在Java/Android上的BigDecimal

Question

编辑：整个计算很复杂，有很多类似这样的操作。我需要尽可能精确的结束。

Answer 1

你需要多少位数的精度？在您的示例中，您只使用了4位数字。如果这是医学和现实世界，你只能测量现实世界中的大多数东西10-13位数的准确性，而double有高达16位数的准确性。

System.out.println(Math.pow(11.11, -1.54));

印刷品

0.024524510581710988

如果您使用这本书的库http://www.apropos-logic.com/nc/，您可以获得

int runs = 10000;

long start = System.nanoTime();
double x1 = 0;
for (int i = 0; i < runs; i++)
    x1 = Math.pow(11.11, -1.54);
long time = System.nanoTime() - start;
System.out.println(x1 + " took " + time / runs / 1e3 + " us avg.");

long start2 = System.nanoTime();
BigDecimal x2 = null;
for (int i = 0; i < runs; i++)
    x2 = exp(ln(BigDecimal.valueOf(11.11), 20).multiply(BigDecimal.valueOf(-1.54)), 20);
long time2 = System.nanoTime() - start2;
System.out.println(x2 + " took " + time2 / runs / 1e3 + " us avg.");

0.024524510581710988 took 0.478 us avg.
0.02452451058171098739 took 603.769 us avg.

0.0245245105817109873886495555036930857940 took 1409 us avg.