hcl/hclsyntax: Return DynamicVal when splatting a DynamicVal

Since the result type from a splat is derived from the source value type,
we can't predict our result type when the source type isn't known.

Previously we were incorrectly returning cty.Tuple(cty.DynamicPseudoType)
in this case because of the automatic tuple promotion logic, but now we'll
correctly just give up even before we do _that_ when given a DynamicVal.

hcl/hclsyntax/expression.go

@@ -1228,6 +1228,12 @@ func (e *SplatExpr) Value(ctx *hcl.EvalContext) (cty.Value, hcl.Diagnostics) {
 	}
 
 	sourceTy := sourceVal.Type()
+	if sourceTy == cty.DynamicPseudoType {
+		// If we don't even know the _type_ of our source value yet then
+		// we'll need to defer all processing, since we can't decide our
+		// result type either.
+		return cty.DynamicVal, diags
+	}
 
 	// A "special power" of splat expressions is that they can be applied
 	// both to tuples/lists and to other values, and in the latter case

hcl/hclsyntax/expression_test.go

@@ -759,6 +759,16 @@ upper(
 			cty.UnknownVal(cty.Tuple([]cty.Type{cty.DynamicPseudoType})),
 			1, // a string has no attribute "name"
 		},
+		{
+			`dyn.*.name`,
+			&hcl.EvalContext{
+				Variables: map[string]cty.Value{
+					"dyn": cty.DynamicVal,
+				},
+			},
+			cty.DynamicVal,
+			0,
+		},
 		{
 			`unkobj.*.name`,
 			&hcl.EvalContext{