Not*_*bug 2 java lambda java-8 method-reference
考虑下面的代码,
interface TestInter {
public void abc();
}
class DemoStatic {
public static void testStatic(String abc) {
System.out.println(abc);
}
public void runTest () {
// Using lambda expression.
String str = "demo string" ;
TestInter demo1 = () -> DemoStatic.testStatic(str);
demo1.abc();
// Using method reference.
TestInter demo2 = DemoStatic::testStatic; // This line is not compiling.
demo2.abc();
}
}
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如果这样的答案中描述的方法的参数将被消除,我们可以调用该testStatic方法作为TestInter接口实体的主体.abc()testStaic()
但在这种情况下,我们如何为参数化方法编写方法参考testStatic?
Hon*_*dek 10
您的功能界面TestInter没有相应的testStatic(String)方法签名.如果要参考testStatic()使用::表示法,则应添加参数:
interface TestInter2 {
public void abc(String abc);
}
public void runTest () {
TestInter2 demo2 = DemoStatic::testStatic;
demo2.abc(str);
}
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根据Oracle Java教程,有四种方法参考:

我准备了3个接口 - 用于0,1和2参数.然后我们有3个静态和3个实例方法:
interface F0 {
void f0();
}
interface F1 {
void f1(MetRef i1);
}
interface F2 {
void f2(MetRef i1, MetRef i2);
}
public class MetRef {
public static void stat0() {;}
public static void stat1(MetRef a) {;}
public static void stat2(MetRef a, MetRef b) {;}
public void inst0() {;}
public void inst1(MetRef a) {;}
public void inst2(MetRef a, MetRef b) {;}
}
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现在看看每种方法如何在各种组合中用作方法参考,并将它们与前一个表进行比较.查看操作中的不同方法引用类型,并注意参数的来源.
public class Test {
public static void main(String[] args) {
final MetRef mr = new MetRef();
final F0 mr01 = MetRef::stat0; // 1: f0() ~ MetRef.stat0()
final F0 mr02 = mr::inst0; // 2: f0() ~ mr.inst0()
final F0 mr04 = MetRef::new; // 4: f0() ~ new MetRef()
final F1 mr11 = MetRef::stat1; // 1: f1(i1) ~ MetRef.stat1(i1)
final F1 mr12 = mr::inst1; // 2: f1(i1) ~ mr.inst1(i1)
final F1 mr13 = MetRef::inst0; // 3: f1(i1) ~ i1.inst0() <== NOTICE!
final F2 mr21 = MetRef::stat2; // 1: f2(i1, i2) ~ MetRef.stat2(i1, i2)
final F2 mr22 = mr::inst2; // 2: f2(i1, i2) ~ mr.inst2(i1, i2)
final F2 mr23 = MetRef::inst1; // 3: f2(i1, i2) ~ i1.inst1(i2) <== NOTICE!
}
}
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