Question

I know Haskell isn’t OO so it isn’t strictly a ‘member variable’.

data Foo = Foo {
    bar :: Int,
    moo :: Int,
    meh :: Int,
    yup :: Int
}

modifyBar (Foo b m me y) = (Foo b' m me y)
    where b' = 2 

This is how my code looks at the moment. The problem is I am now making data types with 16 or more members. When I need to modify a single member it results in very verbose code. Is there a way around this?

Answer 1

modifyBar foo = foo { bar = 2 }

---

This syntax will copy foo, and then modify the bar field of that copy to 2. This could be naturally extended to more fields, so you don’t need to write that modifyBar function at all.

(See http://book.realworldhaskell.org/read/code-case-study-parsing-a-binary-data-format.html#id625467)