Question

在下面的示例代码中，它显示可以从第一个模板参数隐式创建boost :: tuple。因此，我无法编写<<运算符，因为它变得模棱两可。

我也不明白为什么ostringstream& << float也含糊不清。这没有任何隐含的结构。为什么这也会给出含糊不清的错误？

#include <iostream>
#include <boost/tuple/tuple.hpp>
#include <sstream>
#include <string>

using namespace std;

class Myclass
{
};

typedef boost::tuple<int,float,Myclass> Mytuple;

ostringstream& operator<<(ostringstream& os_, Mytuple tuple_)
{
  float f = tuple_.get<1>();
  //os_ << (int)tuple_.get<0>(); // Error because int is implicitly converted into Mytuple. WHYY?
  //os_ << tuple_.get<1>();      // No Clue Why this is ambiguous.
  //os_ << tuple_.get<2>();      // Error because no matching operator. Fine.
  return os_;
}

int main()
{
  Mytuple t1;
  t1 = 3;      // Working because int is implicitly converted into Mytuple!! WHY?
  //t1 = 3.0f; // Error because no matching constructor. Fine.
  return 0;
}

错误Mesasge：

tupleTest2.C：18：错误：ISO C ++说这些都是模棱两可的虽然第一次转化的最差转化率要好于最差转化率转换为第二个：

Answer 1

问题不在于元组，而在于您的运营商。这很好用：

ostream& operator<<(ostream& os_, Mytuple tuple_)
{
    os_ << tuple_.get<0>(); // Error because int is implicitly converted into Mytuple. WHYY?
    os_ << tuple_.get<1>();      // No Clue Why this is ambiguous.
    //os_ << tuple_.get<2>();      // Error because no matching operator. Fine.
    return os_;
}

问题是ostringstream从ostream继承operator<<，它具有此签名：ostringstream& operator<<(ostringstream& os_, Mytuple tuple_)是允许的。那么

ostream& operator<<(ostream& os, T t)

（使用c ++中的所有可用类型更改T，请参阅operator<< reference page

修改

这是一个简化的例子（没有元组）：

ostringstream& operator<<(ostringstream& os_, Mytuple tuple_) { const int i = tuple_.get<0>(); os_ << i; // error in this line return os_; }

现在错误：

dfg.cpp: In function ‘std::ostringstream& operator<<(std::ostringstream&, Mytuple)’: dfg.cpp:18: error: ISO C++ says that these are ambiguous, even though the worst conversion for the first is better than the worst conversion for the second: /usr/lib/gcc/i386-redhat-linux/4.3.0/../../../../include/c++/4.3.0/bits/ostream.tcc:111: note: candidate 1: std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(int) [with _CharT = char, _Traits = std::char_traits<char>] dfg.cpp:14: note: candidate 2: std::ostringstream& operator<<(std::ostringstream&, Mytuple)

以上错误消息说：无法在两个运算符operator<<(ostream&,...)和operator<<(ostringstream&,...). This also raises another question : why on earth do you need运算符＆lt;＆lt;（ostringstream＆amp;，...）`之间进行选择？

Answer 2

写作时

 os << tuple_.get<0>();

没有匹配这两个参数的函数。相反，编译器可以选择对任一参数

应用隐式转换

std::ostream << int

或

std::ostringstream << MyTuple

后者将发生在boost::tuple构造函数中，该构造函数可以获取任意数量的参数，直到元组元素的数量。（并且float失败，因为float可以转换为int。）

重载流运算符时，请使用基类作为左侧（ostream甚至basic_ostream<CharT, Traits>。

修改：您可以通过转换第一个参数来消除歧义歧义。

ostringstream& operator<<(ostringstream& os_, Mytuple tuple_)
{
  static_cast<std::ostream&>(os_) << tuple_.get<0>(); 
  static_cast<std::ostream&>(os_)  << tuple_.get<1>();      
  static_cast<std::ostream&>(os_)  << tuple_.get<2>();      // Error because no matching operator. Fine.
  return os_;
}

然而，使用ostringstream重载运算符仍然是一个坏主意，因为它不适用于运算符链接。

MyTuple a, b;
ostringstream ss;
ss << a << ' ' << b;

将调用：

1）ostringstream& operator<<(ostringstream& os_, Mytuple tuple_)

2）ostream& ostream::operator<<(char)

3） ostream& operator<<(ostream&&, boost::tuple<int,float,Myclass>

Answer 3

所有告诉您使用::std::ostream代替::std::ostringstream的人都是绝对正确的。你不应该那样使用::std::ostringstream。

但我对你的代码的主要负担是令人痛苦的缺乏普遍性。它只适用于一种特定的元组类型，而不是全部类型。

所以我在C ++ 0x中为operator <<写了一个::std::tuple，它适用于任何使用operator <<单独编写成员的元组。使用Boost的元组类型可以相对容易地翻译它。这是：

template < ::std::size_t fnum, typename tup_type>
void print_fields(::std::ostream &os, const tup_type &val)
{
   if (fnum < ::std::tuple_size<tup_type>::value) {
      ::std::cerr << "Fred " << fnum << '\n';
      os << ::std::get<fnum, tup_type>(val);
      if (::std::tuple_size<tup_type>::value > (fnum + 1)) {
         os << ", ";
      }
      print_fields<fnum + 1, tup_type>(os, val);
   }
}

template < ::std::size_t fnum, typename... Elements>
class field_printer;

template <typename... Elements>
class field_printer<0, Elements...> {
 public:
   typedef ::std::tuple<Elements...> tup_type;

   static void print_field(::std::ostream &os, const tup_type &val) {
   }
};

template < ::std::size_t fnum, typename... Elements>
class field_printer {
 public:
   typedef ::std::tuple<Elements...> tup_type;

   static void print_field(::std::ostream &os, const tup_type &val) {
      constexpr auto tupsize = ::std::tuple_size<tup_type>::value;
      os << ::std::get<tupsize - fnum, Elements...>(val);
      if (fnum > 1) {
         os << ", ";
      }
      field_printer<fnum - 1, Elements...>::print_field(os, val);
   }
};

template <class... Types>
::std::ostream &operator <<(::std::ostream &os, const ::std::tuple<Types...> &val)
{
   typedef ::std::tuple<Types...> tup_type;
   os << '(';
   field_printer< ::std::tuple_size<tup_type>::value, Types...>::print_field(os, val);
   return os << ')';
}

这会将元组打印为"(element1, element2, ...elementx)"。

如何为boost :: tuple写一个`＆lt;＆lt;`运算符？

3 个答案: