Java等价于c#列表声明
我想知道java中是否有与此等效的:Java等价于c#列表声明,java,c#,object,variables,Java,C#,Object,Variables,我想知道java中是否有与此等效的: List<Person> People = new List<Person>(){ new Person{ FirstName = "John", LastName = "Doe" }, new Person{ FirstName = "Someone", LastName = "Special" } }; List People=新列表
List<Person> People = new List<Person>(){
new Person{
FirstName = "John",
LastName = "Doe"
},
new Person{
FirstName = "Someone",
LastName = "Special"
}
};
List People=新列表(){
新人{
FirstName=“约翰”,
LastName=“Doe”
},
新人{
FirstName=“某人”,
LastName=“特殊”
}
};
当然,假设。。。有一个名为Person的类,它的FirstName和LastName字段都是{get;set;}可以从Java9编写的
// immutable list
List<Person> People = List.of(
new Person("John", "Doe"),
new Person("Someone", "Special"));
// mutable list
List<Person> People = new ArrayList<Person>() {{
add(new Person("John", "Doe"));
add(new Person("Someone", "Special"));
}};
//不可变列表
列出人(
新人(“约翰”、“Doe”),
新人(“某人”、“特别”);
在Java5.0中,您可以编写
// cannot add or remove from this list but you can replace an element.
List<Person> People = Arrays.asList(
new Person("John", "Doe"),
new Person("Someone", "Special"));
//无法在此列表中添加或删除元素,但可以替换元素。
List People=Arrays.asList(
新人(“约翰”、“Doe”),
新人(“某人”、“特别”);
从Java1.4可以编写
// immutable list
List<Person> People = List.of(
new Person("John", "Doe"),
new Person("Someone", "Special"));
// mutable list
List<Person> People = new ArrayList<Person>() {{
add(new Person("John", "Doe"));
add(new Person("Someone", "Special"));
}};
//可变列表
List People=new ArrayList(){{
添加(新人员(“John”、“Doe”);
添加(新的人(“某人”、“特殊”));
}};
您可以从Java 9编写
// immutable list
List<Person> People = List.of(
new Person("John", "Doe"),
new Person("Someone", "Special"));
// mutable list
List<Person> People = new ArrayList<Person>() {{
add(new Person("John", "Doe"));
add(new Person("Someone", "Special"));
}};
//不可变列表
列出人(
新人(“约翰”、“Doe”),
新人(“某人”、“特别”);
在Java5.0中,您可以编写
// cannot add or remove from this list but you can replace an element.
List<Person> People = Arrays.asList(
new Person("John", "Doe"),
new Person("Someone", "Special"));
//无法在此列表中添加或删除元素,但可以替换元素。
List People=Arrays.asList(
新人(“约翰”、“Doe”),
新人(“某人”、“特别”);
从Java1.4可以编写
// immutable list
List<Person> People = List.of(
new Person("John", "Doe"),
new Person("Someone", "Special"));
// mutable list
List<Person> People = new ArrayList<Person>() {{
add(new Person("John", "Doe"));
add(new Person("Someone", "Special"));
}};
//可变列表
List People=new ArrayList(){{
添加(新人员(“John”、“Doe”);
添加(新的人(“某人”、“特殊”));
}};
注意,List.of
返回一个不可变列表。感谢您的快速回复!它就像一个符咒请注意,List.of
返回一个不可变的列表。感谢您的快速回复!它就像一个符咒