The error message “inferred type ‘s’ for type parameter ‘s’ is not within its bound; should extend” occurs when the type argument provided for a generic parameter does not meet the specified constraints of that parameter. In simple terms, it means that the type used as an argument should be a subclass or implement a specific interface as a constraint on the generic parameter.
Let’s consider an example to understand this error better. Suppose we have a generic class called Container with a type parameter T that should extend the Comparable interface. The Comparable interface is used to define a natural ordering for objects of the generic type T.
public class Container<T extends Comparable<T>> {
private T element;
// Constructor and other methods...
}
Now, let’s say we try to create an instance of Container with a type argument that does not implement the Comparable interface, such as String. It will result in the mentioned error.
Container<String> container = new Container<>(); // Error: inferred type 'java.lang.String' for type parameter 'T' is not within its bound; should extend 'java.lang.Comparable<java.lang.String>'
In this case, the error message is indicating that the type String does not extend or implement the Comparable interface, which is required by the Container class’s type parameter T.
To resolve this error, you need to provide a type argument that satisfies the bound constraint. For example, if you want to use the String type, you can use the String class that implements the Comparable interface:
Container<String> container = new Container<>(new String()); // No error
In this updated code, we explicitly specify the type argument as String, and the error is resolved.
It is crucial to understand the constraints imposed on the type parameters of generic classes or methods to avoid such errors. Reading the error message and checking the constraints specified on the type parameter can guide you in finding the appropriate type arguments.