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鬯ッ�ッ�ス�ッ�ス�ス�ス�ゥ鬮ッ譎「�ス�キ�ス�ス�ス�「�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�「鬯ッ�ョ�ス�ォ�ス�ス�ス�エ鬯ョ�ョ隲幢スカ�ス�ス�ス�」�ス�ス�ス�ス�ス�ス�ス�ソ�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ァ�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ュ鬯ッ�ッ�ス�ゥ髫ー�ウ�ス�セ�ス�ス�ス�ス�ス�ス�ス�オ�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�コ�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ケ鬯ッ�ョ�ス�ォ�ス�ス�ス�エ鬮ッ�キ�ス�ソ鬯ョ�「�つ€�ス�ス�ス�セ髫エ莨夲スス�ヲ�ス�ス�ス�ソ�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�」�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ケ�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス鬯ッ�ッ�ス�ッ�ス�ス�ス�ゥ鬮ッ譎「�ス�キ�ス�ス�ス�「�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�「�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ァ�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ス�ク

【問題4】 符号付き2進数と符号なし2進数完全マスター! 電子回路ドリル II(4)

「high」と「low」のいわゆる“ビットデータ”で表現されるデジタル回路の世界で、負の数をどのように扱えばよいのでしょうか?

» 2008年02月07日 00時00分 公開
[横田一弘 埼玉県立新座総合技術高等学校 教諭,@IT MONOist]

【問題3】の解答

 前回の宿題【問題3】は、“負の数の2進表現”に関する問題でした。皆さん解けましたでしょうか?

 解けた方も解けなかった方も答え合わせをして、次項の解説までぜひ読んでみてください。毎週コツコツ問題を解いて、デジタル回路の基礎知識を身に付けましょう。

 それでは、解答を発表します!


問題3

「-5」を2進数で表すと? 8ビットの2の補数で表してください。

答え.(11111011)2


【問題3】の解説

 【問題3】は2進数で負の数をどのように取り扱うかという問題です。

 われわれが負の数を扱うとき、-5のように「-(マイナス記号)」を数の前に付けて表します。しかし、デジタル回路では、H(high)とL(low)のいわゆるビットデータしか扱うことができません。

 ここで負の数のビット表現を考えてみましょう。

 「ある負の数と対応する正の数の和は“ゼロ”」です。従って-5では、

5 + ( -5 ) = 0


という式が成り立ちます。

 これを基に、-5の2進数(ビット表現)を考えると、

( 101 )2 + ( x )2 = 0


となり、「x」の部分が-5に相当します。

 以上を踏まえたうえで、詳しい解説に入りますが、先に【問題3】の答えを明かします。

 答えは「(11111011)2」です。

 では、この答えで本当に「5 + ( -5 ) = 0」が成立するかどうかを確認してみましょう。

 図1の計算式を見ると、最上位にけた上がりが生じますが、続く8ビットはゼロになり「5 + ( -5 ) = 0」の計算が成り立ちます。

8ビットがゼロになることを確認 図1 8ビットがゼロになることを確認

 以上のような負の数の表現方法を「2の補数表現」といい、2の補数は2進数の加減算に適合することからコンピュータの世界でよく使われます。

 それではここで、2の補数の求め方について解説します。

 最初に対応する正の数を2進数に変換します。-5の場合、5を2進数に変換します。これを8ビットで表現すると、

5 = (00000101)2


となります。

 次に、各ビットの「0」を「1」に、「1」を「0」に反転させます。すると

(11111010)2


となります。

 最後に、これに「1」(つまり「(00000001)2」)を加えれば終了です(図2)。

2の補数の求め方 図2 2の補数の求め方

 図2から-5は「(11111011)2」と求められます。

次回までの宿題 ― 【問題4】

問題4

8ビットの2進数で表現できる数の最大値と最小値はいくつですか? また16ビット、32ビットではそれぞれいくつですか?


答え. 解答はこちら(←クリック)


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