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社区首页 >问答首页 >在值下面查找浮点数

在值下面查找浮点数
EN

Stack Overflow用户
提问于 2013-01-11 12:29:57
回答 2查看 1.8K关注 0票数 10

假设我有一个浮点数。我想要找到小于X的最大数字,并且可以无损地存储在浮点数中。

IEEE标准IIRC说,您可以通过将浮点数转换为int表示,减去一个表示,然后将其转换回浮点数。

(编辑:对于不是NaN或inf的正数,这是正确的。对于负数,您必须添加。请参阅Rawling's answer获取更多信息。)

要在表示之间进行更改,我只知道C# (cast)操作符,它会截断,这不是我想要的。

在C#中有办法做到这一点吗?

EN

回答 2

Stack Overflow用户

回答已采纳

发布于 2013-01-11 12:36:53

下面是如何简单地将float转换为int,更改它,然后将其转换回float

代码语言:javascript
复制
float myFloat = 10.3f;
// Get the bytes making up the float
byte[] bytes = BitConverter.GetBytes(myFloat);
// Make an int out of them
int myInt = BitConverter.ToInt32(bytes, 0);
// Change it
myInt--;
// Get the bytes making up the int
bytes = BitConverter.GetBytes(myInt);
// Make a float out of them
myFloat = BitConverter.ToSingle(bytes, 0);
// gives 10.2999992 or so

BitConverter甚至为64位等价物内置了此功能:

代码语言:javascript
复制
double myDouble = 10.3;
long myLong = BitConverter.DoubleToInt64Bits(myDouble);
myLong--;
myDouble = BitConverter.Int64BitsToDouble(myLong); // gives 10.2999999...

然而,正如Peter Ruderman所指出的,简单地减少底层int并不能可靠地给出最小的float

特别是,对于负数,您需要增加整数,以使浮点数更负。对于float 0,下一个最小的int实际上对应于NaN,所以这里需要一个特例。

这里有几个函数,我把它们拼凑在一起,这些函数一般都能处理这些情况;它看起来也能理智地在大数和正/负无穷大之间传播!我使用了不安全的转换来减少代码长度,但是如果您愿意,可以坚持上面的字节转换:

代码语言:javascript
复制
static unsafe float Increment(float f)
{
    int val = *(int*)&f;
    if (f > 0)
        val++;
    else if (f < 0)
        val--;
    else if (f == 0)
        return float.Epsilon;
    return *(float*)&val;
}
static unsafe float Decrement(float f)
{
    int val = *(int*)&f;
    if (f > 0)
        val--;
    else if (f < 0)
        val++;
    else if (f == 0)
        return -float.Epsilon; // thanks to Sebastian Negraszus
    return *(float*)&val;
}

正如Jeppe所指出的,您可能也想

  • 使用if (float.IsNaN(f)) return f;启动每个函数,这样就不会意外地增加或减少NaN并给出一个数字。
  • 还可以考虑对照float.PositiveInfinity.NegativeInfinity检查输入,因为从数学上讲,在增量/减少的情况下,输入应该保持不变。
票数 10
EN

Stack Overflow用户

发布于 2013-01-11 14:16:17

这里有一个库例程,nexttowardf(x, -INFINITY);

如果你想用你自己的代码来完成它,你可以使用本机(在IEEE 754浮点操作中,而不需要访问浮点编码或组件),如下所示(在C中)。

提供了两个版本,一个在每种情况下都使用一个低于正常值的值(在某些处理器上可能比较慢),另一个版本只在输入较小的情况下才使用低于正常值(但它有分支)。+无穷大不支持作为输入,尽管可以通过简单的测试添加支持。这是为double编写的,但是对float的更改很简单。

如果定义了CompileMain,它还包括一个测试程序。

代码语言:javascript
复制
#include <float.h>
#include <math.h>


/*  Return the next floating-point value before the finite value q.

    This was inspired by Algorithm 3.5 in Siegfried M. Rump, Takeshi Ogita, and
    Shin'ichi Oishi, "Accurate Floating-Point Summation", _Technical Report
    05.12_, Faculty for Information and Communication Sciences, Hamburg
    University of Technology, November 13, 2005.
*/
double NextBefore(double q)
{
    // SmallestPositive is the smallest positive floating-point number.
    static const double SmallestPositive = DBL_EPSILON * DBL_MIN;

    /*  Scale is .625 ULP, so multiplying it by any significand in [1, 2)
        yields something in [.625 ULP, 1.25 ULP].
    */
    static const double Scale = 0.625 * DBL_EPSILON;

#if 0
    /*  This version has a branch but uses subnormal values only if q is so
        small that q * Scale is subnormal.
    */
    double increment = fabs(q)*Scale;

    if (0. == increment)
        return q - SmallestPositive;

    return q - increment;
#else
    /*  This version uses a subnormal, SmallestPositive, in each case.
        This might cause poor performance on some processors.
    */
    return q - fmax(SmallestPositive, fabs(q)*Scale);
#endif
}


#if defined CompileMain


#include <stdio.h>
#include <stdlib.h>


#define NumberOf(a) (sizeof (a) / sizeof *(a))


int main(void)
{
    int status = EXIT_SUCCESS;

    static const struct { double in, out; } cases[] =
    {
        { -INFINITY,                -INFINITY                },
        { -0x1.fffffffffffffp1023,  -INFINITY                },
        { -0x1.ffffffffffffep1023,  -0x1.fffffffffffffp1023  },
        { -0x1.ffffffffffffdp1023,  -0x1.ffffffffffffep1023  },
        { -0x1.ffffffffffffcp1023,  -0x1.ffffffffffffdp1023  },
        { -0x1.0000000000003p1023,  -0x1.0000000000004p1023  },
        { -0x1.0000000000002p1023,  -0x1.0000000000003p1023  },
        { -0x1.0000000000001p1023,  -0x1.0000000000002p1023  },
        { -0x1.0000000000000p1023,  -0x1.0000000000001p1023  },

