英文:
how return args works in Java 8?
问题
I'm new to Java8
Can anyone share what return args is returning in below code snippet? How it looks like in Java 7 to understand what's actually happening?
public ApplicationRunner initializeConnection(
RsvpsWebSocketHandler rsvpsWebSocketHandler) {
return args -> {
WebSocketClient rsvpsSocketClient = new StandardWebSocketClient();
rsvpsSocketClient.doHandshake(rsvpsWebSocketHandler, MEETUP_RSVPS_ENDPOINT);};}
英文:
I'm new to Java8
Can anyone share what return args is returning in below code snippet ? How it looks like in Java 7 to understand what's actually happening ?
public ApplicationRunner initializeConnection(RsvpsWebSocketHandler rsvpsWebSocketHandler) {return args -> {WebSocketClient rsvpsSocketClient = new StandardWebSocketClient();rsvpsSocketClient.doHandshake(rsvpsWebSocketHandler, MEETUP_RSVPS_ENDPOINT);};}
答案1
得分: 2
这是一个 lambda，只是匿名类的简写初始化。在 Java 1.7 中，它会像这样：
return new ApplicationRunner() {@Overridepublic void run(ApplicationArguments args) throws Exception {WebSocketClient rsvpsSocketClient = new StandardWebSocketClient();rsvpsSocketClient.doHandshake(rsvpsWebSocketHandler, MEETUP_RSVPS_ENDPOINT);}};
英文:
It's a lambda, which is just a shorthand initialization of an anonymous class.
In Java 1.7 it would look like:
return new ApplicationRunner() {@Overridepublic void run(ApplicationArguments args) throws Exception {WebSocketClient rsvpsSocketClient = new StandardWebSocketClient();rsvpsSocketClient.doHandshake(rsvpsWebSocketHandler, MEETUP_RSVPS_ENDPOINT);}};
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