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();