在C ++中找到最接近的小于特定整数值的浮点值？ [英] Finding the closest floating point value less than a specific integer value in C++?

问题描述

我有一个输入浮点值是0.0f< =值< 1.0f（注意小于1）。

当把这个值乘上一个更大的范围时，浮点精度自然会下降，这意味着这个值最终会超出等价范围。

例如，如果我从一个值开始，例如：

0.99999983534521f

>

然后乘以100，得到：

100.000000000000f

这是好的，但是我怎样才能减少浮点表示成为最接近的浮点值仍然小于100？

我发现这个小手工诀窍：

$ $ $


$ b $
浮动浮动;
};

测试值;

value.floating = 1.0f;

printf（％x\\\
，value.integer）;

然后我把这个十六进制值减一十六进制数字，然后明确地设置它：

  unsigned int almost_one = 0x3f7fffff; 
 
 float value = 1.0f; 
 
 if（value> = 1.0f）std :: memcpy（& value，&; almost_one，sizeof（float））; 
  

对于这个特定的值，这个效果很好，但是我可以使用更通用的方法吗？

我希望有一个神奇的指令，我不知道，我可以用它来实现这一点！

<编辑：这里的答案很好，std :: nextafter看起来像我之后。不幸的是我还不能使用C ++ 11数学库，所以这对我不起作用。为了节省复杂的事情，我将用C ++ 11标记这个问题，并接受Mike的答案。

我为C ++ 03开始了一个新的问题：替代C ++ 11的std :: nextafter和std :: nexttoward for C ++ 03？

我希望有一个神奇的指令，我不知道，我可以用它来实现这个！

如果您有一个C ++ 11（或C99）标准库，那么 std :: nextafter（value，0.0f）从< cmath> （或 nextafter from < math。 h> ）会给你最大的可表示值，它小于 value 。

在第一个参数之后给出下一个不同的值，在第二个参数的方向上;所以在这里，下一个截然不同的值接近零。

I have an input floating point value that is 0.0f <= value < 1.0f (note less than one).

When multiplying this value up to a larger range, naturally the floating point precision is decreased meaning the value can end up outside of the equivalent range.

For example if I start off with a value such as:

0.99999983534521f

Then multiply it by 100, I get:

100.000000000000f

Which is fine, but how do I then reduce the floating point representation to be the nearest floating point value that is still less than 100?

I found this little manual trick:

union test
{
    int integer;
    float floating;
};

test value;

value.floating = 1.0f;

printf("%x\n", value.integer);

Then I take that hex value and reduce it by one hex digit, then set it explicitly like so:

unsigned int almost_one = 0x3f7fffff;

float value = 1.0f;

if (value >= 1.0f)      std::memcpy(&value, &almost_one, sizeof(float));

That works well for this specific value, but is there a more general approach I can use instead?

I'm hoping there's a magic instruction I'm not aware of that I can use to achieve this!

Edit: Great set of answers here, std::nextafter looks like what I'm after. Unfortunately I can't yet use C++11 math libraries so this won't work for me. To save complicating things, I'll tag this question with C++11 and accept Mike's answer below.

I've started a new question for C++03 : Alternative to C++11's std::nextafter and std::nexttoward for C++03?

解决方案

I'm hoping there's a magic instruction I'm not aware of that I can use to achieve this!

If you've got a C++11 (or C99) standard library, then std::nextafter(value, 0.0f) from <cmath> (or nextafter from <math.h>) will give you the largest representable value smaller than value.

It gives the "next" distinct value after the first argument, in the direction of the second; so here, the next distinct value closer to zero.

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