thrown

Numerical operations may only throw an Exception in these cases:

NullPointerException, if unboxing conversion of a null reference is required
ArithmeticException, if the right hand side is zero for integer divide/remainder operations
OutOfMemoryError, if boxing conversion is required and there is not sufficient memory

They are ordered by importance, with regards to being common source for pitfalls. Generally speaking:

Be especially careful with box types, as just like all other reference types, they may be null
Be especially careful with the right hand side of an integer division/remainder operations
Arithmetic overflow/underflow DOES NOT cause an exception to be thrown
Loss of precision DOES NOT cause an exception to be thrown
A mathematically indefinite floating point operation DOES NOT cause an exception to be thrown

On division by zero

For integer operation:

Division and remainder operations throws ArithmeticException if the right hand side is zero

For floating point operation:

If the left operand is NaN or 0, the result is NaN.
If the operation is division, it overflows and the result is a signed infinity
If the operation is remainder, the result is NaN

The general rule for all floating point operation is as follows:

An operation that overflows produces a signed infinity.
An operation that underflows produces a denormalized value or a signed zero.
An operation that has no mathematically definite result produces NaN.
All numeric operations with NaN as an operand produce NaN as a result.

Appendix

There are still many issues not covered by this already long answer, but readers are encouraged to browse related questions and referenced materials.

Related questions

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Why null == 0 throws NullPointerException in Java? (beware the danger of boxed primitives!)
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Because floats actually have a representation for the "number" you're trying to calculate. So it uses that. An integer has no such representation.

Java (mostly) follows IEEE754 for its floating point support, see here for more details.

It is because integer arithmetic always wraps it's result except for the case of (Division/Remainder By Zero).
In case of float, when there is an overflow or underflow, the wrapping goes to 0, infinity or NaN.
During the overflow, it gives infinity and during underflow, it gives 0.
Again there are positive & negative overflow/underflow.
Try:

float a = -1000;
float b = a/0;
System.out.println("b=" +b);

This gives a negative overflow

Output

b=-Infinity

Similarly positive underflow will result in 0 and negative underflow in -0.
Certain operations can also result in returning a NaN(Not a Number) by float/double.
For eg:

float a = -1000;
double b = Math.sqrt(a);
System.out.println("b=" +b);

Output

b=NaN

It's a programming and math standard for representing / by zero values. float has support for representing such values in JAVA. int (integer) data type doesn't have way to represent same in JAVA.

Check :

http://en.wikipedia.org/wiki/Division_by_zero

http://www.math.utah.edu/~pa/math/0by0.html