问题
I have been reading Real World Haskell, and I am nearing the end, but a matter of style has been niggling at me to do with the (.) and ($) operators.
When you write a function that is a composition of other functions you write it like:
f = g . h
But when you apply something to the end of those functions I write it like this:
k = a $ b $ c $ value
But the book would write it like this:
k = a . b . c $ value
Now, to me they look functionally equivalent, they do the exact same thing in my eyes. However, the more I look, the more I see people writing their functions in the manner that the book does: compose with (.) first and then only at the end use ($) to append a value to evaluate the lot (nobody does it with many dollar compositions).
Is there a reason for using the books way that is much better than using all ($) symbols? Or is there some best practice here that I am not getting? Or is it superfluous and I shouldn't be worrying about it at all?
回答1:
I guess I can answer this from authority.
Is there a reason for using the books way that is much better than using all ($) symbols?
There's no special reason. Bryan and I both prefer to reduce line noise. . is quieter than $. As a result, the book uses the f . g . h $ x syntax.
回答2:
They are indeed equivalent: Keep in mind that the $ operator does, essentially, nothing. f $ x evaluates to f x. The purpose of $ is its fixity behavior: right-associative and minimal precedence. Removing $ and using parentheses for grouping instead of infix precedence, the code snippets look like this:
k = a (b (c (value)))
and
k = (a . b . c) value
The reason for preferring the . version over the $ version is the same reason for preferring both over the very parenthesized version above: aesthetic appeal.
Although, some might wonder if using infix operators instead of parentheses is based on some subconscious urge to avoid any possible resemblance to Lisp (just kidding... I think?).
回答3:
I'd add that in f . g $ x, f . g is a meaningful syntactic unit.
Meanwhile, in f $ g $ x, f $ g is not a meaningful unit. A chain of $ is arguably more imperative -- first get the result of g of x, then do f to it, then do foo to it, then etc.
Meanwhile a chain of . is arguably more declarative, and in some sense closer to a dataflow centric view -- compose a series of functions, and ultimately apply them to something.
回答4:
For me, I think the answer is (a) the neatness, as Don said; and (b) I find that when I'm editing code, my function may end up in point-free style, and then all I have to do is delete the last $ instead of going back and changing everything. A minor point, certainly, but a nicety.
回答5:
There's an interesting discussion of this question on this haskell-cafe thread. Apparently there's a minority viewpoint that holds that the right associativity of $ is "just plain wrong", and choosing f . g . h $ x over f $ g $ h $ x is one way of side-stepping the issue.
回答6:
It's just a matter of style. However, the way the book does it makes more sense to me. It composes all the functions, and then applies it to the value.
Your method just looks strange, and the last $ is unnecessary.
However, it really doesn't matter. In Haskell, there are usually many, many, correct ways to do the same thing.
回答7:
I realize this is a very old question, but I think there is another reason for this that hasn't been mentioned.
If you are declaring a new point-free function f . g . h, the value you pass in will be automatically be applied. However, if you write f $ g $ h, it will not work.
I think the reason the author prefers the composition method is because it leads to a good practice of building up functions.
来源：https://stackoverflow.com/questions/3030675/haskell-function-composition-and-function-application-idioms-correct-us