Is there a difference between floor and truncate in Haskell

问题

Does there exist a difference in functionality between floor and truncate in Haskell?

They seem to perform the same functionality and they have the same type signature:

truncate :: (Integral b, RealFrac a) => a -> b
floor :: (Integral b, RealFrac a) => a -> b

回答1:

Yes, for negative numbers. If we read the documentation, we see:

truncate :: Integral b => a -> b

truncate x returns the integer nearest x between zero and x

and:

floor :: Integral b => a -> b

floor x returns the greatest integer not greater than x

So if we enter a negative number, like -3.5 we obtain:

Prelude> truncate (-3.5)
-3
Prelude> floor (-3.5)
-4

回答2:

It's not haskell specific, but there's a difference between those functions. Floor means the highest integer no higher than a given number. Truncate means to remove at some length, in this case the fractional part. Those have the same effect for zero and positive numbers, but not negative.

Here's a quick comparison in Python:

>>> for i in range(-5,6):
...   j=0.5*i
...   print(j,floor(j),ceil(j),trunc(j),round(j))
...
-2.5 -3 -2 -2 -2
-2.0 -2 -2 -2 -2
-1.5 -2 -1 -1 -2
-1.0 -1 -1 -1 -1
-0.5 -1 0 0 0
0.0 0 0 0 0
0.5 0 1 0 0
1.0 1 1 1 1
1.5 1 2 1 2
2.0 2 2 2 2
2.5 2 3 2 2

Essentially trunc() goes towards zero and floor() towards negative infinity.

回答3:

Looking at the source code the difference emerges quite quickly:

truncate x          =  m  where (m,_) = properFraction x

and

floor x             =  if r < 0 then n - 1 else n
                       where (n,r) = properFraction x

we see the difference will only emerge on negative numbers and so:

Prelude> floor (negate 2.1)
-3
Prelude> truncate (negate 2.1)
-2

回答4:

Truncate:

1    -->  1 
3.1  -->  3
3.9  -->  3
-2.1 --> -2
-2.9 --> -2

Floor:

1    -->  1 
3.1  -->  3
3.9  -->  3
-2.1 --> -2
-2.9 --> -3 (Different!...)

来源：https://stackoverflow.com/questions/45811770/is-there-a-difference-between-floor-and-truncate-in-haskell

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haskell