我有一个简单的Person
课程,其中getName()
会返回String
:
public class Person {
public String getName() {...}
}
如何在运行时使用LambdaMetafactory
为非静态方法getName()
创建lambda?
这是我到目前为止所得到的:
public class MyMain {
public static void main(String[] args) throws Throwable {
GetterFunction getterFunction;
MethodHandles.Lookup lookup = MethodHandles.lookup();
String invokedMethodName = "getName";
MethodType invokedType = MethodType.methodType(GetterFunction.class);
MethodType methodType = MethodType.methodType(Object.class);
MethodHandle virtual = lookup.findVirtual(Person.class, "getName", MethodType.methodType(String.class));
CallSite site = LambdaMetafactory.metafactory(lookup,
invokedMethodName,
invokedType,
methodType,
virtual,
methodType);
getterFunction = (GetterFunction) site.getTarget().invokeExact();
System.out.println(getterFunction.getName(new Person("Ann")));
}
@FunctionalInterface
private interface GetterFunction {
String getName(Person person);
}
}
引发:
java.lang.invoke.LambdaConversionException: Incorrect number of parameters for instance method invokeVirtual foo.Person.getName:()String; 0 captured parameters, 0 functional interface method parameters, 0 implementation parameters
at java.lang.invoke.AbstractValidatingLambdaMetafactory.validateMetafactoryArgs(AbstractValidatingLambdaMetafactory.java:193)
at java.lang.invoke.LambdaMetafactory.metafactory(LambdaMetafactory.java:303)
答案 0 :(得分:2)
这有效:
public class MyMain {
public static void main(String[] args) throws Throwable {
GetterFunction getterFunction;
final MethodHandles.Lookup lookup = MethodHandles.lookup();
MethodType methodType = MethodType.methodType(String.class, Person.class);
final CallSite site = LambdaMetafactory.metafactory(lookup,
"invoke",
MethodType.methodType(GetterFunction.class),
methodType,
lookup.findVirtual(Person.class, "getName", MethodType.methodType(String.class)),
methodType);
getterFunction = (GetterFunction) site.getTarget().invokeExact();
System.out.println(getterFunction.invoke(new Person("Ann")));
}
@FunctionalInterface
interface GetterFunction {
String invoke(final Person callable);
}
}
答案 1 :(得分:1)
只需扩展Geoffrey De Smet的答案,以显示使用Function会是什么样子,因为需要进行一些小的调整:
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