開発環境
- macOS Sierra - Apple (OS)
- Emacs (Text Editor)
- JavaScript (プログラミング言語)
- Node.js, Safari(JavaScript エンジン)
- Learning JavaScript [邦訳](参考書籍)
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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