简单递归帮助
我有一个简单的递归函数，使用 height1:height2 的比率 0.98 来计算简单的摆锤摆动衰减。
该函数的基本情况为 0.0，但由于某种原因它变成了无限的自调用！
谁能发现我缺少什么吗？
代码：
float swingDecay (float value) {
if ( value == 0.00 ) {
return value;
}
else {
return swingDecay (value * 0.98); }
}
mIL3S www.milkdrinkingcow.com
I have a simple recursive function that calculates a simple pendulum swing decay, using the ratio of height1:height2 of 0.98.
The function has a base case of 0.0, but for some reason it turns into infinite self-calls!
Can anyone spot what I'm missing?
Code:
float swingDecay (float value) {
if ( value == 0.00 ) {
return value;
}
else {
return swingDecay (value * 0.98); }
}
mIL3S
www.milkdrinkingcow.com
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您应该始终在浮点计算中使用“近似”比较。例如，
if (abs(value) < EPS)而不是if ( value == 0.00 )。EPS这里是一个小常量（取决于您的要求和数据类型）。我怀疑这就是实际发生的事情。您得到数据类型中可能的最小正值，例如
1 * 2^(-10000)（10000 来自我的脑海），现在value * 0.98 = value代码>.例如，它必须四舍五入到0或total并且0.98*total显然更接近total.但这只是猜测。对于浮点计算，你永远无法确定:)
You should always use 'approximate' comparisons in floating point calculations. E.g.,
if (abs(value) < EPS)instead ofif ( value == 0.00 ).EPShere is a small constant (depends on your requirements and datatype).I suspect this is what's actually happening. You get to the smallest possible positive value in your datatype, like
1 * 2^(-10000)(10000 comes from the top of my head) and nowvalue * 0.98 = value. E.g., it has to be rounded either to0or tototaland0.98*totalis obviously closer tototal.But that's only speculations, though. With floating point computations, you can never be sure :)
由于浮点计算在浮点数学中永远不会精确，因此您永远不会得到 value == 0.00。您可能想尝试类似 value < 0.0000001 或类似的东西并在它起作用的地方进行调整。
Due to floating point calculations never being exact in floating point math you never get to value == 0.00. You might want to try something like value < 0.0000001 or something like that and tweak it where it works.
不要直接比较浮点数；你的“价值”可能永远不会真正是 0.0（零）。
做类似的事情：
Do not directly compare floating point numbers; your "value" will probably never really be 0.0 (zero).
do something like :
( value == 0.00 )
永远不会成立。
或者，该函数运行了太多次，以至于它遇到了堆栈溢出：P
你应该再看看你是如何创建你的函数的。现在它甚至没有用，它只能返回 0。
( value == 0.00 )
Never gets true.
Or, it takes so many runs of the function that the it runs into, well, stack overflow :P
You should take another look at how you made your function. Right now it is not even useful, it can only ever return 0.
您可以检查
if (value * 0.98 == value)而不是if (value == 0)。当value变得如此小（次正规）以至于它的精度位数太少而无法乘以0.98来产生不同的结果时，就会恰好满足此条件。You could check
if (value * 0.98 == value)instead ofif (value == 0). This condition will be met exactly whenvaluebecomes so small (subnormal) that it has too few bits of precision for multiplication by0.98to yield a different result.使用这个（看起来你想要 2 位精度。
你不应该对浮点值使用相等。
use this (as seems you are going for 2 digit accuracy.
You should not use equality for floating point values.
不要将浮点值与常量进行比较，始终检查它们是否处于下限范围内。
例如，将值 == 0.00 更改为值 <= 0.0001
don't compare float values with constants, always check if they fall under a lower-bound.
change your value == 0.00 to value <= 0.0001 for example
哇，谢谢大家的快速回答！
显然，我的课上跳过了那个小浮点细节......
因此，既然每个人都在说同样的事情（不要将浮点数与相等进行比较，因为它们从来都不是真正精确的），那么如果我使用整数或双精度数，同样的情况也成立吗？
最初我的测试就像（ value <= 0.0 ）一样，但这给了我同样的结果。
只是用 <= 0.005 的测试来运行它，这似乎很好！
谢谢大家！
mIL3S
www.milkdrinkingcow.com
Wow, thanks for the quick answer everyone!
Apparently that little floating point detail was skipped in my class...
So since everyone is pretty much saying the same thing (don't compare floating points with equality since they're never really exact), does the same hold true if I used ints or doubles?
Originally I had the test as if ( value <= 0.0 ), but that gave me the same thing.
Just ran it with the test as <= 0.005 and that seemed to be just fine!
Thanks all!
mIL3S
www.milkdrinkingcow.com