当我使用Java 8的新语法糖迭代一个集合时,例如
myStream.forEach(item -> {
// do something useful
});
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这不等同于下面的"旧语法"片段吗?
myStream.forEach(new Consumer<Item>() {
@Override
public void accept(Item item) {
// do something useful
}
});
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这是否意味着Consumer每次迭代集合时都会在堆上创建一个新的匿名对象?这需要多少堆空间?它有什么性能影响?这是否意味着我在迭代大型多级数据结构时应该使用旧样式for循环?
我怎样才能优雅地序列化lambda?
例如,下面的代码抛出一个NotSerializableException.如何在不创建SerializableRunnable"虚拟"界面的情况下修复它?
public static void main(String[] args) throws Exception {
File file = Files.createTempFile("lambda", "ser").toFile();
try (ObjectOutput oo = new ObjectOutputStream(new FileOutputStream(file))) {
Runnable r = () -> System.out.println("Can I be serialized?");
oo.writeObject(r);
}
try (ObjectInput oi = new ObjectInputStream(new FileInputStream(file))) {
Runnable r = (Runnable) oi.readObject();
r.run();
}
}
Run Code Online (Sandbox Code Playgroud) 假设我有一个使用lambda表达式(闭包)定义的对象列表.有没有办法检查它们,以便可以进行比较?
我最感兴趣的代码是
List<Strategy> strategies = getStrategies();
Strategy a = (Strategy) this::a;
if (strategies.contains(a)) { // ...
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完整的代码是
import java.util.Arrays;
import java.util.List;
public class ClosureEqualsMain {
interface Strategy {
void invoke(/*args*/);
default boolean equals(Object o) { // doesn't compile
return Closures.equals(this, o);
}
}
public void a() { }
public void b() { }
public void c() { }
public List<Strategy> getStrategies() {
return Arrays.asList(this::a, this::b, this::c);
}
private void testStrategies() {
List<Strategy> strategies = getStrategies();
System.out.println(strategies);
Strategy a = (Strategy) …Run Code Online (Sandbox Code Playgroud) Loop.times(5, () -> {
System.out.println("looping");
});
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哪些有效编译到?
for(int i = 0; i < 5; i++)
System.out.println("looping");
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或类似的东西
new CallableInterfaceImpl(){
public void call(){
for(int i = 0; i < 5; i++)
System.out.println("looping");
}
}.call();
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它会替换(内联类),还是实际创建一个匿名类?
请考虑以下代码片段:
public static Object o = new Object();
public static Callable x1() {
Object x = o;
return () -> x;
}
public static Callable x2() {
return () -> o;
}
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方法x2()将始终返回相同的lamba对象,而x1()始终会创建新的对象:
System.out.println(x1());
System.out.println(x1());
System.out.println(x2());
System.out.println(x2());
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会打印出这样的东西:
TestLambda$$Lambda$1/821270929@4a574795
TestLambda$$Lambda$1/821270929@f6f4d33
TestLambda$$Lambda$2/603742814@7adf9f5f
TestLambda$$Lambda$2/603742814@7adf9f5f
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哪里(在JVM规范中我猜?)是描述了这种lambda重用规则?JVM如何确定重用与否?
以下测试失败
@Test
public void test() {
Function<String, Integer> foo = Integer::parseInt;
Function<String, Integer> bar = Integer::parseInt;
assertThat(foo, equalTo(bar));
}
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有没有办法让它通过?
编辑:我会尝试更清楚地说明我正在尝试做什么.
可以说我有这些课程:
class A {
public int foo(Function<String, Integer> foo) {...}
}
class B {
private final A a; // c'tor injected
public int bar() {
return a.foo(Integer::parseInt);
}
}
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现在让我说我想为B编写单元测试:
@Test
public void test() {
A a = mock(A.class);
B b = new B(a);
b.bar();
verify(a).foo(Integer::parseInt);
}
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问题是测试失败,因为方法引用不相等.
Java 8引入了lambda表达式,这是一件好事.但现在考虑重写这段代码:
class B implements PropertyChangeListener {
void listenToA(A a) {
a.addPropertyChangeListener(this);
}
void propertyChange(PropertyChangeEvent evt) {
switch(evt.getPropertyName()) {
case "Property1":
doSomething();
break;
case "Property2":
doSomethingElse(); case "Property1":
doSomething;
break;
break;
}
void doSomething() { }
void doSomethingElse() { }
}
class A {
final PropertyChangeSupport pcs = new PropertyChangeSupport(this);
void addPropertyChangeListener(PropertyChangeListener listener) {
pcs.addPropertyChangeListener(listener);
}
void removePropertyChangeListener(PropertyChangeListener listener) {
pcs.removePropertyChangeListener(listener);
}
}
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随着lambda表达式和方法的引用,它已不再是需要有B落实PropertyChangeListner,我们可以写
class B {
void listenToA(A a) {
// using method …Run Code Online (Sandbox Code Playgroud)