确保参数是控制台的输出流 [英] Ensure argument is an output stream for the console

问题描述

我试图做一个流操纵器的颜色使用输出到控制台。它的工作原理，改变文本和背景的颜色：

I'm trying to make a stream manipulator for colour for use with output to the console. It works, changing the colour of text and the background:

std::cout << ConColor::Color::FgBlue << 123 << "abc"; //text is blue, sticky

问题是签名：

std::ostream &FgBlue(std::ostream &);

此签名允许派生类，例如 std :: ostringstream ，但是没有办法改变字符串流的颜色。该函数将改变控制台的颜色，不管它是否使用这样的参数调用。

This signature allows for derived classes, such as std::ostringstream as well, but there is no way to change the colour of a string stream. The function would change the colour of the console regardless if it was called with such an argument.

因此，我想确保参数是 std :: cout ， std :: wcout 等。我更喜欢一般在更多 std :: ostream 对象添加到未来标准。

Therefore, I want to ensure the argument is something along the lines of std::cout, std::wcout, etc. I would prefer it be general in the case that more std::ostream objects are added in a future standard.

我尝试了许多涉及 std :: is_same 和 std :: is_base_of ，当前者不工作，只是最终意识到这是无意义的，因为任何参数类型继承自 std :: basic_ostream<> 将被转换为我传递给函数时比较的类型，给出假阳性。

I tried many things involving std::is_same and std::is_base_of, when the former wouldn't work, just to eventually realize that it was pointless because any argument type inheriting from std::basic_ostream<> will be casted to the type I'm comparing against when passed to the function, giving false positives.

这最终导致我的回答如下（variadic模板模板参数但是，还有一些问题：

This eventually led me to my answer below (variadic template template arguments? Wow, that's a mouthful!) There are a couple problems, however:

• 编译器必须支持可变参数模板。


• 如果派生类拥有不同数量的模板参数（例如 std：），编译器会产生隐藏错误。 ：ostringstream ，它有3而不是2），因为它没有通过函数签名。


• 可以重定向stdout，到文件，因此即使参数 std :: cout ，也会发生与stringstream情况相同的情况。

• The compiler must support variadic templates. I would prefer the solution work on MSVC.
• The compiler gives cryptic errors in the case that a derived class with a different number of template arguments (such as std::ostringstream, which has 3 instead of 2) is used, as it doesn't get past the function signature.
• It's possible to redirect stdout, say, to a file, so even if the argument is std::cout, the same thing as the stringstream case happens.

我鼓励人们发布任何其他的解决方案，希望比我更好，真正希望至少VS11工作的东西。

I encourage people to post any other solutions, hopefully better than mine, and really hopefully something that works with at least VS11.

推荐答案

这里是检测 std :: basic_ostream 实例化的特征：

Here's a trait for detecting std::basic_ostream instantiations:

template<typename T> struct is_basic_ostream {
  template<typename U, typename V>
  static char (&impl(std::basic_ostream<U, V> *))[
    std::is_same<T, std::basic_ostream<U, V>>::value ? 2 : 1];
  static char impl(...);
  static constexpr bool value = sizeof(impl((T *)0)) == 2;
};

使用方法：

template<typename T>
void foo(T &) {
  static_assert(is_basic_ostream<T>::value,
    "Argument must be of type std::basic_ostream<T, U>.");
}

我们使用模板参数推导来推断模板参数） basic_ostream 基类（如果有）。作为更一般的解决方案，用单个可变参数替换 U 和 V 将允许写一个通用 is_instantiation_of trait在支持可变参数模板参数的编译器上。

We use template argument deduction to infer the template parameters on the (non-proper) basic_ostream base class, if any. As a more general solution, replacing U and V with a single variadic parameter would allow writing a generic is_instantiation_of trait on compilers that support variadic template parameters.

要检测stdout是否管道到一个文件（只能在运行时检测，当然）使用 isatty ;请参阅如何在cout上使用isatty（），或者我可以假设cout ==文件描述符1？

To detect whether stdout is piped to a file (which can only be detected at runtime, of course) use isatty; see how to use isatty() on cout, or can I assume that cout == file descriptor 1?

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