相关疑难解决方法(0)

如果我使用以下main()方法运行我的C++应用程序,一切正常:

int main(int argc, char *argv[]) 
{
   cout << "There are " << argc << " arguments:" << endl;

   // Loop through each argument and print its number and value
   for (int i=0; i<argc; i++)
      cout << i << " " << argv[i] << endl;

   return 0;
}

我得到了我的期望,我的论点被打印出来了.

但是,如果我使用_tmain:

int _tmain(int argc, char *argv[]) 
{
   cout << "There are " << argc << " arguments:" << endl;

   // Loop through each argument and print its number and value
   for …