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 usedouble
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
Comments
Post a Comment