Assigning multiple variables in Haskell

Question

Just getting started in Haskell, and I'm trying to figure out the best way to assign multiple variables based on a single condition. So far I've just been packing and unpacking Tuples. Is there a better/more idiomatic way?

(var1, var2, var3) = 
    if foo > 0
        then ("Foo", "Bar", 3)
        else ("Bar", "Baz", 1)

Also curious about the cost of packing and unpacking the Tuples. If I'm reading this correctly, it seems like this gets optimized away in functions, but not sure if that's the case with an assignment.

You could use lists as well, just so you know. You could even pattern match it in greater chunks using (a:b:c:rest) = [1,2,3,4,5,6] where a = 1, b = 2, c = 3, and rest = [4,5,6]. Furthermore, this is only necessary in let and where statements, not in base expression level. — AJF
– AJF, Commented Jan 25, 2015 at 17:45
Since we do not have mutable variables in Haskell, we call these "bindings" or "definitions" rather than "assignments". Your code looks fine, and I wouldn't worry about the cost unless profiling points at it. Also, if you worry about this tuple, you don't want to know how "Bar" is actually represented at runtime... ;-) — chi
– chi, Commented Jan 25, 2015 at 17:46

dfeuer · Accepted Answer · 2015-01-25 18:17:39Z

6

Yes, that's perfectly fine. If you compile with optimizations enabled, the tuples will indeed by "unboxed", so they incur no extra cost. The code will be transformed to something vaguely like this:

(# var1, var2, var3 #) = 
    case foo > 0 of
        False -> (# "Bar", "Baz", 1 #)
        True -> (# "Foo", "Bar", 3 #)

An unboxed 3-tuple is actually just three values—it doesn't have any sort of extra structure. As a result, it can't be stored in a data structure, but that's okay. Of course, if foo is known at compile time, the case will be optimized away too, and you'll just get three definitions.

answered Jan 25, 2015 at 18:17

dfeuer

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