Werner van Mook - 1 year ago 61

Java Question

I have the following two calculation using Math.round(...):

`double x = 0.57145732;`

x = x * 100;

x = Math.round(x * 10);

x = x / 10;

If I now print the value of x it will show me: 57.1.

`double x = 0.57145732;`

x = (Math.round((x * 100) * 10)) / 10;

// x = (Math.round(x * 1000)) / 10; //Also gives me 57.0.

If I now print the value of x it will show me: 57.0.

Why is there this difference in the outcome?

Recommended for you: Get network issues from **WhatsUp Gold**. **Not end users.**

Answer Source

The `Math.round()`

method returns an `int`

(or `long`

. Ole V.V corrected my mistake). Many people think that it returns `float`

or `double`

which gives rise to confusions as these.

In the second calculation,

```
Math.round((x * 100) * 10)
```

returns `571`

. Now, this value and `10`

both are integers (571 is long, 10 is int). So when the calculation takes the form

```
x = 571 / 10
```

where x is double, `571/10`

returns `57`

instead of `57.1`

since it is `int`

. Then, `57`

is converted to double and it becomes `57.0`

If you do

```
x = (double)Math.round((x * 100) * 10) / 10.0;
```

its value becomes `57.1`

.

**Edit**: There are two versions of the `Math.round()`

function. The one you used accepts a double (since x is double) and returns `long`

. In this case, long and int make no difference though.

Recommended from our users: **Dynamic Network Monitoring from WhatsUp Gold from IPSwitch**. ** Free Download**