Java Streams concat Streams with Supplier 其次是不同的（惰性评估行为）

Question

我有一个像这样的简单代码

import java.util.function.Supplier;
import java.util.stream.IntStream;
import java.util.stream.Stream;
public class StreamSupplierVersusConcat {
    public static void main(String[] args) {
         final StreamSupplierVersusConcat clazz = new StreamSupplierVersusConcat();
         clazz.doConcat();                
    }

    private void doConcat(){        
         System.out.println(Stream.concat(buildStreamFromRange(0,1000).get()
                ,buildStreamFromRange(1000,2000).get())
                .anyMatch("1"::equals));                
}

    private Supplier<Stream<String>>buildStreamFromRange(final int start,final int end){
        return ()->IntStream.range(start, end)
                .mapToObj(i->{
                    System.out.println("index At: "+i);
                    return String.valueOf(i);
                });
    }    
}

我知道 concat 是惰性的，所以当我运行我看到的代码只生成 2 个值，这很好，但知道 distinct 是一个有状态的操作，我认为将该方法放在 Stream 管道上Stream 会生成所有值然后执行 anyMatch 方法但是如果我这样说

    private void doConcat(){        
         System.out.println(Stream.concat(buildStreamFromRange(0,1000).get()
                ,buildStreamFromRange(1000,2000).get())
                .distinct()//ARE ALL THE VALUES GENERATED NOT REQUIRED HERE???
                .anyMatch("1"::equals));                
}

但是无论有无 distinct，我都会得到相同的响应。

index At: 0
index At: 1
true

我错过了什么？ 我认为 distinct 会在 anyMatch 看到之前看到所有项目。 在 Java 8.

上测试

非常感谢。

在简历中我的理解是 我认为 distinct 会在 anyMatch 看到任何之前看到所有项目。 不正确这个例子解释了它。

private void distinctIsNotABlockingCall(){
    final boolean match = Stream.of("0","1","2","3","4","5","6","7","8","8","8","9","9","9","9","9","9","9","9","9","10","10","10","10")
            .peek(a->System.out.println("before: "+a))
            .distinct()//I THOUGHT THAT NOT ANYMATCH WAS CALLED AFTER DISTINCT HANDLE ALL THE ITEMS BUT WAS WRONG.
            .peek(a->System.out.println("after: "+a))
            .anyMatch("10"::equals);
    System.out.println("match? = " + match);                
}

before: 0
after: 0
before: 1
after: 1
before: 2
after: 2
before: 3
after: 3
before: 4
after: 4
before: 5
after: 5
before: 6
after: 6
before: 7
after: 7
before: 8
after: 8
before: 8 distinct working
before: 8 distinct working
before: 9
after: 9 
before: 9 distinct working
before: 9 distinct working
before: 9 distinct working
before: 9 distinct working
before: 9 distinct working
before: 9 distinct working
before: 9 distinct working
before: 9 distinct working
before: 10
after: 10
match? = true

您可以看到 distinct 收到了重复值和非重复值，而且 anyMatch 收到了那些非重复值，并且 distinct 和 anyMatch 正在同时工作，非常感谢。

Answer 1

Streams are lazy because intermediate operations are not evaluated unless terminal operation is invoked. check SO answer here

据我所知，Stream Api 流式传输每个元素，直到应用终端操作，然后流式传输下一个元素。

这也说明了这里的情况。元素 "0" 被流式传输，不满足终端操作。还有一个需要推流，"1"现在，终端操作 .anyMatch("1"::equals));满足。无需流式传输更多元素。 Distinct 将在两者之间被调用，而无需更改流元素。

因此，如果您在 "0" 之后还有另一个 "0" 要进行流式传输，则它根本不会到达终端操作。

 private void doConcat(){        
     
System.out.println(Stream.concat(buildStreamFromRange(0,1000).get()
                ,buildStreamFromRange(1000,2000).get())
                .distinct()
                .peek( e -> System.out.println(e))
                .anyMatch("1"::equals));

尝试添加 peek 并尝试在开头流式传输 2 "0" 个元素。其中只有 1 个会通过流程并从 peek 打印出来。

Peek 也可用于调试目的，并在您不确定时查看流程的行为，因此在将来使用它对您有利。

供未来读者参考的简单示例：

下面是一个更简单的例子，未来的读者将能够理解流中的惰性运算符是如何工作的：

Stream.of("0","0","1","2","3","4")
                .distinct()
                .peek(a->System.out.println("after distinct: "+a))
                .anyMatch("1"::equals);

将打印

after distinct: 0
after distinct: 1

首先"0"一直走到终端操作却不满足。必须流式传输另一个元素。

第二个"0"通过.distinct()过滤，永远不会到达终端操作

终端操作还没满足，流下一个元素

"1"经过终端操作，满足

不需要流式处理更多元素。

Java Streams concat Streams with Supplier 其次是不同的（惰性评估行为）

Java Streams concat Streams with Supplier followed by distinct (lazy evaluation behavior)

java

java-8

java-stream

distinct-values