奇怪的算术与swi-prolog [英] strange arithmetic with swi-prolog

问题描述

我发现结果很奇怪。为什么不是0.3？有人可以告诉我为什么这个结果？有没有可能解决这个问题。

 ？ -  X是5.3-5。 
 X = 0.2999999999999998。 
 $ b $  -  
  

我的第二个问题是如何从'hour 'notation '13 .45'---->'15.30'换成小时数？例如，上面计算的15.30-13.45的时间将是1.85。但是我需要在一小时的时间里进行操作，而不是数字的剩余部分。像15 1/2 - 13/4，这样比较好。我尝试

 ？ -  X是（5.3-5）* 100/60。 
 X = 0.4999999999999997。 
 
？ -  X是（5.3-5）* 100 // 60。 
错误：/// 2：输入错误：`integer'expected，found`29.999999999999982'
  

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解决方案

你从SWI得到的答案是正确的。

在许多编程语言，而且几乎所有的Prolog系统，浮点实现都基于二进制（radix = 2）系统。数字 5.3 不能在这个系统中精确表示，所以选择一些近似值。

 ？ -  X是5.3-5-0.3。 
 X = -1.6653345369377348e-16。 
  

只要你留下的数字是2的幂（包括2 ^-1 ，2 ^-2 ，...），只要后面的浮点数的精度允许，你将得到精确的结果：

 ？ -  X是5.000244140625-5-0.000244140625。 
 X = 0.0。 
  

符合Prolog标准的系统需要做的是确保在写入一个带写入选项的浮点数引用（true），浮动数据写入的方式使得它可以被精确地读取。

至于第二个问题：（//）/ 2 仅在整数上定义。如果你想将一个float转换为一个在ISO Prolog中的正常的LIA-1函数：
$ b

floor / 1，truncate / 1 ，一轮/ 1，天花板/ 1 。

 ？ -  X是round（5.3）。 
 X = 5. 
  

但是，我宁愿推荐使用（div）/ 2 来代替（//）/ 2 。 LIA-1：2012（Standard ISO / IEC 10967-1：2012）不再支持（//）/ 2 的含义。有充分的理由。请参阅此和此答案了解详情。

I find the result very strange. Why not 0.3? Can somebody tell me why this result? Is it possible to fix this.

 ?- X is 5.3-5.
    X = 0.2999999999999998.

    ?- 

My second question is how would I transform from 'hour' notation '13.45' ---->'15.30' into numbers of hours ? For example the period above calculated 15.30-13.45 would be 1.85. But I need to operate on parts of the hour and not the remainders of the numbers. Like 15 1/2 - 13 /4, this way is better. I try

?- X is (5.3-5)*100/60.
X = 0.4999999999999997.

?- X is (5.3-5)*100//60.
ERROR: ///2: Type error: `integer' expected, found `29.999999999999982'

Any suggestions?

解决方案

The answer you get from SWI is correct.
In many programming languages, and practically all Prolog systems, floating point implementations are based on a binary (radix = 2) system. The number 5.3 cannot be represented precisely in this system, so some approximation is chosen. Subtracting is particularly well suited to expose such inaccuracies.

?- X is 5.3-5-0.3.
X = -1.6653345369377348e-16.

As long as you are staying with numbers that are the sum of powers of 2 (including 2^-1, 2^-2, ...) you will get precise results as far as the precision of the floats behind permits it:

?- X is 5.000244140625-5-0.000244140625.
X = 0.0.

What ISO conforming Prolog systems have to do is to ensure that when writing a float with write-option quoted(true), the float is written in such a manner that it can be read back precisely.

As to your second question: (//)/2 is defined on integers only. If you want to convert a float to an integer you have in ISO Prolog the usual LIA-1 functions:

floor/1, truncate/1, round/1, ceiling/1.

?- X is round(5.3).
X = 5.

However, I'd rather recommend using (div)/2 in place of (//)/2. The meaning of (//)/2 is no longer supported by LIA-1:2012 (Standard ISO/IEC 10967-1:2012) — for good reason. See this and this answer for details.

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