关于Lambda详见lambda描述。关于函数式接口详见函数式接口
首先我们来看一下BiFunction类,代码如下:
package java.util.function;
import java.util.Objects;
/**
* Represents a function that accepts two arguments and produces a result.
* This is the two-arity specialization of {@link Function}.
*
* <p>This is a <a href="package-summary.html">functional interface</a>
* whose functional method is {@link #apply(Object, Object)}.
*
* @param <T> the type of the first argument to the function
* @param <U> the type of the second argument to the function
* @param <R> the type of the result of the function
*
* @see Function
* @since 1.8
*/
//这个注解表示是函数式接口
@FunctionalInterface
public interface BiFunction<T, U, R> {
/**
* Applies this function to the given arguments.
*
* @param t the first function argument
* @param u the second function argument
* @return the function result
*/
R apply(T t, U u);
/**
* Returns a composed function that first applies this function to
* its input, and then applies the {@code after} function to the result.
* If evaluation of either function throws an exception, it is relayed to
* the caller of the composed function.
*
* @param <V> the type of output of the {@code after} function, and of the
* composed function
* @param after the function to apply after this function is applied
* @return a composed function that first applies this function and then
* applies the {@code after} function
* @throws NullPointerException if after is null
*/
//这里面的default是java8的特性,通过这个关键字来设置函数的默认实现,这样实现类就可以不用实现这个接口,而采用接口中默认的 //实现
default <V> BiFunction<T, U, V> andThen(Function<? super R, ? extends V> after) {
Objects.requireNonNull(after);
return (T t, U u) -> after.apply(apply(t, u));
}
}
//下面的一行代码返回的是BitFunction<T,U,V>类型对象。
return (T t, U u) -> after.apply(apply(t, u));
首先,执行this.apply(t,u),这里面的this是调用andThen函数的BitFunction类对象。
然后,将返回值传到after类对象的apply函数中。
最后,返回BiFunction类对象。
这行代码等价于:
return new BiFunction<T,U,V>(){
V apply(T t, U u){
return after.apply(value);//这里面的value为this调用apply返回的结果。
}
}
举一个例子:
public int compute2(int a, b,BitFunction<Integer, Integer,Integer> function1, Function<Integer, Integer> function2) {
return function1.andThen(function2).apply(a);
}
调用这个方法:
test.compute2(2,3, (v1,v2) -> v1+ v2, value -> value * value);
(value,value) -> value + value是一个BitFunction<Integer, Integer,Integer>类对象,等价于
return new BiFunction<T,U,V>(){
V apply(T t, U u){
return after.apply(V);
}
}值就是function1
同样value -> value * value等价于
return new Function<R,V>(){
V apply(T t, U u){
return after.apply(V);
}
}值就是function2
所以return function1.andThen(function2).apply(a);执行步骤:
第一步:执行function1.apply(2,3)得到2+3=5。返回一个BiFunction类对象,该类对象中的apply函数执行function2.apply(5)
第二步:执行function2.apply(5)得到5*5=25。
第三步:BiFunction类对象的apply函数中返回25。
详细的描述请看详细介绍