您可以使用ctime重现或解释此Visual C ++错误吗? [英] Can you reproduce or explain this Visual C++ bug with ctime?

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.com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.com, 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 <的调用之前./code>.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.

这篇关于您可以使用ctime重现或解释此Visual C ++错误吗?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！