VC++ 6.0 中的调试与发布行为之谜

Question

对于下面的程序，根据运行在 Windows 上 VC++ 6.0 中处于 Debug 模式还是 Release 模式，我会得到不同的结果 7. Debug 的差异和 Release 行为几乎总是表明在处理指针和循环时存在错误，但我无法发现错误。

在调试模式下，我得到了预期的结果：

Entered loop with i == 0, RecordCountNew == 0
RecordCountNew = 1 is positive.
Entered loop with i == 1, RecordCountNew == 1
Adding record with i == 1, RecordCountNew == 1
Added record with i == 1, RecordCountNew == 2
RecordCountNew = 3 is positive.
Entered loop with i == 2, RecordCountNew == 3
RecordCountNew = 4 is positive.
Finished loop with i == 3, RecordCountNew == 4

在发布模式下，我得到相同的结果，除了 RecordCountNew 为正的断言：

Entered loop with i == 0, RecordCountNew == 0
RecordCountNew = 1 is positive.
Entered loop with i == 1, RecordCountNew == 1
Adding record with i == 1, RecordCountNew == 1
Added record with i == 1, RecordCountNew == 2
RecordCountNew = 3 is positive.
Entered loop with i == 2, RecordCountNew == 3
Finished loop with i == 3, RecordCountNew == 4

任何人都可以在他们的机器上复制这个，或者更好的是，解释一下吗？

#include <stdio.h>
#include <algorithm>

using namespace std;

struct record {
    int ID;
};

int RecordLimit;
record* Records = NULL;
record** RecordIndex = NULL;
record** RecordIndexNew = NULL;

int main(int argc, char* argv[]) {

    RecordLimit = 10;
    Records = new (nothrow) record[RecordLimit];
    RecordIndex = new (nothrow) record*[RecordLimit];
    RecordIndexNew = new (nothrow) record*[RecordLimit];

    int i;
    for (i = 0; i < RecordLimit; i++) {
        RecordIndex[i] = NULL;
        RecordIndexNew[i] = NULL;
    }

    int RecordCount = 0;
    for (i = 0; i < 3; i++) {
        Records[i].ID = i;
        RecordCount++;
    }

    int RecordCountNew = 0;
    for (i = 0; i < RecordCount; i++) {

        printf("Entered loop with i == %d, RecordCountNew == %d\n", i, RecordCountNew);

        RecordIndexNew[RecordCountNew] = RecordIndex[i];

        bool AddNewRecord = (i == 1);

        if (AddNewRecord) {
            printf("Adding record with i == %d, RecordCountNew == %d\n", i, RecordCountNew);
            Records[RecordCount + (RecordCountNew - i)].ID = RecordCount + (RecordCountNew - i);
            RecordIndexNew[RecordCountNew + 1] = RecordIndexNew[RecordCountNew];
            RecordIndexNew[RecordCountNew] = &Records[RecordCount + (RecordCountNew - i)];
            RecordCountNew++;
            printf("Added record with i == %d, RecordCountNew == %d\n", i, RecordCountNew);
        }

        RecordCountNew++;
        if (RecordCountNew > 0) printf("RecordCountNew == %d is positive.\n", RecordCountNew);
    }

    printf("Finished loop with i == %d, RecordCountNew == %d\n", i, RecordCountNew);

    delete[] Records;
    delete[] RecordIndex;
    delete[] RecordIndexNew;

    return 0;
}

Answer 1

(Correction replacing prior comment): Similar results in VC6++ SP6, but I get no "is positive" output at all. I'm going to take a look. We'll see if I can find anything. No promises ( Euro Micelli )

我已经复制了相同的结果（发布时根本没有输出）@EuroMicelli 找到了。但是，如果您将 RecordCountNew 声明为易变的，则输出会出现：

volatile int RecordCountNew = 0;

供您参考，volatile 是一个关键字，它告诉编译器变量可以在外部随机修改（例如在 CPU 中断期间）并防止编译器积极优化周围的代码.

tldr：MSVC6 错误地优化了 RecordCountNew。

PS ：将 RecordCountNew 声明为 short 而不是 int 使打印输出重新出现。你永远不知道 20 岁编译器的大脑里在想什么。

PPS : 因为我被要求解释错误，这里是正确输出的反汇编版本 :

edi 寄存器存储 RecordCountNew 值，test 指令命令跳转到 printf。但是，这是 OP 的编译版本：

test条件是在基指针寄存器ebp上完成的，与RecordCountNew无关。根据 ebp 的值，程序每次都输出该行，或者从不输出。

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