如何编写允许表达语法的cout函数包装器？

问题描述

我想包装 std :: cout 进行格式化，例如：

I'd like to wrap std::cout for formatting, like so:

mycout([what type?] x, [optional args]) {
    ... // do some formatting on x first
    std::cout << x;
}

仍然可以使用像

mycout("test" << i << endl << somevar, indent)

而不是被强迫变得更像

mycout(std::stringstream("test") << i ...)

我该如何实施？哪种类型制作 x ？

How can I implement this? What type to make x?

编辑：增加了可选参数的考虑

推荐答案

如何解决：

struct MyCout {};

extern MyCout myCout;

template <typename T>
MyCout& operator<< (MyCout &s, const T &x) {
  //format x as you please
  std::cout << x;
  return s;
}

并放入 MyCout myCout; 到任何一个.cpp文件中。

And put MyCout myCout; into any one .cpp file.

然后您可以像这样使用 myCout ：

You can then use myCout like this:

myCout << "test" << x << std::endl;

它将调用模板 operator<< 可以进行格式化。

And it will call the template operator<< which can do the formatting.

当然，如果愿意，还可以为特定类型的特殊格式提供运算符的重载。

Of course, you can also provide overloads of the operator for special formatting of specific types if you want to.

编辑

显然（感谢@soon），对于标准操纵器来说，它还有更多功能超载是必要的：

Apparently (thanks to @soon), for standard manipulators to work, a few more overloads are necessary:

MyCout& operator<< (MyCout &s, std::ostream& (*f)(std::ostream &)) {
  f(std::cout);
  return s;
}

MyCout& operator<< (MyCout &s, std::ostream& (*f)(std::ios &)) {
  f(std::cout);
  return s;
}

MyCout& operator<< (MyCout &s, std::ostream& (*f)(std::ios_base &)) {
  f(std::cout);
  return s;
}

---

编辑2

我可能会误解了您的原始要求。怎么样（加上上面相同的操纵器重载）：

I may have misunderstoor your original requirements slightly. How about this (plus the same manipulator overloads as above):

struct MyCout
{
  std::stringstream s;

  template <typename T>
  MyCout& operator << (const T &x) {
    s << x;
    return *this;
  }

  ~MyCout() {
    somehow_format(s);
    std::cout << s.str();
  }
};

int main() {
  double y = 1.5;
  MyCout() << "test" << y;
}

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