Fix tuple concatenation containing lists, tuples, refs

Bump to 0.5.3

records.nimble

@@ -1,7 +1,7 @@
 import os
 
 # Package
-version = "0.5.2"
+version = "0.5.3"
 author = "Dan Rose"
 description = "Operations on tuples as heterogeneous record types a la Relational Algebra"
 license = "MIT"

src/records/tupleops.nim

@@ -13,17 +13,17 @@ proc concat*(t1: tuple, t2: tuple): tuple =
       expectKind(arg.getTypeImpl(), {nnkTupleConstr, nnkTupleTy})
       for i, d in pairs(arg.getTypeImpl):
         case kind(d):
-          of nnkSym:
-            # un-named tuple field
-            result.add newTree(nnkBracketExpr, arg, newLit(i))
           of nnkIdentDefs:
             # named tuple field
             let prop = d[0]
             result.add newColonExpr(prop,
               newTree(nnkDotExpr, arg, prop)
             )
           else:
-            error("Unexpected field kind: `" & $kind(d) & "`")
+            # un-named tuple field
+            # usually d is of kind nnkSym
+            # but can be nnkBracketExpr, nnkTupleConstr, nnkRefTy
+            result.add newTree(nnkBracketExpr, arg, newLit(i))
   concatImpl()
 
 proc `&` *(t1: tuple, t2: tuple): auto = concat(t1, t2)

tests/testTuple.nim

@@ -45,6 +45,36 @@ test "concatenation":
   let c = concat( (1, 2), (3, "a"))
   check c == (1, 2, 3, "a")
 
+  check concat((),()) == ()
+  check concat((),(1,)) == (1,)
+  check concat((1,),()) == (1,)
+
+test "concatenateWithLists":
+    let a = ([1],[2,3])
+    let b = ([4],)
+    check a & b == ([1],[2,3],[4])
+
+    let d = (x:[1],y:[2,3])
+    let e = (z:[4],)
+    check d & e == (x:[1],y:[2,3],z:[4])
+
+test "concatenateWithTuples":
+  let a = ((),)
+  let b =  ((1,2,3),(4))
+  check a & b == ((),(1,2,3),(4))
+
+test "concatenateWithObjects":
+  type Foo = object
+    x:int
+  let a = Foo(x:1)
+  let b = Foo(x:2)
+  check concat((a,),(b,)) == (a,b)
+  check concat((a,),(b,)) != (b,a)
+
+  let ra = (ref Foo)(x:1)
+
+  check concat((ra,),(a,)) == (ra,a)
+
 test "join":
   let x = (a: 1, b: 2)
   check join(x, x) == some(x)