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在使用 Visual C++ Express 2010 处理 C++ 项目时，我发现了一个我想了解的有趣问题。问题是如果我在 Debug 或 Release 模式下编译，我的程序的结果会有所不同。我做了一个小程序来重现它：

#include <stdio.h>


int firstarg(int i)
{
    printf("First argument called with i = %d\n", i);

    return i;
}

int secondarg(int i)
{
    printf("Second argument called with i = %d\n", i);

    return i;
}

void function(int i, int j)
{
    printf("Function called with %d, %d\n", i,j);
}

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

    // Line with the problem!
    for (int i = 0; i < 5; ) function(firstarg(i), secondarg(i++)); 

    return 0;
}

// Result on RELEASE:
Second argument called with i = 0
First argument called with i = 0
Function called with 0, 0
Second argument called with i = 1
First argument called with i = 1
Function called with 1, 1
Second argument called with i = 2
First argument called with i = 2
Function called with 2, 2
Second argument called with i = 3
First argument called with i = 3
Function called with 3, 3
Second argument called with i = 4
First argument called with i = 4
Function called with 4, 4

// Result on DEBUG
Second argument called with i = 0
First argument called with i = 1
Function called with 1, 0
Second argument called with i = 1
First argument called with i = 2
Function called with 2, 1
Second argument called with i = 2
First argument called with i = 3
Function called with 3, 2
Second argument called with i = 3
First argument called with i = 4
Function called with 4, 3
Second argument called with i = 4
First argument called with i = 5
Function called with 5, 4

如您所见，在这两种情况下，第二个参数都在第一个参数之前进行评估（如果参数是在某种 LIFO 堆栈中处理的，那是我所期望的）；但是在 release 中，变量 i 的增量被“优化掉”并延迟到循环的下一次迭代。这是出乎意料的，我真的很想了解发生了什么。

当然，我可以通过将循环更改为轻松“修复”我的代码

    for (int i = 0; i < 5; ++i) function(firstarg(i+1), secondarg(i));

无论编译参数如何，它都会给出相同的结果。但是，我仍然很想了解这种增量优化背后的原因。

PS。顺便说一句，我无法在 linux 下使用 gcc 重现此问题（使用 -O0 标志进行调试，使用 -O3 进行发布）。

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3 回答 3

3

你误解了结果。增量不会“优化掉”或延迟到循环的下一次迭代。i在下一次迭代之前，您无法查看 的值是多少。试试这个：

for (int i = 0; i < 5; )
{
   function(firstarg(i), secondarg(i++)); 
   printf("At end of iteration, i = %d\n");
}

你会发现它根本没有延迟。

您的代码是 UB，因为您可以访问变量并修改同一变量，而无需插入序列点。虽然你可能会得到非常疯狂的结果，但实际上你会得到两个“预期”的结果，这取决于你的优化。未指定评估函数参数的顺序。

于 2012-09-03T09:51:12.950 回答
2

您的程序修改和读取变量i两次，没有中间的序列点（在函数调用表达式中function），这只是未定义的行为。

于 2012-09-03T09:51:48.143 回答
0

我认为您存在导致未定义行为的序列点问题：

无论如何，解决方案是不要编写这种代码。

于 2012-09-03T09:51:23.883 回答