Java 8 function that always return the same value without regarding to parameter

问题

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;
   };
}

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));
    }
}

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?

回答1:

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

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

来源：https://stackoverflow.com/questions/23665502/java-8-function-that-always-return-the-same-value-without-regarding-to-parameter