Why does Java compile this equation: 78 / 10 * 10 + 10 - 78 equal to 2?

Question

I've been trying to understand java operator precedence but I can't wrap my head around this equation someone has given to me. By adding parentheses myself I get (((78 / 10) * 10) + 10) - 78 but this shouldn't equate to 2. Can someone explain this to me? Thanks

Because it is, int division will truncate. The parenthesis are unnecessary. If you want the mathematically correct answer use `float` or `double`: `78.0 / 10 * 10 + 10 - 78`, the `.0` in the first term will ensure it's done using `double` (or `float`?, check documentation) instead of `int`. — Everyone, Dec 03 '20 at 22:10

DCR · Accepted Answer · 2020-12-03T22:25:17.443

1

78 / 10 = 7
7 * 10 = 70
70 + 10 = 80
80 - 78 = 2

remember an int /int will be an int

a float / int will be a float so you could do the following:

(float)78 / 10 * 10 + 10 - 78 or 78.0 / 10 * 10 + 10 - 78

edited Dec 03 '20 at 22:25

answered Dec 03 '20 at 22:06

DCR

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I'm such an idiot, I completely forgot it stay an int, thanks so much. – Joe Swindells Dec 03 '20 at 22:09
no problem, accept answer if you're clear on what's going on – DCR Dec 03 '20 at 22:10
Add to the answer how to go around this by casting (implicit or explicit). – Everyone Dec 03 '20 at 22:12

score 0 · Answer 2 · answered Dec 03 '20 at 22:07

0

It's because in integer math 78 / 10 is 7, not 7.8

Then:

 7 * 10 = 70
   + 10 = 80
   - 78 =  2

answered Dec 03 '20 at 22:07

Alnitak

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Why does Java compile this equation: 78 / 10 * 10 + 10 - 78 equal to 2?

2 Answers2