sau*_*mum 5 java lambda list java-8
我写了一个简单的程序来迭代List
使用java 8 lambda.
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.function.Consumer;
public class FirstLamdaExpression {
public static void main(String[] args) {
List<Integer> list = Arrays.asList(1, 2, 3, 4, 5, 6, 7, 8, 9, 10);
//Way 1 : old way
list.forEach(new Consumer<Integer>() {
@Override
public void accept(Integer t) {
System.out.print(t + " ");
}
});
//Way 2
System.out.println(" ");
list.forEach((Integer t) -> System.out.print(t + " "));
//Way 3
System.out.println(" ");
list.forEach((t) -> System.out.print(t + " "));
//Way 4
System.out.println(" ");
list.forEach(System.out::print);
}
}
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在下面的程序中,我在lambda中执行了两个以上的逻辑.我面临的问题是如何更新第四种方式即System.out::print
?
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.function.Consumer;
public class SecondLamdaExpression {
public static void main(String[] args) {
List<Integer> list = Arrays.asList(1, 2, 3, 4, 5, 6, 7, 8, 9, 10);
//Way 1 : Old way
list.forEach(new Consumer<Integer>() {
@Override
public void accept(Integer t) {
System.out.print(t + " Twice is : ");
System.out.print(t*2 + " , ");
}
});
//Way 2
System.out.println(" ");
list.forEach((Integer t) ->
{System.out.print(t + " Twice is : ");
System.out.print(t*2 + " , ");
});
//Way 3
System.out.println(" ");
list.forEach((t) -> {System.out.print(t + " Twice is : ");
System.out.print(t*2 + " , ");
});
//Way 4
//System.out.println(" ");
//list.forEach((t)-> System.out::print{(t + " Twice is : ");});
}
}
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看来你问的是如何传递t + " Twice is : " + t*2 + " , "
给方法引用.您不能将显式参数传递给方法引用,也不能将方法引用与lambda表达式组合在一起.
您可以使用Stream
管道首先map
t
为每个值打印任何内容t
,然后使用forEach
with方法引用来打印它:
list.stream().map(t -> t + " Twice is : " + t*2 + " , ").forEach(System.out::print);
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