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！
我为C ++ 03开始了一个新的问题：
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?
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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