問82
頻出度★☆☆
2台のPCから一つのファイルを並行して更新した。ファイル中の同一データ（データ1）に対する処理が①～④の順に行われたとき，データ1はどの値になるか。ここで，データ1の初期値は5であった。
| PC−A 【処理内容】 | PC−B 【処理内容】 |
|---|---|
| ① データ1を読み込み，10を足す。 | ② データ1を読み込み，1を引く。 |
| ③ ①の結果を，データ1に書き込む。 | ④ ②の結果を，データ1に書き込む。 |
平成22年度 春期 ITパスポート試験 問82
テクノロジ系データベース › トランザクション処理
This question's explanation and your answers so far.
解説
問 82
- ア
正解 Follow the order (1) to (4) and track when each process reads and when it writes, and the answer falls out. The initial value is 5. In (1), PC-A reads Data 1, but the value in the file at that moment is 5, so A computes 5 + 10 = 15 on its own side. In (2), PC-B reads Data 1, but the write in (3) has not been done yet, so the value in the file is still 5, and B computes 5 - 1 = 4 on its own side. In (3), PC-A writes the result of (1), 15, to the file, so Data 1 becomes 15. In (4), PC-B writes the result of (2), 4, so the 15 that was just there is overwritten and disappears, and Data 1 finally becomes 4. What is happening here is a lost update: when the same data is updated concurrently without exclusive control, the update that was applied first (+10) is lost by the later write. Therefore the correct answer is ア (4).
Incorrectイ
5 is the initial value itself, the value before any update. Writes to the file are actually carried out in (3) and (4), so the value cannot stay at the initial value. In addition, +10 and −1 do not cancel each other out, so the reading that the two updates offset each other and bring the value back to 5 does not hold either.
Incorrectウ
14 is 5 + 10 - 1, the value that ought to result when the updates from both PC-A and PC-B are correctly applied. That is the outcome when exclusive control (locking) is in effect and (2) reads after the write in (3); in the order (1) to (2) to (3) to (4) given in this question, (2) reads the value 5 from before the write, so B's computed result is 4, not 14.
Incorrectエ
15 is the value right after PC-A writes in (3), but in the last step (4) PC-B writes its own computed result, 4, so the 15 is overwritten and does not remain. Note that what is being asked is the final value after step (4) has finished.
ここがポイント
- If several users update the same data concurrently without exclusive control, the write performed later overwrites the earlier update and that earlier update is lost. This is called a lost update. What each process uses in its calculation is the value as of the moment it read the data, and a value written by the other side afterwards is not reflected in it.
- The mechanism that prevents this is exclusive control (locking). If the series of operations from reading through writing is handled as a single transaction and the other side is made to wait, both updates are correctly applied, giving 5 + 10 - 1 = 14.
用語