I did clumsy,
double d= Double.valueOf("5.0596"); int i= (int) d; double d2= d- i; // дробная часть How can you do differently?
1 answer
As already advised, d%1 performs exactly what is needed, with the most syntactically short method. On the other hand, it is not necessarily the cheapest method in the calculation; Your "clumsy" method in this sense can be even better; if speed is important, it is better to make a benchmark compared to subtracting the whole part.
Note that int may not hold all integer values for double within the limits where double represents all integers. For long, this is at least for a range where the step of double values is 1 or less. And the conversion of larger values works in such a way that it silently produces the corresponding extreme integer, so the result should be checked.
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