c++ - Can you pass a manipulator to a function?

Question

I'd like to pass a list of manipulators to a function, something like this:

void print(const vector<std::smanip>& manips) {
  // ...
  for (auto m : manips)
    cout << m;
  // ...
}

which would ideally be called by code something like this:

some_object.print({std::fixed, std::setprecision(2)}));

g++ 4.7.0 says:

error: ‘std::smanip’ has not been declared

Apparently, smanip isn't really defined in the standard, and C++11 compilers don't need to provide an explicit name for the type of manipulators. I tried declaring a type by leeching off of a known manipulator, like this:

typedef decltype(std::fixed) manip;

This opened up a host of new error messages, including this one:

error: ‘const _Tp* __gnu_cxx::new_allocator< <template-parameter-1-1> 
>::address(__gnu_cxx::new_allocator< <template-parameter-1-1> >::const_reference)
const [with _Tp = std::ios_base&(std::ios_base&); __gnu_cxx::new_allocator<
<template-parameter-1-1> >::const_pointer = std::ios_base& (*)(std::ios_base&);
__gnu_cxx::new_allocator< <template-parameter-1-1> >::const_reference =
std::ios_base& (&)(std::ios_base&)]’ cannot be overloaded

Should I just give up now, or is there a way to do this?

score 7 · Accepted Answer

输出操纵器只是为某些实例os << m定义的任何类型。basic_ostream操纵器可以是一个函数（受的operator<<重载basic_ostream），但它也可以是定义自己的任何类型operator<<。因此，我们需要执行类型擦除以捕获operator<<适当的basic_ostream实例化；最简单的方法是使用std::functionlambda：

#include <iostream>
#include <iomanip>
#include <functional>
#include <vector>

template<typename S>
struct out_manipulator: public std::function<S &(S &)> {
   template<typename T> out_manipulator(T &&t): std::function<S &(S &)>(
      [=](S &i) -> S &{ return i << t; }) {}
   template<typename T> out_manipulator(T *t): std::function<S &(S &)>(
      [=](S &i) -> S &{ return i << t; }) {}    // for g++
   template<typename U> friend U &operator<<(U &u, out_manipulator &a) {
      return static_cast<U &>(a(u));
   }
};

void print(const std::vector<out_manipulator<std::ostream>> &manips) {
   for (auto m: manips)
      std::cout << m;
}

int main() {
   print({std::fixed, std::setprecision(2)});
   std::cout << 3.14159;
}

score 6 · Accepted Answer

您的操纵器可以具有几乎任意类型，因此您必须使用模板来处理它们。为了使用固定类型的指针或引用访问它们，您必须为所有这些模板使用公共基类。这种多态性仅适用于指针和引用，但您可能需要值语义，特别是用于将这些存储在包含中。所以最简单的方法是让一个shared_ptr处理内存管理，并使用另一个类向用户隐藏所有丑陋的细节。

结果可能如下所示：

#include <memory>
#include <iostream>

// an abstract class to provide a common interface to all manipulators
class abstract_manip {
public:
  virtual ~abstract_manip() { }
  virtual void apply(std::ostream& out) const = 0;
};

// a wrapper template to let arbitrary manipulators follow that interface
template<typename M> class concrete_manip : public abstract_manip {
public:
  concrete_manip(const M& manip) : _manip(manip) { }
  void apply(std::ostream& out) const { out << _manip; }
private:
  M _manip;
};

// a class to hide the memory management required for polymorphism
class smanip {
public:
  template<typename M> smanip(const M& manip)
    : _manip(new concrete_manip<M>(manip)) { }
  template<typename R, typename A> smanip(R (&manip)(A))
    : _manip(new concrete_manip<R (*)(A)>(&manip)) { }
  void apply(std::ostream& out) const { _manip->apply(out); }
private:
  std::shared_ptr<abstract_manip> _manip;
};

inline std::ostream& operator<<(std::ostream& out, const smanip& manip) {
  manip.apply(out);
  return out;
}

这样，您的代码在对命名空间进行一些细微更改后即可工作：

void print(const std::vector<smanip>& manips) {
  for (auto m : manips)
    std::cout << m;
}

int main(int argc, const char** argv) {
  print({std::fixed, std::setprecision(2)});
}

score 1 · Accepted Answer

由于操纵器是函数，因此取决于它们的签名。这意味着，您可以创建具有相同签名的操纵器向量。

例如：

#include <iomanip>
#include <vector>
#include <iostream>

typedef std::ios_base& (*manipF)( std::ios_base& );

std::vector< manipF > getManipulators()
{
    std::vector< manipF > m =
    {
        std::showpos,
        std::boolalpha
    };
    return m;
}

int main()
{
  auto m = getManipulators();

  for ( auto& it : m )
  {
    std::cout<<it;
  }
  std::cout<<"hi " << true << 5.55555f << std::endl;
}

另一种方法是使用 lambdas ：

#include <iomanip>
#include <vector>
#include <iostream>
#include <functional>

typedef std::function< std::ios_base& ( std::ios_base& ) > manipF;

std::vector< manipF > getManipulators()
{
    std::vector< manipF > m =
    {
        std::showpos,
        std::boolalpha,
        [] ( std::ios_base& )->std::ios_base&
        {
          std::cout << std::setprecision( 2 );
          return std::cout;
        }
    };
    return m;
}

int main()
{
  auto m = getManipulators();

  for ( auto& it : m )
  {
    it(std::cout);
  }
  std::cout<<"hi " << true << 5.55555f << std::endl;
}

score 0 · Accepted Answer

标准 C++17 解决方案，可能基于 std::tuple。这是 POC

int main()
{
// quick p.o.c.
auto ios_mask_keeper = [&](auto mask) {
    // keep tuple here
    static auto mask_ = mask;
    return mask_;
};

// make required ios mask and keep it
auto the_tuple = ios_mask_keeper(
    // please note we can mix ios formaters and iomanip-ulators
    std::make_tuple(std::boolalpha, std::fixed, std::setprecision(2) ) 
);

// apply iomanip's stored in a tuple
std::apply([&](auto & ...x) 
{
    // c++17 fold 
    (std::cout << ... << x);
}, the_tuple);

return 0;

}

可以将元组保留在一个类中，等等。在 MSVC CL 编译器版本 19.14.26431.0 和使用 clang的强制性 Wand Box中都可以使用。

c++ - Can you pass a manipulator to a function?

4 回答 4

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