Passing just a type as a parameter in C#

Question

There are two common approaches. First, you can pass System.Type

object GetColumnValue(string columnName, Type type)
{
    // Here, you can check specific types, as needed:

    if (type == typeof(int)) { // ...

This would be called like: int val = (int)GetColumnValue(columnName, typeof(int));

The other option would be to use generics:

T GetColumnValue<T>(string columnName)
{
    // If you need the type, you can use typeof(T)...

This has the advantage of avoiding the boxing and providing some type safety, and would be called like: int val = GetColumnValue<int>(columnName);

foo.GetColumnValues(dm.mainColumn, typeof(string))

Alternatively, you could use a generic method:

public void GetColumnValues<T>(object mainColumn)
{
    GetColumnValues(mainColumn, typeof(T));
}

and you could then use it like:

foo.GetColumnValues<string>(dm.mainColumn);

You can pass a type as an argument, but to do so you must use typeof:

foo.GetColumnValues(dm.mainColumn, typeof(int))

The method would need to accept a parameter with type Type.

---

where the GetColumns method will call a different method inside depending on the type passed.

If you want this behaviour then you should not pass the type as an argument but instead use a type parameter.

foo.GetColumnValues<int>(dm.mainColumn)

foo.GetColumnValues(dm.mainColumn, typeof(int));
foo.GetColumnValues(dm.mainColumn, typeof(string));

Or using generics:

foo.GetColumnValues<int>(dm.mainColumn);
foo.GetColumnValues<string>(dm.mainColumn);

You can do this, just wrap it in typeof()

foo.GetColumnValues(typeof(int))

public void GetColumnValues(Type type)
{
    //logic
}