Why precision lost for double casting in Java?

Question

I am surprised to see below difference in Java:

 (double)(int1/int2)  // return 0.0 - precision lost.

 (double)int1/int2    // return 9.3123325E-10 - precision kept.

Is it by design that the bracket in the first expression makes the evaluation lose the precision?

You're using integer maths in the first example (then casting to double), double maths in the second. This is indeed by design — Richard Tingle, Oct 18 '15 at 18:54
@RomanC: Sorry, that dupe doesn't answer the question *at all*. There's no mention of order-of-operations, which is *exactly* what's happening here. — Makoto, Oct 18 '15 at 18:56
Possible duplicate of [Dividing two integers to a double in java](http://stackoverflow.com/questions/19090526/dividing-two-integers-to-a-double-in-java) — Richard Tingle, Oct 18 '15 at 18:56
@RichardTingle: That's not a valid dupe *either*. Seriously, the issue is just order of operations. The OP already identifies that there's integer division occurring, which is why they're trying to cast. — Makoto, Oct 18 '15 at 19:13
@Makoto The answers to that question exactly cover that point — Richard Tingle, Oct 18 '15 at 19:14

score 2 · Accepted Answer · answered Oct 18 '15 at 18:53

Yes, it's by design. By putting the expression in parens, you're modifying the normal precedence, so int1/int2 will execute first, and because both are (presumably) integers, use integer division, which will truncate the result to an integer. In the second case, the normal precedence applies, so int1 is being converted to a double first, then the division done using floating point division.

Makoto · Answer 2 · 2015-10-18T19:08:39.273

2

It's a matter of order-of-operations. Parentheses have a higher priority than casts, so your first statement will perform the operations in the parentheses first - it casts the quotient to a double.

Casts have a higher priority than division, so in your second statement, the cast is applied before the division operation.

edited Oct 18 '15 at 19:08

answered Oct 18 '15 at 18:53

Makoto

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score 1 · Answer 3 · answered Oct 18 '15 at 18:54

1

Because of the brackets (double)(int1/int2) first performs integer division (precision lose included) and then casts the result to double.

The (double)int1/int2 performs a double division because the first operand is double.

answered Oct 18 '15 at 18:54

vlp

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score 1 · Answer 4 · answered Oct 18 '15 at 18:54

Absolutely. In statement 1 you do the division and then cast. In statement 2 you first cast int1 to a double, then you do division which promotes int2 to a double and precision is kept.

I don't intend this to sound snarky, but the language (and compiler) is doing what you asked it to do. :)

score 0 · Answer 5 · answered Oct 18 '15 at 18:53

In the first one, you're doing integer division, then casting to double. Since integer division truncates any decimals, you're losing the decimal bit of the answer, and casting it to double doesn't bring that info back.

In the second, you're casting the top one to double, then doing double division. This makes the answer automatically a double, since it takes the higher-precision unit when dividing, and ensures that you're getting the decimal bit too.

The second is equivalent to ((double) int1) / int2, since casts have higher precedence than /.

score 0 · Answer 6 · answered Oct 18 '15 at 18:54

I believe this has to do with the fact that parenthesis (per PEMDAS) causes the calculation to happen within the parenthesis before the cast occurs and as such, the decimal value if the division is less than 1 is 0.

Example:

(double) (3/5) = (double) 0; (Since Int can't be .6)

But (double) 3/5 --> (double) .6

Why precision lost for double casting in Java?

6 Answers6