Return actual type if the case expr returns the same types (#1237)

src/visitor/DeduceTypeVisitor.cpp

@@ -556,6 +556,7 @@ void DeduceTypeVisitor::visit(CaseExpression *expr) {
         if (!ok()) return;
     }
 
+    std::unordered_set<Value::Type> types;
     for (const auto &whenThen : expr->cases()) {
         whenThen.when->accept(this);
         if (!ok()) return;
@@ -567,10 +568,14 @@ void DeduceTypeVisitor::visit(CaseExpression *expr) {
         }
         whenThen.then->accept(this);
         if (!ok()) return;
+        types.emplace(type_);
     }
 
-    // Will not deduce the actual value type returned by case expression.
-    type_ = Value::Type::__EMPTY__;
+    if (types.size() == 1) {
+        type_ = *types.begin();
+    } else {
+        type_ = Value::Type::__EMPTY__;
+    }
 }
 
 void DeduceTypeVisitor::visit(PredicateExpression *expr) {

tests/tck/features/expression/Case.feature

@@ -288,3 +288,13 @@ Feature: Case Expression
       | "Boris Diaw" | 2008       | 2012     | "old" | "Hornets" |
       | "Boris Diaw" | 2012       | 2016     | "bad" | "Spurs"   |
       | "Boris Diaw" | 2016       | 2017     | "old" | "Jazz"    |
+
+  Scenario: Using the return value of case expr as an input
+    When executing query:
+      """
+      RETURN CASE WHEN true THEN "Tim Duncan" ELSE "ABC" END AS a | GO FROM $-.a OVER like;
+      """
+    Then the result should be, in order:
+      | like._dst       |
+      | "Manu Ginobili" |
+      | "Tony Parker"   |