我正在使用Crypto ++的加密应用程序 作为此应用程序的一个不起眼的部分，我需要确定特定数值范围内可存在的唯一浮点值的最大数量。
显然，0和1之间存在无限数字 - 但并非所有数字都可以用唯一的浮点值表示。
我有一个最小浮点值和一个最大浮点值 我需要确定此范围内可能的浮点值的数量。
这很棘手，因为浮点值间隔更远，距离0越远。
例如，0和1之间的可能浮点值的数量与100,000和100,001之间的浮点值的数量非常不同
就我的目的而言，我希望计数也包括最小值和最大值
但是，生成独占计数的算法同样有用，因为我可以根据需要添加1或2。
其他问题：
如果0在范围内怎么办？
例如，如果最小值为-2.0，最大值为正2.0，我不想计算0两次（一次用于0，再次用于-0） 。
另外，如果最小值或最大值是+/-无穷大会出现什么问题？
（如果最小值或最大值是NaN，我可能会抛出异常。）
uint32_t RangeValueCount ( float fMin , float fMax )
{
if ( fMin > fMax )
swap ( fMin , fMax ) ; // Ensure fMin <= fMax
// Calculate the number of possible floating-point values between fMin and fMax.
return ( *reinterpret_cast < uint32_t* > ( &fMax ) -
*reinterpret_cast < uint32_t* > ( &fMin ) ) + 1 ;
// This algorithm is obviously unsafe, assumes IEEE 754
// How should I account for -0 or infinity?
}
如果可以解决这个问题，我认为解决方案同样适用于double值（可能是long double值，但由于80位整数值等，这可能会稍微复杂一点。）
答案 0 :(得分：5)
这是处理所有有限数字的代码。它期望IEEE 754算术。我用更简单，更清晰的代码替换了我以前的版本。这不是两次距离计算的实现，而是有两种将浮点数转换为其编码的实现（一种是通过复制位，一种是通过数学方式对其进行操作）。之后，距离计算相当简单（必须调整负值，然后距离只是减法）。
#include <ctgmath>
#include <cstdint>
#include <cstdlib>
#include <iostream>
#include <limits>
typedef double Float; // The floating-point type to use.
typedef std::uint64_t UInt; // Unsigned integer of same size as Float.
/* Define a value with only the high bit of a UInt set. This is also the
encoding of floating-point -0.
*/
static constexpr UInt HighBit
= std::numeric_limits<UInt>::max() ^ std::numeric_limits<UInt>::max() >> 1;
// Return the encoding of a floating-point number by copying its bits.
static UInt EncodingBits(Float x)
{
UInt result;
std::memcpy(&result, &x, sizeof result);
return result;
}
// Return the encoding of a floating-point number by using math.
static UInt EncodingMath(Float x)
{
static constexpr int SignificandBits = std::numeric_limits<Float>::digits;
static constexpr int MinimumExponent = std::numeric_limits<Float>::min_exponent;
// Encode the high bit.
UInt result = std::signbit(x) ? HighBit : 0;
// If the value is zero, the remaining bits are zero, so we are done.
if (x == 0) return result;
/* The C library provides a little-known routine to split a floating-point
number into a significand and an exponent. Note that this produces a
normalized significand, not the actual significand encoding. Notably,
it brings significands of subnormals up to at least 1/2. We will
adjust for that below. Also, this routine normalizes to [1/2, 1),
whereas IEEE 754 is usually expressed with [1, 2), but that does not
bother us.
*/
int xe;
Float xf = std::frexp(fabs(x), &xe);
// Test whether the number is subnormal.
if (xe < MinimumExponent)
{
/* For a subnormal value, the exponent encoding is zero, so we only
have to insert the significand bits. This scales the significand
so that its low bit is scaled to the 1 position and then inserts it
into the encoding.
*/
result |= (UInt) std::ldexp(xf, xe - MinimumExponent + SignificandBits);
}
else
{
/* For a normal value, the significand is encoded without its leading
bit. So we subtract .5 to remove that bit and then scale the
significand so its low bit is scaled to the 1 position.
*/
result |= (UInt) std::ldexp(xf - .5, SignificandBits);
/* The exponent is encoded with a bias of (in C++'s terminology)
MinimumExponent - 1. So we subtract that to get the exponent
encoding and then shift it to the position of the exponent field.
Then we insert it into the encoding.
