### Moving the decimal place in a double

This is taken from a popular answer to the question Moving decimal places over in a double

So to do this I multiply .1 to 1234 two times, kinda like this

Is there a way, without simply formatting it to two decimal places, to have the double store 12.34 correctly?

The problem you have is that 0.1 is not an exact representation, and by performing the calculation twice, you are compounding that error.

However, 100 can be represented accurately, so try:

which prints:

This works because

prints:

In short, rounding is unavoidable for sensible answers in floating point whether you are doing this explicitly or not.

Note:

prints

## Question

So I have a double set to equal 1234, I want to move a decimal place over to make it 12.34So to do this I multiply .1 to 1234 two times, kinda like this

This will print the result, "12.340000000000002"`double x = 1234; for(int i = 1; i <= 2; i++) { x = x*.1; } System.out.println(x);`

Is there a way, without simply formatting it to two decimal places, to have the double store 12.34 correctly?

## Answer

If you use`double`

or `float`

, you should use rounding or expect to see some rounding errors. If you can't do this, use `BigDecimal`

.The problem you have is that 0.1 is not an exact representation, and by performing the calculation twice, you are compounding that error.

However, 100 can be represented accurately, so try:

```
double x = 1234;
x /= 100;
System.out.println(x);
```

which prints:

`12.34`

This works because

`Double.toString(d)`

performs a small amount of rounding on your behalf, but it is not much. If you are wondering what it might look like without rounding:```
System.out.println(new BigDecimal(0.1));
System.out.println(new BigDecimal(x));
```

prints:

```
0.100000000000000005551115123125782702118158340454101562
12.339999999999999857891452847979962825775146484375
```

In short, rounding is unavoidable for sensible answers in floating point whether you are doing this explicitly or not.

Note:

`x / 100`

and `x * 0.01`

are not exactly the same when it comes to rounding error. This is because the round error for the first expression depends on the values of x, whereas the `0.01`

in the second has a fixed round error.```
for(int i = 0; i < 200; i++) {
double d1 = (double) i / 100;
double d2 = i * 0.01;
if (d1 != d2)
System.out.println(d1 + " != "+d2);
}
```

prints

```
0.35 != 0.35000000000000003
0.41 != 0.41000000000000003
0.47 != 0.47000000000000003
0.57 != 0.5700000000000001
0.69 != 0.6900000000000001
0.7 != 0.7000000000000001
0.82 != 0.8200000000000001
0.83 != 0.8300000000000001
0.94 != 0.9400000000000001
0.95 != 0.9500000000000001
1.13 != 1.1300000000000001
1.14 != 1.1400000000000001
1.15 != 1.1500000000000001
1.38 != 1.3800000000000001
1.39 != 1.3900000000000001
1.4 != 1.4000000000000001
1.63 != 1.6300000000000001
1.64 != 1.6400000000000001
1.65 != 1.6500000000000001
1.66 != 1.6600000000000001
1.88 != 1.8800000000000001
1.89 != 1.8900000000000001
1.9 != 1.9000000000000001
1.91 != 1.9100000000000001
```

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