The following statement throws java.lang.ArithmeticException: / by zero as obvious.
System.out.println(0/0);
because the literal 0 is considered to be an int literal and divide by zero is not allowed in integer arithmetic.
The following case however doesn’t throw any exception like java.lang.ArithmeticException: / by zero.
int a = 0;
double b = 6.199;
System.out.println((b/a));
It displays Infinity.
The following statement produces NaN (Not a Number) with no exception.
System.out.println(0D/0); //or 0.0/0, or 0.0/0.0 or 0/0.0 - floating point arithmetic.
In this case, both of the operands are considered to be double.
Similarly, the following statements don’t throw any exception.
double div1 = 0D/0; //or 0D/0D
double div2 = 0/0D; //or 0D/0D
System.out.printf("div1 = %s : div2 = %s%n", div1, div2);
System.out.printf("div1 == div2 : %b%n", div1 == div2);
System.out.printf("div1 == div1 : %b%n", div1 == div1);
System.out.printf("div2 == div2 : %b%n", div2 == div2);
System.out.printf("Double.NaN == Double.NaN : %b%n", Double.NaN == Double.NaN);
System.out.printf("Float.NaN == Float.NaN : %b%n", Float.NaN == Float.NaN);
They produce the following output.
div1 = NaN : div2 = NaN
div1 == div2 : false
div1 == div1 : false
div2 == div2 : false
Double.NaN == Double.NaN : false
Float.NaN == Float.NaN : false
They all return false. Why is this operation (division by zero) allowed with floating point or double precision numbers?
By the way, I can understand that floating point numbers (double precision numbers) have their values that represent positive infinity, negative infinity, not a number (NaN)…
In short, that’s the way it’s specified in the IEEE-754 standard, which is what Java’s Floating-Point Operations are based on.