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In practice, why would different compilers compute different values of int x = ++i + ++i;? Ask Question

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In practice, why would different compilers compute different values of int x = ++i + ++i;? Ask Question

Consider this code:

int i = 1;
int x = ++i + ++i;

We have some guesses for what a compiler might do for this code, assuming it compiles.

  1. both ++i return 2, resulting in x=4.
  2. one ++i returns 2 and the other returns 3, resulting in x=5.
  3. both ++i return 3, resulting in x=6.

To me, the second seems most likely. One of the two ++ operators is executed with i = 1, the i is incremented, and the result 2 is returned. Then the second ++ operator is executed with i = 2, the i is incremented, and the result 3 is returned. Then 2 and 3 are added together to give 5.

However, I ran this code in Visual Studio, and the result was 6. I'm trying to understand compilers better, and I'm wondering what could possibly lead to a result of 6. My only guess is that the code could be executed with some "built-in" concurrency. The two ++ operators were called, each incremented i before the other returned, and then they both returned 3. This would contradict my understanding of the call stack, and would need to be explained away.

What (reasonable) things could a C++ compiler do that would lead to a result of 4 or a result or 6?

Note

This example appeared as an example of undefined behavior in Bjarne Stroustrup's Programming: Principles and Practice using C++ (C++ 14).

See cinnamon's comment.

ベストアンサー1

The compiler takes your code, splits it into very simple instructions, and then recombines and arranges them in a way that it thinks optimal.

The code

int i = 1;
int x = ++i + ++i;

consists of the following instructions:

1. store 1 in i
2. read i as tmp1
3. add 1 to tmp1
4. store tmp1 in i
5. read i as tmp2
6. read i as tmp3
7. add 1 to tmp3
8. store tmp3 in i
9. read i as tmp4
10. add tmp2 and tmp4, as tmp5
11. store tmp5 in x

But despite this being a numbered list the way I wrote it, there are only a few ordering dependenciesここで、1->2->3->4->5->10->11 と 1->6->7->8->9->10->11 は相対的な順序を維持する必要があります。それ以外は、コンパイラは自由に順序を変更でき、冗長性を排除できる可能性があります。

たとえば、リストを次のように順序付けることができます。

1. store 1 in i
2. read i as tmp1
6. read i as tmp3
3. add 1 to tmp1
7. add 1 to tmp3
4. store tmp1 in i
8. store tmp3 in i
5. read i as tmp2
9. read i as tmp4
10. add tmp2 and tmp4, as tmp5
11. store tmp5 in x

なぜコンパイラはこれを実行できるのでしょうか? それは、増分による副作用に順序がないためです。しかし、コンパイラは単純化できます。たとえば、4 には無効なストアがあり、値はすぐに上書きされます。また、tmp2 と tmp4 は実際には同じものです。

1. store 1 in i
2. read i as tmp1
6. read i as tmp3
3. add 1 to tmp1
7. add 1 to tmp3
8. store tmp3 in i
5. read i as tmp2
10. add tmp2 and tmp2, as tmp5
11. store tmp5 in x

これで、tmp1 に関連するものはすべてデッド コードになりました。つまり、使用されません。また、i の再読み込みも削除できます。

1. store 1 in i
6. read i as tmp3
7. add 1 to tmp3
8. store tmp3 in i
10. add tmp3 and tmp3, as tmp5
11. store tmp5 in x

ご覧のとおり、このコードはずっと短くなりました。オプティマイザーは満足です。プログラマーは満足していません。なぜなら、i は 1 回しか増分されていないからです。おっと。

代わりにコンパイラが実行できる別の方法を見てみましょう。元のバージョンに戻りましょう。

1. store 1 in i
2. read i as tmp1
3. add 1 to tmp1
4. store tmp1 in i
5. read i as tmp2
6. read i as tmp3
7. add 1 to tmp3
8. store tmp3 in i
9. read i as tmp4
10. add tmp2 and tmp4, as tmp5
11. store tmp5 in x

コンパイラは次のように順序を変更できます。

1. store 1 in i
2. read i as tmp1
3. add 1 to tmp1
4. store tmp1 in i
6. read i as tmp3
7. add 1 to tmp3
8. store tmp3 in i
5. read i as tmp2
9. read i as tmp4
10. add tmp2 and tmp4, as tmp5
11. store tmp5 in x

そして、i が 2 回読み取られていることに再度注意し、そのうちの 1 つを削除します。

1. store 1 in i
2. read i as tmp1
3. add 1 to tmp1
4. store tmp1 in i
6. read i as tmp3
7. add 1 to tmp3
8. store tmp3 in i
5. read i as tmp2
10. add tmp2 and tmp2, as tmp5
11. store tmp5 in x

それは素晴らしいですが、さらに進んで、tmp1 を再利用できます。

1. store 1 in i
2. read i as tmp1
3. add 1 to tmp1
4. store tmp1 in i
6. read i as tmp1
7. add 1 to tmp1
8. store tmp1 in i
5. read i as tmp2
10. add tmp2 and tmp2, as tmp5
11. store tmp5 in x

すると、6 の i の再読み込みが不要になります。

1. store 1 in i
2. read i as tmp1
3. add 1 to tmp1
4. store tmp1 in i
7. add 1 to tmp1
8. store tmp1 in i
5. read i as tmp2
10. add tmp2 and tmp2, as tmp5
11. store tmp5 in x

現在、4 はデッド ストアです。

1. store 1 in i
2. read i as tmp1
3. add 1 to tmp1
7. add 1 to tmp1
8. store tmp1 in i
5. read i as tmp2
10. add tmp2 and tmp2, as tmp5
11. store tmp5 in x

そして、3 と 7 を 1 つの命令に結合できるようになりました。

1. store 1 in i
2. read i as tmp1
3+7. add 2 to tmp1
8. store tmp1 in i
5. read i as tmp2
10. add tmp2 and tmp2, as tmp5
11. store tmp5 in x

最後の一時的なものを削除します。

1. store 1 in i
2. read i as tmp1
3+7. add 2 to tmp1
8. store tmp1 in i
10. add tmp1 and tmp1, as tmp5
11. store tmp5 in x

そして、Visual C++ が提供する結果が得られます。

どちらの最適化パスでも、何もしない命令が削除されない限り、重要な順序の依存関係が保持されていることに注意してください。

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