编译器是否从另一个中找出一个通用参数列表？
Does compiler figure out one of the generic argument list from the other?
摘自 "Thinking in Java" 书：
One of the complaints about generics is that it adds even more text to
your code. Consider this :
Map<Person, List<? extends Pet>> petPeople =
new HashMap<Person, List<? extends Pet>>();
It appears that you are repeating yourself, and that the compiler
should figure out one of the generic argument lists from the other.
Alas, it cannot, ...
这是否意味着双方的显式类型规范是由编译器强制要求的？
相反，它对我来说似乎工作得很好（没有明确说明）：
Map<Person, List<? extends Pet>> petPeople = new HashMap();
你看的是一本古籍，自从Java-7 加上菱形运算符：
后效果很好
Map<Person, List<? extends Pet>> petPeople = new HashMap<>();
摘自 "Thinking in Java" 书：
One of the complaints about generics is that it adds even more text to your code. Consider this :
Map<Person, List<? extends Pet>> petPeople = new HashMap<Person, List<? extends Pet>>();It appears that you are repeating yourself, and that the compiler should figure out one of the generic argument lists from the other. Alas, it cannot, ...
这是否意味着双方的显式类型规范是由编译器强制要求的？
相反，它对我来说似乎工作得很好（没有明确说明）：
Map<Person, List<? extends Pet>> petPeople = new HashMap();
你看的是一本古籍，自从Java-7 加上菱形运算符：
后效果很好Map<Person, List<? extends Pet>> petPeople = new HashMap<>();