假设我执行此操作：
(X / const) * const
使用由 定义的双精度参数IEEE 754-2008，首先除法，然后是乘法。
const在范围内0 < ABS(const) < 1。
假设操作成功（没有发生溢出），X该操作的不同参数是否保证返回不同的结果？
换句话说，有没有X1，X2等等0 < ABS(const) < 1，X1 <> X2但是(X1 / const) * const = (X2 / const) * const？
问问题
695 次
2 回答
7
是的。
公共类TestDoubleDivision
{
公共静态无效主要（字符串 [] 参数）
{
最终随机随机=新随机（）；
诠释 i = 0;
而 (i < 10)
{
最终双 c = random.nextDouble();
最终双 x1 = 10.0 * random.nextDouble();
最终双 x2 = nextDouble(x1);
如果 (x1 / c * c == x2 / c * c)
{
System.out.printf("x1 = %.20f, x2 = %.20f, c = %.20f\n", x1, x2, c);
我++；
}
}
}
私人静态双 nextDouble（双 d1）
{
返回 Double.longBitsToDouble(Double.doubleToLongBits(d1) + 1);
}
}
印刷
x1 = 5.77383813703796800000，x2 = 5.77383813703796900000，c = 0.15897456707659440000 x1 = 2.97635611350670850000，x2 = 2.97635611350670900000，c = 0.15347615678619309000 x1 = 7.98634439050267450000，x2 = 7.98634439050267500000，c = 0.83202322046715640000 x1 = 0.11618686267768408000，x2 = 0.11618686267768409000，c = 0.09302449134082225000 x1 = 0.98646731978098480000，x2 = 0.98646731978098490000，c = 0.40549842805620606000 x1 = 3.95828649870362700000，x2 = 3.95828649870362750000，c = 0.75526917984495820000 x1 = 1.65404856207794440000，x2 = 1.65404856207794460000，c = 0.14500102367827516000 x1 = 5.72713430182017500000，x2 = 5.72713430182017550000，c = 0.68241935505532810000 x1 = 3.71143195248990980000，x2 = 3.71143195248991000000，c = 0.21294683305890750000 x1 = 5.66441726170857800000，x2 = 5.66441726170857900000，c = 0.69355199625947250000
于 2010-02-10T21:37:07.797 回答
2
（我只是想在 starblue 的答案中添加一些内容——太长了，无法放入评论中。）
我发现当我可以看到 double 的完整二进制值时，更容易看到发生了什么——我希望你也能看到。我将 starblue 的示例放入 C 程序中并将输出转换为二进制（使用我在http://www.exploringbinary.com/converting-floating-point-numbers-to-binary-strings-in-c/上的转换程序）。这是输出，加上计算的结果：
x1 = 101.1100011000011010010000011001001011111001110000111 x2 = 101.11000110000110100100000110010010111110011100001111 c = 0.00101000101100101000111010100110011111010101111100001 r = 100100.01010001101110101101000101101100011111011010101 x1 = 10.111110011111001001111001011010001100001011001111011 x2 = 10.1111100111110010011110010110100011000010110011111 c = 0.0010011101001010001101101010001000011100110010101011 r = 10011.011001001001100010101001001110011100011111011111 x1 = 111.1111110010000001000100001110001111001101010100101 x2 = 111.11111100100000010001000011100011110011010101001011 c = 0.11010100111111110111100101001001011010110100010111011 r = 1001.100110010100010010100101110100000100000110000011 x1 = 0.0001110110111110011011000001011101101100111011010101 x2 = 0.00011101101111100110110000010111011011001110110101010001 c = 0.0001011111010000011100111111110000001001001011101001 r = 1.00111111101111011111001110101010100101010101010101 x1 = 0.1111110010001001000111110100110100001000001101111111 x2 = 0.11111100100010010001111101001101000010000011011111111 c = 0.01100111110011101011111010110111000101001011000000111 r = 10.011011101100011101000000101000110110101011010011111 x1 = 11.1111010101010010010000111001010000100001011000111 x2 = 11.111101010101001001000011100101000010000101100011101 c = 0.110000010101100101010010001010110001110001011111111 r = 101.00111101101010110100110000011111101001010010101111 x1 = 1.1010011101101111101110100000000000011110110111110111 x2 = 1.1010011101101111101110100000000000011110110111111 c = 0.00100101000111101100100101111110100101011010111111001 r = 1011.011010000011101100001011000110000010011111110001 x1 = 101.10111010001001010111100100111110000111100001000011 x2 = 101.101110100010010101111001001111100001111000010001 c = 0.101011101011001100001000111011000001111010111011011 r = 1000.0110010001110100001001010000000101111000011111011 x1 = 11.101101100010000001100111100010010100011000001001111 x2 = 11.10110110001000000110011110001001010001100000101 c = 0.0011011010000011101011110000001111000110010101111111 r = 10001.01101101110011010100011111101110101011001010001 x1 = 101.10101010000101110011111111101001111011111010101111 x2 = 101.1010101000010111001111111110100111101111101011 c = 0.1011000110001100100111111010011000000010100011 r = 1000.0010101011010001010101111000111101110100001000001
（顺便说一句，表达式的“* const”部分是不必要的：仅除以 const 表明 X1 / const == X2 / const。）
当您将双精度值与真实的任意精度值进行比较时，您可以真正看到发生了什么。以第一个例子为例：
x1/c = x2/c (双) = 100100.01010001101110101101000101101100011111011010101 x1/c (真) = 100100.01010001101110101101000101101100011111011010100 1011... x2/c (真) = 100100.01010001101110101101000101101100011111011010101 0111...
我在有效位 53 和 54 之间放置了一个空格，其中舍入以双精度位进行。x1/c 向上舍入，x2/c 向下舍入（截断），成为相同的值。
于 2010-02-12T02:41:52.257 回答