为什么mod在表达式中给出的结果与函数调用中给出的结果不同? [英] Why does mod give a different result in an expression than in a function call?

问题描述

假设要计算该函数:

f (x,y) = ((x `mod` 3)+(y `mod` 3)) `mod` 2

然后，如果有人手动展开f (-1,0)，就会得到:

Then, if one expands f (-1,0) manually, one gets:

((-1 `mod` 3)+(0 `mod` 3)) `mod` 2
1

但是如果使用内联函数，则结果为:

If one however uses an inline function, the result is:

let f (x,y) = ((x `mod` 3)+(y `mod` 3)) `mod` 2 in f (-1,0)
0

存储无法产生预期结果的函数时会发生什么?

What happens when storing the function that yields not the expected result?

我认为这是因为f使用Integral而不是Int吗?

I assume this is because f uses Integral instead of Int?

推荐答案

好像是解析问题. -1 `mod` 3被解析为-(1 `mod` 3)而不是(-1) `mod` 3.

Looks like it's a matter of parsing. -1 `mod` 3 gets parsed as -(1 `mod` 3) and not (-1) `mod` 3.

*Main> -(1 `mod` 3)
-1
*Main> (-1) `mod` 3
2

老实说，一元-在Haskell中的工作方式有点像个hack，我个人觉得很困惑.如果我真的需要一个否定的文字，我通常只需添加额外的括号即可.

Honestly, the way unary - works in Haskell is a bit of a hack that I personally find confusing. If I really need a negative literal, I usually just add the extra parentheses to be sure.

要考虑的另一件事是，Haskell具有两个模函数，分别是mod和rem，它们对负数的处理方式有所不同.有关更多详细信息，请参见这些

Another thing to consider is that Haskell has two modulo functions, mod and rem, that treat negative numbers differently. For more details, look to these two other questions.

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