我应该如何正确地将cout分配给静态ostream引用变量？

Question

我正在定义一个这样的类:

class StaticRuntimeContext {
 public:
  enum Verbosity {
    kHIGH,
    kMEDIUM,
    kLOW,
    kSILENT
  };
  static void Construct();
  static std::ostream& stdout1() {return stdout1_;}
  static std::ostream& stdout2() {return stdout2_;}
  static std::ostream& stdout3() {return stdout3_;}
  static std::ostream& stderr() {return stderr_;}
 protected:
 private:
  static std::ostream& stdout1_;
  static std::ostream& stdout2_;
  static std::ostream& stdout3_;
  static std::ostream& stderr_;
};

我将构造函数定义为:

void StaticRuntimeContext::Construct() {
  std::ostream& test = cout;
  stdout1_ = cout;
  stdout2_ = cout;
  //stdout3_ = NULL;
  stderr_ = cerr;
}

我无法理解为什么将cout分配给test(std :: ostream&)可以编译,但编译器会为其余部分生成错误消息,如"stdout1_ = cout".错误消息是:

/usr/lib/gcc/x86_64-redhat-linux/4.6.2/../../../../include/c++/4.6.2/bits/ios_base.h:791:5: error: ‘std::ios_base& std::ios_base::operator=(const std::ios_base&)’ is private

我想知道我应该怎么做才能正确地将cout分配给这些ostream引用变量.谢谢!

Answer 1

这是因为引用具有值语义,而operator =正在复制对象而不是分配新的引用.

您应该定义静态指针而不是引用,在Construct中分配它们,并在访问器中返回引用

  static std::ostream& stdout1() {return *stdout1_;}
  static std::ostream& stdout2() {return *stdout2_;}
  static std::ostream& stdout3() {return *stdout3_;}
  static std::ostream& stderr()  {return *stderr_;}
 protected:
 private:
  static std::ostream* stdout1_;
  static std::ostream* stdout2_;
  static std::ostream* stdout3_;
  static std::ostream* stderr_;

void StaticRuntimeContext::Construct() {
  stdout1_ = &cout;
  stdout2_ = &cout;
  stdout3_ = &cout;
  stderr_ = &cerr;
}

编辑:你必须在你的.cpp文件中添加它

std::ostream* StaticRuntimeContext::stdout1_ = NULL;
std::ostream* StaticRuntimeContext::stdout2_ = NULL;
std::ostream* StaticRuntimeContext::stdout3_ = NULL;
std::ostream* StaticRuntimeContext::stderr_ = NULL;