The difference between T and ? in java generics

T represents a type.

Add to the class:

class SuperClass<A>{}

Add to the method:

public <T>void fromArrayToCollection(T[] a, Collection<T> c){}

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The 8742468051c85b06f0a0af9e3e506b5c on the method means that generic parameters are used in the brackets. If generics are passed in the class, they do not need to be passed here. The generic parameters on the calling type are provided, provided that they are used in the method. The generic type is consistent with the generic type passed in the class.

class People<T>{
public void show(T a) {
   }
}

T extends T2 means that the passed parameter is T2 or a subtype of T2.

? is a wildcard character and refers to all types.

is generally used to define a reference variable. The advantage of this is that, as shown below, defining a sup reference variable can point to multiple objects.

SuperClass<?> sup = new SuperClass<String>("lisi");
sup = new SuperClass<People>(new People());
sup = new SuperClass<Animal>(new Animal());

If you don’t use ?, and use a fixed type, then:

SuperClass<String> sup1 = new SuperClass<String>("lisi");
SuperClass<People> sup2 = new SuperClass<People>("lisi");
SuperClass<Animal> sup3 = new SuperClass<Animal>("lisi");

This is the benefit of ? wildcard.

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