使用预处理器取消std :: cout代码行 [英] Cancelling std::cout code lines using preprocessor

问题描述

可以使用 #define printf 删除对 printf（）的所有调用。如果我有很多调试打印如 std :: cout<< x < endl; ？如何使用预处理程序快速关闭单个文件中的 cout 语句？

One can remove all calls to printf() using #define printf. What if I have a lot of debug prints like std::cout << x << endl; ? How can I quickly switch off cout << statements in a single file using preprocessor?

推荐答案

如果你正在寻找快速删除调试语句的东西，NullStream可以是一个很好的解决方案。但是，我建议创建自己的调试类，当需要更多的调试功能时，可以根据需要进行扩展：

NullStream can be a good solution if you are looking for something quick that removes debug statements. However I would recommend creating your own class for debugging, that can be expanded as needed when more debug functionality is required:

class MyDebug
{
    std::ostream & stream;
  public:
    MyDebug(std::ostream & s) : stream(s) {}
#ifdef NDEBUG
    template<typename T>
    MyDebug & operator<<(T& item)
    {
      stream << item;
      return *this;
    }
#else
    template<typename T>
    MyDebug & operator<<(T&)
    {
      return *this;
    }
#endif
};

这是一个简单的设置，可以做你最初想要的，您添加了调试级别等功能。

This is a simple setup that can do what you want initially, plus it has the added benefit of letting you add functionality such as debug levels etc..

更新：
现在由于操纵器被实现为函数，如果你想接受操纵器（endl）你可以添加：

Update: Now since manipulators are implemented as functions, if you want to accept manipulators as well (endl) you can add:

MyDebug & operator<<(std::ostream & (*pf)(std::ostream&))
{
  stream << pf;
  return *this;
}

并且对于所有的机械手类型所有机械手类型）：

And for all manipulator types (So that you don't have to overload for all manipulator types):

template<typename R, typename P>
MyDebug & operator<<(R & (*pf)(P &))
{
  stream << pf;
  return *this;
}

注意这最后一个，因为它也接受常规函数指针。

Be careful with this last one, because that will also accept regular functions pointers.

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