有没有一种打印精度达到指数级的方法?我需要用微小的数字填充一些数据,然后将它们进行比较……
我有这样的想法:
(0) - Mesured estimatedTime for multithread in block: 1.4917951E16 s
(1) - Mesured estimatedTime for multithread in block: 7.531891E15 s
(2) - Mesured estimatedTime for multithread in block: 2.9287E13 s
(3) - Mesured estimatedTime for multithread in block: 3.28478435E17 s
(4) - Mesured estimatedTime for multithread in block: 6.038E12 s我想这样打印:
(0) - Mesured estimatedTime for original in block: 0000.15595175E15 s
(1) - Mesured estimatedTime for original in block: 0007.335638E15 s
(2) - Mesured estimatedTime for original in block: 0416.66E15 s
(3) - Mesured estimatedTime for original in block: 0000.0390156852E15 s
(4) - Mesured estimatedTime for original in block: 6642.0E15 s我知道你可以像这样强迫它:
// Force minimum number of digits to left and right of decimal point
formatter = new DecimalFormat("0.0E0");
s = formatter.format(-1234.567); // -1.2E3但我不知道如何才能将它强制到Expo-Precision:( help!!:/
发布于 2012-08-30 23:03:49
看起来你想暗示1e15的精度。
for (double d : new double[]{1.4917951E16, 7.531891E15, 2.9287E13, 3.28478435E17, 6.038E12})
System.out.printf("%11fE15%n", d/1e15);打印
14.917951E15
7.531891E15
0.029287E15
328.478435E15
0.006038E15如果范围大得多,可以使用DecimalFormat
DecimalFormat df = new DecimalFormat();
df.setMaximumFractionDigits(300);
df.setGroupingUsed(false);
for(double d = 1 ; d <= 1e30; d *= 1e5) {
System.out.println(df.format(d/1e15)+"E15");
}打印
0.000000000000001E15
0.0000000001E15
0.00001E15
1E15
100000E15
10000000000E15https://stackoverflow.com/questions/12199815
复制相似问题