2017年5月26日金曜日

開発環境

Think Perl 6: How to Think Like a Computer Scientist (Laurent Rosenfeld(著)、Allen B. Downey(著)、Oreilly & Associates Inc)の Part 1(Starting with the basics)、Chapter 11(Case Study: Data Structure Selection)の Constructing New Operators、Exercise: Constructing New Operators.を JavaScript で取り組んでみる。

Exercise: Constructing New Operators.

コード(Emacs)

HTML5

<pre id="output0"></pre>
<label for="n0">n = </label>
<input id="n0" type="number" min="0" step="1" value="20">
<button id="run0">run</button>
<button id="clear0">clear</button>

<script src="sample02.js"></script>

JavaScript

let btn0 = document.querySelector('#run0'),
    btn1 = document.querySelector('#clear0'),
    pre0 = document.querySelector('#output0'),
    input_n = document.querySelector('#n0'),
    p = (x) => pre0.textContent += x + '\n',
    range = (start, end, step=1) => {
        let result = [];
        for (let i = start; i < end; i += 1) {
            result.push(i);
        }
        return result;
    };

// JavaScript では新しい演算子を定義する方法がないから、階乗を計算する関数を再帰とかいくつかの方法で定義してみる。
let factorial1 = (n) => n <= 1 ? 1 : n * factorial1(n - 1);

let factorial2 = (n) => {
    let inner = (n, result) => {
        if (n === 0) {
            return result;
        }
        return inner(n - 1, n * result);
    };
    return inner(n, 1);
};
let factorial3 = (n) => range(1, n + 1).reduce((x, y) => x * y, 1),
    fns = [factorial1, factorial2, factorial3];

let output = () => {
    p(
        range(0, 11)
            .map((n) =>
                 fns
                 .map((fn) => fn(n))
                 .join('\n'))
            .join('\n')
    );
    
    let n = parseInt(input_n.value, 10);
    
    p(fns.map((fn) => fn(n)).join('\n'));
};

let clear = () => pre0.textContent = '';

input_n.onchange = output;
btn0.onclick = output;
btn1.onclick = clear;

output();















						

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