在Java中返回null时,如何评估下一个语句？

我想要做的是继续评估语句或方法,直到它得到一些东西而不是null.

我希望调用者代码简单有效.

你可以为它创建一个方法.

public static  T coalesce(Supplier... ts) {
    return asList(ts)
        .stream()
        .map(t -> t.get())
        .filter(t -> t != null)
        .findFirst()
        .orElse(null);
}

代码取自:http://benjiweber.co.uk/blog/2013/12/08/null-coalescing-in-java-8/

编辑正如评论中所述.在下面找到一个小片段如何使用它.使用StreamAPI比使用vargs方法参数更有优势.如果方法返回的值很昂贵而且简单的getter不返回,那么vargs解决方案将首先评估所有这些方法.

import static java.util.Arrays.asList;
import java.util.function.Supplier;
...
static class Person {
    String name;
    Person(String name) {
        this.name = name;
    }
    public String name() {
        System.out.println("name() called for = " + name);
        return name;
    }
}

public static  T coalesce(Supplier... ts) {
    System.out.println("called coalesce(Supplier... ts)");
    return asList(ts)
            .stream()
            .map(t -> t.get())
            .filter(t -> t != null)
            .findFirst()
            .orElse(null);
}

public static  T coalesce(T... ts) {
    System.out.println("called coalesce(T... ts)");
    for (T t : ts) {
        if (t != null) {
            return t;
        }
    }
    return null;
}

public static void main(String[] args) {
    Person nobody = new Person(null);
    Person john = new Person("John");
    Person jane = new Person("Jane");
    Person eve = new Person("Eve");
    System.out.println("result Stream API: " 
            + coalesce(nobody::name, john::name, jane::name, eve::name));
    System.out.println();
    System.out.println("result vargas    : " 
            + coalesce(nobody.name(), john.name(), jane.name(), eve.name()));
}

产量

called coalesce(Supplier... ts)
name() called for = null
name() called for = John
result Stream API: John

name() called for = null
name() called for = John
name() called for = Jane
name() called for = Eve
called coalesce(T... ts)
result vargas    : John

如输出中所示.在Stream解决方案中,将在coalesce方法内部评估返回值的方法.只执行两次,因为第二次调用返回预期non-null值.在vargs解决方案中,在coalesce调用方法之前,将评估返回值的所有方法.