Re: How to create an instance of type T?

Daniel Pitts <>
Sun, 24 Feb 2008 22:51:43 -0800
prowyh wrote:

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

        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:

    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();

However you *can* do
Integer o = GClass.getValue(Integer.class);
Foo foo = GClass.getValue(Foo.class);

What's wrong with that?

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