How to assure the return StringList will be ordered : Scala - scala

Question

I am using Scala 2.11.8
I am trying to read queries from my Property File. Each Query Set has multiple parts (explained below)
And i have certain sequence in which these queries must execute.
Code:
import com.typesafe.config.ConfigFactory
object ReadProperty {
def main(args : Array[String]): Unit = {
val queryRead = ConfigFactory.load("testqueries.properties").getConfig("select").getStringList("caseInc").toArray()
val localRead = ConfigFactory.load("testqueries.properties").getConfig("select").getStringList("caseLocal").toArray.toSet
queryRead.foreach(println)
localRead.foreach(println)
}
}
PropertyFile Content :
select.caseInc.2 = Select emp_salary, emp_dept_id from employees
select.caseLocal.1 = select one
select.caseLocal.3 = select three
select.caseRemote.2 = Select e1.emp_name, d1.dept_name, e1.salary from emp_1 e1 join dept_1 d1 on(e1.emp_dept_id = d1.dept_id)
select.caseRemote.1 = Select * from departments
select.caseInc.1 = Select emp_id, emp_name from employees
select.caseLocal.2 = select two
select.caseLocal.4 = select four
Output:
Select emp_id, emp_name from employees
Select emp_salary, emp_dept_id from employees
select one
select two
select three
select four
As we can see in output, The result is Sorted . In the property if you see i have tried numbering the queries in the sequence it should run.(passing the caseInc, caseLocal as arguments).
With getStringList() i am always getting the Sorted List on the basis of the sequence number i am providing.
Even when i tried using toArray() & toArray().toSet i am getting sorted output.
So far its Good
But how to be sure that it will always return in Sorted Order which i have provided in the property file. I am confused because somehow i am not able to find the API which says that the returned List will be Sorted.

Answer 1

I think you can rely on this fact. Looking into the code of DefaultTransformer you can see following piece of logic:
} else if (requested == ConfigValueType.LIST && value.valueType() == ConfigValueType.OBJECT) {
// attempt to convert an array-like (numeric indices) object to a
// list. This would be used with .properties syntax for example:
// -Dfoo.0=bar -Dfoo.1=baz
// To ensure we still throw type errors for objects treated
// as lists in most cases, we'll refuse to convert if the object
// does not contain any numeric keys. This means we don't allow
// empty objects here though :-/
AbstractConfigObject o = (AbstractConfigObject) value;
Map<Integer, AbstractConfigValue> values = new HashMap<Integer, AbstractConfigValue>();
for (String key : o.keySet()) {
int i;
try {
i = Integer.parseInt(key, 10);
if (i < 0)
continue;
values.put(i, o.get(key));
} catch (NumberFormatException e) {
continue;
}
}
if (!values.isEmpty()) {
ArrayList<Map.Entry<Integer, AbstractConfigValue>> entryList = new ArrayList<Map.Entry<Integer, AbstractConfigValue>>(
values.entrySet());
// sort by numeric index
Collections.sort(entryList,
new Comparator<Map.Entry<Integer, AbstractConfigValue>>() {
#Override
public int compare(Map.Entry<Integer, AbstractConfigValue> a,
Map.Entry<Integer, AbstractConfigValue> b) {
return Integer.compare(a.getKey(), b.getKey());
}
});
// drop the indices (we allow gaps in the indices, for better or
// worse)
ArrayList<AbstractConfigValue> list = new ArrayList<AbstractConfigValue>();
for (Map.Entry<Integer, AbstractConfigValue> entry : entryList) {
list.add(entry.getValue());
}
return new SimpleConfigList(value.origin(), list);
}
}
Note how keys are parsed as integer values and then sorted using Integer.compare