*/
result |= ((UInt) xe - MinimumExponent + 1) << (SignificandBits-1);
}
return result;
}
/* Return the encoding of a floating-point number. For illustration, we
get the encoding with two different methods and compare the results.
*/
static UInt Encoding(Float x)
{
UInt xb = EncodingBits(x);
UInt xm = EncodingMath(x);
if (xb != xm)
{
std::cerr << "Internal error encoding" << x << ".\n";
std::cerr << "\tEncodingBits says " << xb << ".\n";
std::cerr << "\tEncodingMath says " << xm << ".\n";
std::exit(EXIT_FAILURE);
}
return xb;
}
/* Return the distance from a to b as the number of values representable in
Float from one to the other. b must be greater than or equal to a. 0 is
counted only once.
*/
static UInt Distance(Float a, Float b)
{
UInt ae = Encoding(a);
UInt be = Encoding(b);
/* For represented values from +0 to infinity, the IEEE 754 binary
floating-points are in ascending order and are consecutive. So we can
simply subtract two encodings to get the number of representable values
between them (including one endpoint but not the other).
Unfortunately, the negative numbers are not adjacent and run the other
direction. To deal with this, if the number is negative, we transform
its encoding by subtracting from the encoding of -0. This gives us a
consecutive sequence of encodings from the greatest magnitude finite
negative number to the greatest finite number, in ascending order
except for wrapping at the maximum UInt value.
Note that this also maps the encoding of -0 to 0 (the encoding of +0),
so the two zeroes become one point, so they are counted only once.
*/
if (HighBit & ae) ae = HighBit - ae;
if (HighBit & be) be = HighBit - be;
// Return the distance between the two transformed encodings.
return be - ae;
}
static void Try(Float a, Float b)
{
std::cout << "[" << a << ", " << b << "] contains "
<< Distance(a,b) + 1 << " representable values.\n";
}
int main(void)
{
if (sizeof(Float) != sizeof(UInt))
{
std::cerr << "Error, UInt must be an unsigned integer the same size as Float.\n";
std::exit(EXIT_FAILURE);
}
/* Prepare some test values: smallest positive (subnormal) value, largest
subnormal value, smallest normal value.
*/
Float S1 = std::numeric_limits<Float>::denorm_min();
Float N1 = std::numeric_limits<Float>::min();
Float S2 = N1 - S1;
// Test 0 <= a <= b.
Try( 0, 0);
Try( 0, S1);
Try( 0, S2);
Try( 0, N1);
Try( 0, 1./3);
Try(S1, S1);
Try(S1, S2);
Try(S1, N1);
Try(S1, 1./3);
Try(S2, S2);
Try(S2, N1);
Try(S2, 1./3);
Try(N1, N1);
Try(N1, 1./3);
// Test a <= b <= 0.
Try(-0., -0.);
Try(-S1, -0.);
Try(-S2, -0.);
Try(-N1, -0.);
Try(-1./3, -0.);
Try(-S1, -S1);
Try(-S2, -S1);
Try(-N1, -S1);
Try(-1./3, -S1);
Try(-S2, -S2);
Try(-N1, -S2);
Try(-1./3, -S2);
Try(-N1, -N1);
Try(-1./3, -N1);
// Test a <= 0 <= b.
Try(-0., +0.);
Try(-0., S1);
Try(-0., S2);
Try(-0., N1);
Try(-0., 1./3);
Try(-S1, +0.);
Try(-S1, S1);
Try(-S1, S2);
Try(-S1, N1);
Try(-S1, 1./3);
Try(-S2, +0.);
Try(-S2, S1);
Try(-S2, S2);
Try(-S2, N1);
Try(-S2, 1./3);
Try(-N1, +0.);
Try(-N1, S1);
Try(-N1, S2);
Try(-N1, N1);
Try(-1./3, 1./3);
Try(-1./3, +0.);
Try(-1./3, S1);
Try(-1./3, S2);
Try(-1./3, N1);
Try(-1./3, 1./3);
return 0;
}
答案 1 :(得分：0)
很棘手，可能你可以尝试在循环中使用std::nexttoward(from_starting, to_end);并计算直到它结束。 Haven没有尝试过这个问题，需要很长时间才能完成。如果这样做，请确保检查错误标志，请参阅：http://en.cppreference.com/w/cpp/numeric/math/nextafter