        { -0x1.fffffffffffffp1022,  -0x1.0000000000000p1023  },

        { -0x1.fffffffffffffp1,     -0x1.0000000000000p2     },
        { -0x1.ffffffffffffep1,     -0x1.fffffffffffffp1     },
        { -0x1.ffffffffffffdp1,     -0x1.ffffffffffffep1     },
        { -0x1.ffffffffffffcp1,     -0x1.ffffffffffffdp1     },
        { -0x1.0000000000003p1,     -0x1.0000000000004p1     },
        { -0x1.0000000000002p1,     -0x1.0000000000003p1     },
        { -0x1.0000000000001p1,     -0x1.0000000000002p1     },
        { -0x1.0000000000000p1,     -0x1.0000000000001p1     },

        { -0x1.fffffffffffffp-1022, -0x1.0000000000000p-1021 },
        { -0x1.ffffffffffffep-1022, -0x1.fffffffffffffp-1022 },
        { -0x1.ffffffffffffdp-1022, -0x1.ffffffffffffep-1022 },
        { -0x1.ffffffffffffcp-1022, -0x1.ffffffffffffdp-1022 },
        { -0x1.0000000000003p-1022, -0x1.0000000000004p-1022 },
        { -0x1.0000000000002p-1022, -0x1.0000000000003p-1022 },
        { -0x1.0000000000001p-1022, -0x1.0000000000002p-1022 },
        { -0x1.0000000000000p-1022, -0x1.0000000000001p-1022 },

        { -0x0.fffffffffffffp-1022, -0x1.0000000000000p-1022 },
        { -0x0.ffffffffffffep-1022, -0x0.fffffffffffffp-1022 },
        { -0x0.ffffffffffffdp-1022, -0x0.ffffffffffffep-1022 },
        { -0x0.ffffffffffffcp-1022, -0x0.ffffffffffffdp-1022 },
        { -0x0.0000000000003p-1022, -0x0.0000000000004p-1022 },
        { -0x0.0000000000002p-1022, -0x0.0000000000003p-1022 },
        { -0x0.0000000000001p-1022, -0x0.0000000000002p-1022 },
        { -0x0.0000000000000p-1022, -0x0.0000000000001p-1022 },

        { +0x1.fffffffffffffp1023,  +0x1.ffffffffffffep1023  },
        { +0x1.ffffffffffffep1023,  +0x1.ffffffffffffdp1023  },
        { +0x1.ffffffffffffdp1023,  +0x1.ffffffffffffcp1023  },
        { +0x1.0000000000004p1023,  +0x1.0000000000003p1023  },
        { +0x1.0000000000003p1023,  +0x1.0000000000002p1023  },
        { +0x1.0000000000002p1023,  +0x1.0000000000001p1023  },
        { +0x1.0000000000001p1023,  +0x1.0000000000000p1023  },

        { +0x1.0000000000000p1023,  +0x1.fffffffffffffp1022  },

        { +0x1.0000000000000p2,     +0x1.fffffffffffffp1     },
        { +0x1.fffffffffffffp1,     +0x1.ffffffffffffep1     },
        { +0x1.ffffffffffffep1,     +0x1.ffffffffffffdp1     },
        { +0x1.ffffffffffffdp1,     +0x1.ffffffffffffcp1     },
        { +0x1.0000000000004p1,     +0x1.0000000000003p1     },
        { +0x1.0000000000003p1,     +0x1.0000000000002p1     },
        { +0x1.0000000000002p1,     +0x1.0000000000001p1     },
        { +0x1.0000000000001p1,     +0x1.0000000000000p1     },

        { +0x1.0000000000000p-1021, +0x1.fffffffffffffp-1022 },
        { +0x1.fffffffffffffp-1022, +0x1.ffffffffffffep-1022 },
        { +0x1.ffffffffffffep-1022, +0x1.ffffffffffffdp-1022 },
        { +0x1.ffffffffffffdp-1022, +0x1.ffffffffffffcp-1022 },
        { +0x1.0000000000004p-1022, +0x1.0000000000003p-1022 },
        { +0x1.0000000000003p-1022, +0x1.0000000000002p-1022 },
        { +0x1.0000000000002p-1022, +0x1.0000000000001p-1022 },
        { +0x1.0000000000001p-1022, +0x1.0000000000000p-1022 },

        { +0x1.0000000000000p-1022, +0x0.fffffffffffffp-1022 },
        { +0x0.fffffffffffffp-1022, +0x0.ffffffffffffep-1022 },
        { +0x0.ffffffffffffep-1022, +0x0.ffffffffffffdp-1022 },
        { +0x0.ffffffffffffdp-1022, +0x0.ffffffffffffcp-1022 },
        { +0x0.0000000000004p-1022, +0x0.0000000000003p-1022 },
        { +0x0.0000000000003p-1022, +0x0.0000000000002p-1022 },
        { +0x0.0000000000002p-1022, +0x0.0000000000001p-1022 },
        { +0x0.0000000000001p-1022, +0x0.0000000000000p-1022 },
    };

    for (int i = 0; i < NumberOf(cases); ++i)
    {
        double in = cases[i].in, expected = cases[i].out;
        double observed = NextBefore(in);
        printf("NextBefore(%a) = %a.\n", in, observed);
        if (! (observed == expected))
        {
            printf("\tError, expected %a.\n", expected);
            status = EXIT_FAILURE;
        }
    }

    return status;
}


#endif  // defined CompileMain
票数 3
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/14278248

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