C# and Java : 3 / 2 * 3.2 = 3.2, why? [duplicate]

Question

Conside the following for C# and Java,

double d = 3 / 2 * 3.2;

Java

System.out.println(d); // 3.2

C#

Console.WriteLine(d); //3.2

It skip the 3/2,

We know that the correct answer should be 4.8

if i change to

double d = 3.00 / 2 * 3.2;

I can get 4.8,

So i want to ask, if (3 / 2 * 3.2) is illegal , why eclipse and vs2008 have no error? And how to prevent this problem in both C# and Java?

Answer 1

3/2 is Integer division, whose result is 1.

1 * 3.2 equals 3.2, which is the result you receive.

This is a proper well-defined formula with well-defined expected results, hence no error by the compiler. Use 3.00, it's the best straightforward way to force the compiler to use floating point math.

From Thorbjørn's answer: another option is to standardize all your floating point calculations by starting with 1.0*, in your example it would be 1.0*3/2*3.2.

Answer 2

3 / 2 is considered as an integer division, so the result comes out to be 1.

Then, performing a multiplication between 1 and 3.2 causes the integer 1 to be promoted to floating point 1, resulting in 3.2.

The idea is:

// Both "3" and "2" are integers, so integer division is performed.
3 / 2 == 1

// "3.2" is a floating point value, while "1" is an integer, so "1" is
// promoted to an floating point value.
1 * 3.2  -->  1.0 * 3.2 == 3.2

When typing 2.0, the decimal point tells Java to consider the literal as a floating point value (in this case, a double), so the result of the calculation is 4.8 as expected. Without the decimal point, the value is a integer literal.

It is not an error, but an issue with how the literals are handled.

The following links have more information:

• Primitive Data Types from The Java Tutorials.
• Conversions and Promotions from The Java Language Specification.