Reputation: 33

Java: Variable assignment issue

just a quick question and I'm probably gonna feel stupid for asking but still would like to know why it is so...!

Anyways, quick example:

x is a double.

double conversion = (x-32)*5/9;

This does the maths just fine.

double  conversion = (x-32)*(5/9);

This isn't fine because the (5/9) is being treated as an int, thus result is overall 0.

double  conversion = (x-32)*(5f/9f);

This does the maths just fine, as it explicitly makes the 5/9 values a float.

So my question is: Why does the first equation work perfectly fine? ( double conversion = (x-32)*5/9; )

Why isn't the 5/9 being made a 0 if it were an int supposedly? What makes the 5/9 different from (5/9) ?

Upvotes: 1

Answers (7)

Elliott Frisch

Reputation: 201409

It's a matter of when the conversion takes place. 5/9 consists of one 5 and one 9 (both ints) being divided with integer division. If either is a float (5f/9 or 5/9f) they will divide as floats.

Upvotes: 0

Naveen Ramawat

Reputation: 1445

Your value of x might be double and it is making entire equation in double because brackets execute first.

Upvotes: 0

Chuidiang

Reputation: 1055

The reason is the order of the operations.

(x-32)*5/9 makes first (x-32)*5 and then divides the result by 9

(x-32)*(5/9) makes first (x-32) and (5/9). After both results are multiplied.

Upvotes: 0

Sanjay Manohar

Reputation: 7026

It is the order it's done in. If you multiply by 5, you get a large number. If you then divide by 9, you still get an int, But the remainder is discarded.

Upvotes: 0

Jon Skeet

Reputation: 1499790

The difference is between whether you do the multiplication first or the division first - and what the types of those operations are.

This:

(x - 32) * 5 / 9

is equivalent to:

((x - 32) * 5) / 9

So if the type of x is double, then the type of x - 32 is double, so the 5 is promoted to double, the multiplication is done in double arithmetic, giving a double result, and then the division is also done in double arithmetic.

Even if x is an integer type, you're doing the multiplication first, which will presumably give you a value bigger than 9 (in your test case), leaving you with a non-zero result. For example, if x is 45, then x-32 is 13, (x - 32) * 5 is 65, and the overall result is 7, then converted to 7.0 on assignment. That's not the same result you'll get if x is a double with the value 45.0, but it's still better than multiplying by 0...

Upvotes: 4