问题
I am a beginner in Racket and I am trying to updare a hash table using hash-update! where the value is a mutable set. Below are the code lines:
(hash-update! hash key (curryr set-add! new_val) (mutable-set 1))
However I receive an error
expected: set?
given: #<void>
argument position: 1st
other arguments...:
x: 2
where I tried 2 as the new_val
Any suggestions ?
回答1:
This is because the updater is supposed to be a function that takes a value as input and produces a new value output. Since the set is mutable and you're using set-add! to mutate it, the "updater" isn't returning a new value, just mutating the old one and producing void.
There are two ways to fix this:
- Mutable sets as values, mutate them separately, not inside hash-update!.
- Immutable sets as values, use a functional updater inside hash-update!.
Since you specified you want the values as mutable sets, I will show (1).
The most basic thing you can do is hash-ref to get a mutable-set, and then use set-add! on that.
(set-add! (hash-ref hash key) new-val)
However, this doesn't work when there is no mutable-set value for that key yet. It needs to be added to the table when it doesn't exist yet which why is why you have the (mutable-set 1) failure-result argument. The solution to this isn't hash-update!, it's hash-ref!.
(set-add! (hash-ref! hash key (mutable-set 1)) new-val)
Although it would probably be better if you wrapped the failure-result in a thunk
(set-add! (hash-ref! hash key (λ () (mutable-set 1))) new-val)
来源：https://stackoverflow.com/questions/62089061/update-functions-in-hash-table-in-racket