Groovy 未来get从executor服务提交返回null
在执行Java到Groovy代码迁移的初始阶段时,我遇到了一个问题,Groovy版本从未来返回null,而Java返回正确的整数(123) J1.java和G1.groovy之间的唯一更改是类名和lambda到闭包的转换 //文件:J1.javaGroovy 未来get从executor服务提交返回null,groovy,Groovy,在执行Java到Groovy代码迁移的初始阶段时,我遇到了一个问题,Groovy版本从未来返回null,而Java返回正确的整数(123) J1.java和G1.groovy之间的唯一更改是类名和lambda到闭包的转换 //文件:J1.java import java.util.concurrent.ExecutorService; import java.util.concurrent.Executors; import java.util.concurrent.Future; public
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.Future;
public class J1 {
public static void main (String... args) throws Exception {
ExecutorService executor = Executors.newFixedThreadPool (1);
Future<Integer> future = executor.submit (() -> 123);
System.out.println ("Result: " + future.get ());
executor.shutdown ();
}
}
在Groovy中,闭包是
可运行的
和可调用的
不幸的是,当您调用executor.submit{..}
时,运行库选择调用executor.submit(Runnable)
,而该调用不返回值
您应该显式地将闭包强制转换为可调用的:
def executor = Executors.newFixedThreadPool(1)
def future = executor.submit ({ -> 123 } as Callable)
println "Result: ${future.get()}"
executor.shutdown()
Groovy: 2.4.5
Java: 1.8 64-bit
Platform: Windows 7 64-bit
def executor = Executors.newFixedThreadPool(1)
def future = executor.submit ({ -> 123 } as Callable)
println "Result: ${future.get()}"
executor.shutdown()