Why does (int)(2/0.9) = 2?

Question

I put this into java and I was given the result 2. Was wondering why it isn't an ArithmeticException, as shouldn't (int)(2/0.9) turn into 2/0. All help is appreciated.

Answer 1

Your code is

(int) (2 / 0.9)

So the int-cast is only applied to the result after computing 2 / 0.9. Now 2 / 0.9 divides an int by a double value. In such a case the int will also be interpreted as double (2.0 in this case). The result is thus like

2.0 / 0.9 = 2.22...

This process is known as numeric promotion (it uses widening primitive conversion here JLS§5.1.2). From JLS§5.6:

A numeric promotion is a process by which, given an arithmetic operator and its argument expressions, the arguments are converted to an inferred target type T. T is chosen during promotion such that each argument expression can be converted to T and the arithmetic operation is defined for values of type T.

After that you cast to int which will round the value to the next lower integer, 2 in this case. Thus the result is 2.

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What you expected would be the result of an expression like

2 / (int) 0.9
// or more explicit:
2 / ((int) 0.9)

which first casts the 0.9 to an int before dividing. In this case you would correctly get

2 / 0

yielding the expected ArithmeticException.