奇怪的for循环问题 [英] Strange for loop problem

问题描述

我不确定这是否是一个错误，所以我想也许你们可能想看看.

问题出在这段代码上:

for i=0,1,.05 做打印(一)结尾

输出应该是:

<预><代码>0.05.1--剪断--.951

相反，输出是:

<预><代码>0.05.1--剪断--.95

while 循环也发生了同样的问题:

w = 0而 w <= 1 做打印(宽)w = w + .05结尾- 输出:0.05.1--剪断--.95

w 的值为 1，可以通过循环后的打印语句进行验证.

我已尽可能验证任何小于或等于 0.05 的步长都会产生此错误.任何高于 0.05 的步骤都应该没问题.我验证了 1/19 (0.052631579) 确实打印了 1.(显然，像 19.9 或 10.5 这样的十进制分母不会产生 [0,1] 包含的输出.)这是否有可能不是语言错误?解释器和常规 Lua 文件都会产生此错误.

解决方案

这是一个舍入问题.问题是 0.05 表示为浮点二进制数，并且它没有二进制的精确表示.在基数 2(二进制)中，它是一个重复的小数，类似于像基数 10 中的 1/3 这样的数字.重复相加时，四舍五入的结果是略大于 1 的数字.它只是非常非常略大于 1，所以如果你把它打印出来，它会显示 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+0.05+0.05+0.05+>打印(x)1>打印(1==x)错误的>打印(x-1)2.2204460492503e-16

所以，如您所见，虽然非常接近 1，但实际上略多一些.

当我们有重复分数时，类似的情况会出现在十进制中.如果我们将 1/3 + 1/3 + 1/3 加在一起，但我们必须四舍五入到六位数才能使用，我们将添加 0.333333 + 0.333333 + 0.333333 并得到 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

>打印(0.05*20-1)0>打印(0.05*20==1)真的

因此，您可以重写代码来说明

for i=0,20,1 做打印(i*0.05)结尾

它会正常工作.一般来说，在可以避免的情况下，最好不要使用浮点数(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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