I am reading the following, https://docs.oracle.com/javase/tutorial/java/generics/genTypeInference.html in particular the part "Type Inference and Generic Constructors of Generic and Non-Generic Classes".

I have tried to run the following:

public class MyClass < X > {

X myObject;

<T> MyClass(T t, X x) {
    myObject = x;
    System.out.println("t is " + x.getClass().getName());
    System.out.println("x is " + t.getClass().getName());
    System.out.println("myObject is " + myObject.getClass().getName());

    System.out.println("t value is " + t);
    System.out.println("x value is " + x);
    System.out.println("myObject value is " + myObject);

    myObject = new Integer(t); // 1 
}

public static void main(String[] args) {
    String myString = "1";
    MyClass<Integer> myObject = new MyClass<>(myString, new Integer(myString));
}

}

but I get the following compilation error:

$javac -Xdiags:verbose MyClass.java 
MyClass.java:15: error: no suitable constructor found for Integer(T)
        myObject = new Integer(t); // 1 
                   ^
    constructor Integer.Integer(int) is not applicable
      (argument mismatch; T cannot be converted to int)
    constructor Integer.Integer(String) is not applicable
      (argument mismatch; T cannot be converted to String)
  where T,X are type-variables:
    T extends Object declared in constructor <T>MyClass(T,X)
    X extends Object declared in class MyClass
1 error

If I comment //1 there is no error and the output is

t is java.lang.Integer
x is java.lang.String
myObject is java.lang.Integer
t value is 1
x value is 1
myObject value is 1

Can somebody tell me please what's happening and why the error?

1 Answers

2
luk2302 On Best Solutions

From the compiler's perspective:

T is not a String, which Integer.parseInt(...) expects to be passed in.

And Integer.parseInt(...) would return an int which is not an X.