是否有一个警告自由模板函数将基本类型转换为字符串 [英] Is there a warning free template function to convert basic types to string

问题描述

我想提供一个模板函数，将大多数基本类型转换为字符串。
我到目前为止最好的是以下：

I want to provide a templated function that converts most basic types to string. The best I have come up with so far is the following:

template<typename T> inline std::string anyToString(const T& var) {
  std::ostringstream o;
  o << var;
  return o.str();
}

用于以下操作：

 class TimeError:public std::runtime_error{
     public:
       explicit TimeError(int time):std::runtime_error(anyToString(time)),
       mTime(time){};
     protected:
       int mTime;
     };

anyToString和类似函数的问题是在使用gcc版本4.4.3编译时生成歧义警告-Wall -Wexta -Werror
ISO C ++说这些是不明确的，即使第一个的最坏转换比第二个的最坏转换更好

The problem with anyToString and similar functions is the generation of ambiguity warnings when compiling with gcc version 4.4.3 -Wall -Wexta -Werror "ISO C++ says that these are ambiguous, even though the worst conversion for the first is better than the worst conversion for the second"

据我所知，警告的原因在于调用<<时隐含的转换可能性。

To my knowledge the reason for the warning lies in the implicit conversion possibilities when calling <<.

这些模糊主要是由其他模板生成的，如下所示：

Those ambiguities are mainly generated by other templates as the following:

  template<typename T>
    T& operator<<(T& out, const SymRad& angle){
    return out << angle.deg();
  }

但是这些有其他优点，
所以我想保留他们。如果我把第二个模板变成一个简单的方法，例如。 ostream的歧义被清除，但我在寻找sth。允许保留两个模板。
是否有一个通用函数提供相同的简单性，而不使用描述的选项生成警告？
如果没有，那么本地禁用已发出的警告的最好方法是什么？

But those have other advantages like working for several stream types. So I would like to keep them. If I turn the second template into a plain method for e.g. ostream the ambiguity is cleaned, but I'm looking for sth. that allows keeping both templates. Is there a generic function that does provide the same simplicity without generating warnings using the described options ? If not, what is the best way to locally disable the issued warning ?

推荐答案

看起来你会从这样的场景得到这样的消息：

It seems you'd get such a message from a scenario like this:

#include <sstream>
#include <string>
#include <iostream>

struct Y {};
struct X
{
    operator Y() const {return Y(); }
};

std::ostream& operator<< (std::ostream& os, X) { return os << "X"; }
std::ostream& operator<< (std::ostringstream& os, Y) { return os << "Y"; }

template<typename T> inline std::string anyToString(const T& var) {
  std::ostringstream o;
  o << var;
  return o.str();
}

int main()
{
    std::cout << anyToString(X()) << '\n';
}

我建议使用 code> flag。 GCC编译它在所有感谢编译器扩展，与其他编译器这将是一个直接的错误。

I'd recommend using the -pedantic flag instead. GCC compiles it at all thanks to a compiler extension, with other compilers this will be a straight error.

添加：

  template<typename T>
    T& operator<<(T& out, const SymRad& angle){
    return out << angle.deg();
  }

但是这些有其他优点，流类型。

But those have other advantages like working for several stream types.

这实际上不适用于几种流类型。例如，如果 T 是 stringstream ，则 out<可能会返回一个 ostream 的引用，不能被隐式转换回 stringstream / code>参考。

This actually doesn't work for several stream types. E.g if T is stringstream, then out << angle.deg(); will likely return a reference to an ostream which cannot be implicitly downcasted back to a stringstream reference.

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