奇怪的循环问题 [英] Strange for loop problem

问题描述

我不确定这是不是一个bug，所以我想也许你们可能想看看。

问题出在这个代码：

  for i = 0,1，.05 do 
 print（i）
 end 
  

输出应该是：

  0 
 .05 
 .1 
 --snip-- 
 .95 
 1 
  
  
 
 
相反，输出是：
 
 $ p $  0 
 .05 
 .1 
 --snip-- 
 .95 
  

while循环发生同样的问题：

  w = 0 
 while w < = 1 do 
 print（w）
w = w + .05 
 end 
  - 输出：
 0 
 .05 
 .1 
 --snip-- 
 .95 
  

w的值是1 ，这可以通过循环后的print语句来验证。

我已经尽可能地验证了任何小于或等于.05的步骤都会产生这个错误。上面的任何步骤.05应该没问题。我证实了1/19（0.052631579）确实打印出了1（显然，像19.9或10.5这样的小数分母不会产生[0,1]的输出）。是否有可能这不是语言的错误？解释器和一个常规的L​​ua文件都会产生这个错误。 解决方案

这是一个舍入问题。问题是，0.05被表示为一个浮点二进制数，并没有在二进制中的确切表示。在基数2（二进制）中，它是一个重复的十进制数，类似于基数为10的数字。当反复添加时，舍入结果的数字略大于1.它只是非常略大于1 ，所以如果你打印出来，它显示1作为输出，但它不完全是1。 x = 0.05 + 0.05 + 0.05 + 0.05 + 0.05 + 0.05 + 0.05 + 0.05 + 0.05 + 0.05 + 0.05 + 0.05 + 0.05 + 0.05 + 0.05 + 0.05 + 0.05 + 0.05 + 0.05 + 0.05
> print（x）
1
> print（1 == x）
false
> print（x-1）
2.2204460492503e-16

如你所见，虽然真的接近1，但它实际上稍微多一些。

当我们有重复分数时，类似的情况会出现在小数点。如果我们将1/3 + 1/3 + 1/3加起来，但是我们不得不四舍五入来处理，我们会加上0.333333 + 0.333333 + 0.333333，得到0.999999这实际上不是1.这是一个二进制数学的类似情况。 1/20不能用二进制精确表示。

请注意，乘法的舍入略有不同，因此

 >打印（0.05 * 20-1）
 0 
> print（0.05 * 20 == 1）
 true 
  

因此，您可以重写你的代码要说：

$ p $ for i = 0,20,1 do
print（i * 0.05）
结束


它会正常工作。一般来说，建议不要使用浮点数（即带小数点的数字）来控制循环。

I'm not sure if this is a bug or not, so I thought that maybe you folks might want to take a look.

The problem lies with this code:

for i=0,1,.05 do
    print(i)
end

The output should be:

0
.05
.1
--snip--
.95
1

Instead, the output is:

0
.05
.1
--snip--
.95

This same problem happened with a while loop:

w = 0
while w <= 1 do
    print(w)
    w = w + .05
end
--output:
0
.05
.1
--snip--
.95

The value of w is 1, which can be verified by a print statement after the loop.

I have verified as much as possible that any step that is less than or equal .05 will produce this error. Any step above .05 should be fine. I verified that 1/19 (0.052631579) does print a 1. (Obviously, a decimal denominator like 19.9 or 10.5 will not produce output from [0,1] inclusive.) Is there a possibility that this is not an error of the language? Both the interpreter and a regular Lua file produce this error.

解决方案

This is a rounding problem. The issue is that 0.05 is represented as a floating point binary number, and it does not have an exact representation in binary. In base 2 (binary), it is a repeating decimal similar to numbers like 1/3 in base 10. When added repeatedly, the rounding results in a number which is slightly more than 1. It is only very, very slightly more than 1, so if you print it out, it shows 1 as the output, but it is not exactly 1.

> x=0.05+0.05+0.05+0.05+0.05+0.05+0.05+0.05+0.05+0.05+0.05+0.05+0.05+0.05+0.05+0.05+0.05+0.05+0.05+0.05
> print(x)
1
> print(1==x)
false
> print(x-1)
2.2204460492503e-16

So, as you can see, although really close to 1, it is actually slightly more.

A similar situation can come up in decimal when we have repeating fractions. If we were to add together 1/3 + 1/3 + 1/3, but we had to round to six digits to work with, we would add 0.333333 + 0.333333 + 0.333333 and get 0.999999 which is not actually 1. This is an analogous case for binary math. 1/20 cannot be precisely represented in binary.

Note that the rounding is slightly different for multiplication so

> print(0.05*20-1)
0
> print(0.05*20==1)
true

As a result, you could rewrite your code to say

for i=0,20,1 do
    print(i*0.05)
end

And it would work correctly. In general, it's advisable not to use floating point numbers (that is, numbers with decimal points) for controlling loops when it can be avoided.

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