Java 8函数始终返回相同的值而无需考虑参数 [英] Java 8 function that always return the same value without regarding to parameter

问题描述

Java 8中是否有一个预定义的函数，其功能如下:

Is there a predefined Function in Java 8 that does something like this:

static <T, R> Function<T, R> constant(R val) {
    return (T t) -> {
        return val;
   };
}

要回答人们对我为什么需要此功能的疑问，是当我尝试将整数解析为罗马数字时的实际用法:

To answer people's query on why I need this function here is the real usage when I am trying to parse an integer to an roman numerals:

// returns the stream of roman numeral symbol based
// on the digit (n) and the exponent (of 10)
private static Stream<Symbol> parseDigit(int n, int exp) {
    if (n < 1) return Stream.empty();
    Symbol base = Symbol.base(exp);
    if (n < 4) {
        return IntStream.range(0, n).mapToObj(i -> base);
    } else if (n == 4) {
        return Stream.of(base, Symbol.fifth(exp));
    } else if (n < 9) {
        return Stream.concat(Stream.of(Symbol.fifth(exp)),
            IntStream.range(5, n).mapToObj(i -> base));
    } else { // n == 9 as n always < 10
        return Stream.of(base, Symbol.base(exp + 1));
    }
}

我猜想IntStream.range(0, n).mapToObj(i -> base)可以简化为Stream.of(base).times(n - 1)之类的东西，不幸的是流对象上没有times(int)方法.有人知道怎么做吗?

And I guess the IntStream.range(0, n).mapToObj(i -> base) could be simplified to something like Stream.of(base).times(n - 1), unfortunately there is no times(int) method on stream object. Does anyone know how to make it?

推荐答案

一个简单的lambda，x -> val似乎等同于您的方法；

A simple lambda, x -> val seems to be equivalent to your method;

Function<Integer, Integer> test1 = constant(5);
Function<Integer, Integer> test2 = x -> 5;

...两者都忽略输入，并在应用时输出常量5；

...both ignore the input and output the constant 5 when applied;

> System.out.println(test1.apply(2));
5
> System.out.println(test2.apply(2));
5

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