Re: How to create an instance of type T?
Peter, maybe you can not understand what I mean.
Stefan's solution that use supertype of all possible T's can solve the
problem, but it is achieved by inheritance, not by generic.
In fact, I just want to test the ability of Java generic.
In C#, I can have a perfect solution:
// C# solution
class InputClass<T>
{
public T GetValue()
{
string str = GetInputFromConsole();
T v = default(T); // critical !!!
MethodInfo[] mis = v.GetType().GetMethods();
foreach (MethodInfo m in mis)
{
ParameterInfo[] pis = m.GetParameters();
if (m.Name == "Parse" && pis.Length == 1)
{
object[] o = {str};
v = (T)m.Invoke(v, o); // T has method Parse(string
s)
break;
}
}
return v;
}
}
so, you can make invocations as following:
InputClass<int> o = new InputClass<int>();
int k = o.GetValue();
InputClass<double> oo = new InputClass<double>();
double d = oo.GetValue();
This solution can not be achieved in Java due to its type erasure.
By the way, let's look at Jusha's solution. It can be simplified as:
GClass
{
public static <T> T getValue(Class<T> clazz) throws Exception
{
String p = getInputFromConsole();
return clazz.getConstructor(String.class).newInstance(p);
}
}
so, you can make invokcations as following:
Integer o = GClass.<Integer>getValue(Integer.class);
Double oo = GClass.<Double>getValue(Double.class);
Foo ooo = GClass.<Foo>getValue(Foo.class);
but you can not do:
Integer o = GClass.<Integer>getValue();
Double oo = GClass.<Double>getValue();
Foo ooo = GClass.<Foo>getValue();