假设我执行以下操作:
(X / const) * const
使用双精度参数,其定义由IEEE 754-2008
确定,先进行除法再进行乘法。
const
的范围为0 < ABS(const) < 1
。
假设操作成功(没有溢出发生),对于此操作的不同参数X
是否保证返回不同的结果?
换句话说,是否存在X1
、X2
和0 < ABS(const) < 1
,使得X1 <> X2
,但是(X1 / const) * const = (X2 / const) * const
?
假设我执行以下操作:
(X / const) * const
使用双精度参数,其定义由IEEE 754-2008
确定,先进行除法再进行乘法。
const
的范围为0 < ABS(const) < 1
。
假设操作成功(没有溢出发生),对于此操作的不同参数X
是否保证返回不同的结果?
换句话说,是否存在X1
、X2
和0 < ABS(const) < 1
,使得X1 <> X2
,但是(X1 / const) * const = (X2 / const) * const
?
是的。
public class TestDoubleDivision { public static void main(String[] args) { final Random random = new Random(); int i = 0; while (i < 10) { final double c = random.nextDouble(); final double x1 = 10.0 * random.nextDouble(); final double x2 = nextDouble(x1);
if (x1 / c * c == x2 / c * c) { System.out.printf("x1 = %.20f, x2 = %.20f, c = %.20f\n", x1, x2, c); i++; } } }
private static double nextDouble(double d1) { return Double.longBitsToDouble(Double.doubleToLongBits(d1) + 1); } }
输出:
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
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
顺便说一下,“* const”部分是不必要的:仅通过const除法就可以表明X1 / const == X2 / const。
当您将双精度值与真正的任意精度值进行比较时,您可以真正看到发生了什么。例如,以第一个例子为例:
x1/c = x2/c (double) = 100100.01010001101110101101000101101100011111011010101
x1/c (true) = 100100.01010001101110101101000101101100011111011010100 1011...
x2/c (true) = 100100.01010001101110101101000101101100011111011010101 0111...
我在重要位53和54之间放了一个空格,这是在双倍中进行舍入的位置。x1/c四舍五入,而x2/c向下取整,变成相同的值。
const
小于1
,可能会出现商不同但乘积相同的情况。当我提出这个问题时,我并不确定商可能会有所不同(并且对它们可以有所不同感到惊讶)。 - Quassnoi
64位
双精度值,在比较时应该返回false。+0
和-0
不是不同的。在你的情况下,X1
和X2
将按位相同,因此不是不同的。 - Quassnoi