Result of Math.round() is ignored

Question

Why is it not working?I want to "odleglosc" is 2.2 not 2.214354356

`Math.round` doesn't round in place but returns a value. Since you never save the result to a variable, it is thrown away. — QBrute, Feb 14 '18 at 11:53
you are not storing the value returned by Math.round(). Use codeXXX= Math.round(codeXXX) — akshaya pandey, Feb 14 '18 at 11:54
Another way to state this is that Java is call-by-value, not call-by-reference. For your code to work, `Math.round(x)` would have to be a call-by-reference call. — Stephen C, Feb 14 '18 at 12:00
This not work, because `Math.round` is int type, and it's rounding values by half up to the fully int value (not float/double). Let use `NumberFormat` class instead. — grabarz121, Feb 14 '18 at 12:04

score 2 · Answer 1 · answered Feb 14 '18 at 11:54

2

Math.round(argument) returns a number that is rounded from the argument.

In your example you ignore the returned value.

You probably meant to write:

 odleglosc = Math.round(odleglosc);

answered Feb 14 '18 at 11:54

assylias

score 1 · Answer 2 · answered Feb 14 '18 at 12:28

1

x = Math.round(x);

Otherwise if you just write Math.round(x); Java will make the calculation and have no variable to assign it to, and gets thrown away.

answered Feb 14 '18 at 12:28

Oygen87

score 0 · Answer 3 · answered Feb 14 '18 at 11:54

0

Math.round() does not modify your variable because a double the value is passed to the function (compare all-by-value vs call-by-reference).

To round your value use

a = Math.round(a);

answered Feb 14 '18 at 11:54

Uwe Plonus

And see https://stackoverflow.com/questions/40480/is-java-pass-by-reference-or-pass-by-value – Stephen C Feb 14 '18 at 12:02

3 Answers3