Consider the following code snippet
float num = 281.583f;
int amount = (int) Math.round(num*100f);
float rounded = amount/100.0f;
double dblPrecision = rounded;
double dblPrecision2 = num;
System.out.println("num : " + num + " amount: " + amount + " rounded: " + rounded + " dbl: " + dblPrecision + " dbl2: " + dblPrecision2);
The output I get is
num : 281.583 amount: 28158 rounded: 281.58 dbl: 281.5799865722656 dbl2: 281.5830078125
Why is there the approximation when a float number is assigned to a double variable?
Approximation actually takes place when you convert decimal fraction to float. I might surprise you, but 281.583 can't be represented exactly as floating point number in PC. it happens because floating point numbers are represented as sum of binary fractions in PC. 0.5, 0.25 and 0.125 can be converted precisely, but not 0.583.
Floats (and doubles) are represented as Σ( 1/2^i*Bi ), where Bi is i-th bit (0|1). 0.625 = 1/2 + 1/4 for example. The problem is that not all decimal fraction can be converted to finitie sum of binary fractions.
Here is how this number is converted (first line is columns definition).
i| *2 and trim| Bit value| (2^-1)*bit
0,583
1 1,166 1 0,5
2 0,332 0 0
3 0,664 0 0
4 1,328 1 0,0625
5 0,656 0 0
6 1,312 1 0,015625
7 0,624 0 0
8 1,248 1 0,00390625
9 0,496 0 0
10 0,992 0 0
11 1,984 1 0,000488281
12 1,968 1 0,000244141
13 1,936 1 0,00012207
14 1,872 1 6,10352E-05
15 1,744 1 3,05176E-05
16 1,488 1 1,52588E-05
17 0,976 0 0
18 1,952 1 3,8147E-06
19 1,904 1 1,90735E-06
SUM= 0,582998276