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I don't think it makes sense to characterize Haskell as "big" on this basis, because 1) it is trivial to define an operator in Haskell, so there's bound to be a
by dpratt71 8y ago
I don't think it makes sense to characterize Haskell as "big" on this basis, because 1) it is trivial to define an operator in Haskell, so there's bound to be a lot of them and 2) even the "standard" operators typically have a simple definition (e.g. https://www.stackage.org/haddock/lts-12.9/base-4.11.1.0/src/GHC-Base.html#%24 https://www.stackage.org/haddock/lts-12.9/base-4.11.1.0/src/...).
- wtracy 8y agoEven if the operators themselves don't count as "part of the language", the complex precedence rules around them certainly should.
- dwohnitmok 8y agoIn what way? Those are defined as part of the operator. I usually parenthesize them anyway just like I would for an equation.
- village-idiot 8y agoOn the whole, I consider user-defined infix operators to be a huge mistake. While the few common ones are great, the ability for every single library creator to add their own infix operator turns into a mess in the long run.
- gowld 8y agoThey fine inside a limited domain-specific scope, just don't go importing operators from many libs willy-nilly.
- village-idiot 8y agoIt’s very hard to convince people to keep them in that limited scope.