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I am learning Java Generics. My understanding is that Generics parameterize Collections by type. In the Oracle tutorial there is the following comment:
In generic code, the question mark (?), called the wildcard, represents an unknown type.
On the next page there is the following example of method declaration with an upper-bounded wildcard in the parameters:
public void process(List<? extends Foo> list)
Given that, I am wondering why this method declaration is illegal:
public void process(List<E extends Number> list)
while this one is legal:
public <E extends Number> void process(List<E> list)
293
For static generic methods, the type parameter section must appear before the method's return type. The complete syntax for invoking this method would be: Pair<Integer, String> p1 = new Pair<>(1, "apple"); Pair<Integer, String> p2 = new Pair<>(2, "pear"); boolean same = Util. <Integer, String>compare(p1, p2);
To declare a bounded type parameter, list the type parameter's name, followed by the extends keyword, followed by its upper bound, which in this example is Number . Note that, in this context, extends is used in a general sense to mean either "extends" (as in classes) or "implements" (as in interfaces).
In a generic type or method definition, a type parameter is a placeholder for a specific type that a client specifies when they create an instance of the generic type.
When specifying the method parm types, you're using the generic type, so it has to be defined upfront. In this statement, you use E without definition
public void process(List<E extends Number> list) { /* ... */ }
However, in the second one, it is defined before the method return type (void):
public <E extends Number> void process(List<E> list) { /* ... */ }
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