Question

假设：

ExecutorService service = ...;

// somewhere in the code the executorService is used this way:
service.submit(() -> { ... });

lambda表达式默认为Callable。
有没有办法让它实例化Runnable？

感谢您的帮助。

Answer 1

您可以将其声明为Runnable，或使用强制转换：

Runnable r = () -> { ... };
executorService.submit(r);

或

executorService.submit((Runnable) () -> { ... });

Answer 2

你的前提是错的。此调用不会默认为Callable。选择是通过lambda表达式的形状进行的，即它是否返回值：

ExecutorService service = null;

// somewhere in the code the executorService is used this way:

// invokes submit(Runnable)
service.submit(() -> {  });
// invokes submit(Runnable)
service.submit(() -> { return; });
// invokes submit(Callable)
service.submit(() -> "foo");
// invokes submit(Callable)
service.submit(() -> { return "foo"; });

// in case you really need disambiguation: invokes submit(Runnable,null)
service.submit(() -> { throw new RuntimeException(); }, null);
// dito (ignoring the returned value)
service.submit(this::toString, null);

请注意，如果您不需要返回的Future，则只需使用execute(Runnable)直接将Runnable入队，而不是将其包裹在{{1}中}}

Answer 3

   service.submit((Runnable) () -> {
      ...
   });

Answer 4

Callable和Runnable在签名方面的区别在于Callable正在返回一个值，而Runnable却没有。

所以() -> { }是Runnable，例如() -> ""是Callable。

消除lambda中的功能接口的歧义

4 个答案: