您可以使用ctime重现或解释此Visual C ++错误吗?

本文介绍了您可以使用ctime重现或解释此Visual C ++错误吗?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

此代码示例将输出 time:0 ，而不管 N 的值如何>在发布模式下使用Visual Studio Professional 2013 Update 3进行编译时，可同时使用32位和64位选项:

This code example will output time: 0 regardless of the value of N when compiled with Visual Studio Professional 2013 Update 3 in release mode, both 32 and 64-bit option:

#include <iostream> #include <functional> #include <ctime> using namespace std; void bar(int i, int& x, int& y) {x = i%13; y = i%23;} int g(int N = 1E9) { int x, y; int r = 0; for (int i = 1; i <= N; ++i) { bar(i, x, y); r += x+y; } return r; } int main() { auto t0 = clock(); auto r = g(); auto t1 = clock(); cout << r << " time: " << t1-t0 << endl; return 0; }

在rextester上使用gcc，clang和其他版本的vc ++进行测试时，它的行为正确，并输出大于零的 time .任何线索，这是怎么回事?

When tested with gcc, clang and other version of vc++ on rextester, it behaves correctly and outputs time greater than zero. Any clues what is going on here?

我注意到，内联 g()函数可以恢复正确的行为，但是更改 t0 ， r 和 t1 不会.

I noticed that inlining the g() function restores correct behaviour, but changing declaration and initialization order of t0, r and t1 does not.

推荐答案

如果使用调试器查看反汇编窗口，则可以看到生成的代码.对于处于发布模式的VS2012 Express，您会得到以下信息:

If you look at the disassembly winddow using the debugger you can see the generated code. For VS2012 express in release mode you get this:

00AF1310 push edi auto t0 = clock(); 00AF1311 call dword ptr ds:[0AF30E0h] 00AF1317 mov edi,eax auto r = g(); auto t1 = clock(); 00AF1319 call dword ptr ds:[0AF30E0h] cout << r << " time: " << t1-t0 << endl; 00AF131F push dword ptr ds:[0AF3040h] 00AF1325 sub eax,edi 00AF1327 push eax 00AF1328 call g (0AF1270h) 00AF132D mov ecx,dword ptr ds:[0AF3058h] 00AF1333 push eax 00AF1334 call dword ptr ds:[0AF3030h] 00AF133A mov ecx,eax 00AF133C call std::operator<<<std::char_traits<char> > (0AF17F0h) 00AF1341 mov ecx,eax 00AF1343 call dword ptr ds:[0AF302Ch] 00AF1349 mov ecx,eax 00AF134B call dword ptr ds:[0AF3034h]

在汇编的前4行中，您可以看到对 clock 的两次调用( ds:[0AF30E0h] )发生在对 g .因此，在这种情况下， g 花费多长时间都没有关系，结果将仅显示两次连续调用之间的时间.

from the first 4 lines of assembly you can see that the two calls to clock (ds:[0AF30E0h]) happen before the call to g. So in this case it doesn't matter how long g takes, the result will only show the time take between those two sequential calls.

似乎VS已确定 g 没有任何会影响 clock 的副作用，因此可以安全地移动呼叫.

It seems VS has determined that g doesn't have any side effects that would affect clock so it is safe to move the calls around.

正如Michael Petch在注释中指出的那样，在 r 的声明中添加 volatile 将阻止编译器移动调用.

As Michael Petch points out in the comments, adding volatile to to the declaration of r will stop the compiler from moving the call.