Different answer when converting a Double to an Int - Java vs .Net

Question

In C# if I want to convert a double (1.71472) to an int then I get the answer 2. If I do this in Java using intValue() method, I get 1 as the answer.

Does Java round down on conversions?

Why do the Java API docs have such scant information about their classes i.e.

Returns the value of the specified number as an int. This may involve rounding or truncation.

A bit more info about the rounding would have been helpful!

Answer 1

Java rounds toward zero when narrowing from a floating point to an integer type—and so does C#, when you use the casting conversion. It's Convert.ToInt32 that rounds:

double d = 1.71472;
int x = (int) d; // x = 1
int y = Convert.ToInt32(d); // y = 2

Details can be found in the Java Language Specification. Note that while the documentation cited in Number leaves options open for subclasses, the documentation on the concrete boxing types, like Double, is explicit about the implementation:

Returns the value of this Double as an int (by casting to type int).

When using BigDecimal, you can specify one of eight different rounding policies.

Answer 2

Java rounds toward zero when narrowing from a floating point to an integer type.

There's more documentation about the general rules in the Java Language Specification. Note that while the documentation in Number leaves options open for subclasses, the documentation on the concrete boxing types is more explicit:

Returns the value of this Double as an int (by casting to type int).

When using BigDecimal, you can specify one of eight different rounding policies.

Answer 3

Jon Skeet is correct, but something else to watch for is that .NET uses Banker's Rounding as its rounding algorithm. When you're halfway between round towards the even integer.