Moving decimal places over in a double

So I have a double set to equal 1234, I want to move a decimal place over to make it 12.34

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

double x = 1234;
for(int i=1;i<=2;i++)
{
  x = x*.1;
}
System.out.println(x);

This will print the result, "12.340000000000002"

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

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

NOTE: This has nothing to do with randomness in your system (or your power supply). This is due to a representation error, which will produce the same outcome every time. The precision of double is limited and in base 2 rather than base 10, so numbers which can be precisely represented in decimal often cann't be precisely represented in base 2.

No - if you want to store decimal values accurately, use BigDecimal. double simply can't represent a number like 0.1 exactly, any more than you can write the value of a third exactly with a finite number of decimal digits.

if it's just formatting, try printf

double x = 1234;
for(int i=1;i<=2;i++)
{
  x = x*.1;
}
System.out.printf("%.2f",x);

output

12.34

In financial software it is common to use integers for pennies. In school, we were taught how to use fixed-point instead of floating, but that is usually powers of two. Storing pennies in integers might be called "fixed point" as well.

int i=1234;
printf("%d.%02d\r\n",i/100,i%100);

In class, we were asked in general what numbers can be exactly represented in a base.

For base=p1^n1*p2^n2... you can represent any N where N=np1^m1p2^m2.

Let base=14=2^1*7^1... you can represent 1/7 1/14 1/28 1/49 but not 1/3

I know about financial software -- I converted Ticketmaster's financial reports from VAX asm to PASCAL. They had their own formatln() with codes for pennies. The reason for the conversion was 32 bit integers were no longer enough. +/- 2 billion pennies is $20 million and that overflowed for the World Cup or Olympics, I forgot.

I was sworn to secrecy. Oh well. In academea, if it's good you publish; in industry, you keep it secret.

you can try integer number representation

int i =1234;
int q = i /100;
int r = i % 100;

System.out.printf("%d.%02d",q, r);

This is caused by the way computers store floating-point numbers. They don't do so exactly. As a programmer, you should read http://floating-point-gui.de/">this floating-point guide to familiarize yourself with the trials and tribulations of handling floating-point numbers.

Funny that numerous posts mention to use BigDecimal but no-one bothers to give the correct answer based on BigDecimal? Because even with BigDecimal, you can still go wrong, as demonstrated by this code

String numstr = "1234";
System.out.println(new BigDecimal(numstr).movePointLeft(2));
System.out.println(new BigDecimal(numstr).multiply(new BigDecimal(0.01)));
System.out.println(new BigDecimal(numstr).multiply(new BigDecimal("0.01")));

Gives this output

12.34
12.34000000000000025687785232264559454051777720451354980468750
12.34

The BigDecimal constructor specifically mentions that it is better to use String constructor than a numeric constructor. Ultimate precision is also influenced by the optional MathContext.

According to the BigDecimal Javadoc it is possible to create a BigDecimal which is exactly equal to 0.1, provided you use the String constructor.

Yes, there is. With each double operation you may lose accuracy but the amount of accuracy differs for each operation and can be minimized by choosing the right sequence of operations. For example when multiplying set of numbers, it is best to sort set by exponent before multiplying.

Any decent book on number crunching describes this. For example: http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html

And to answer your question:

Use divide instead of multiply, this way you get correct result.

double x = 1234;
for(int i=1;i<=2;i++)
{
  x =  x / 10.0;
}
System.out.println(x);

No, as Java floating point types (indeed all floating point types) are a trade-off between size and precision. While they're very useful for a lot of tasks, if you need arbitrary precision, you should use BigDecimal.