程序在Visual Studio 2012中运行,但不在ideone.com中运行

Question

我有一种直觉VS2012这个错了,但我不确定.

看完这个问题之后,我觉得要尝试实现类似的东西.

我的版本在Visual Studio 2012上运行良好,但甚至不在Ideone上编译.

这是我的主界面:

#include <iostream>
#include <string>

template <class In, class Out>
struct Pipe
{
    typedef In in_type ;
    typedef Out out_type ;

    In in_val ;

    Pipe (const in_type &in_val = in_type()) : in_val (in_val)
    {
    }

    virtual auto operator () () const -> out_type
    {
        return out_type () ;
    }
};

template <class In, class Out, class Out2>
auto operator>> (const Pipe <In, Out> &lhs, Pipe <Out, Out2> &rhs) -> Pipe <Out, Out2>&
{
    rhs = lhs () ;
    return rhs ;
}

template <class In, class Out>
auto operator>> (const Pipe <In, Out> &lhs, Out &rhs) -> Out&
{
    rhs = lhs () ;
    return rhs ;
}

以下是一些测试类:

struct StringToInt : public Pipe <std::string, int>
{
    StringToInt (const std::string &s = "") : Pipe <in_type, out_type> (s)
    {
    }

    auto operator () () const -> out_type
    {
        return std::stoi (in_val) ;
    }
};

struct IntSquare : public Pipe <int, int>
{
    IntSquare (int n = 0) : Pipe <in_type, out_type> (n)
    {
    }

    auto operator () () const -> out_type
    {
        return in_val * in_val ;
    }
};

struct DivideBy42F : public Pipe <int, float>
{
    DivideBy42F (int n = 0) : Pipe <in_type, out_type> (n)
    {
    }

    auto operator () () const -> out_type
    {
        return static_cast <float> (in_val) / 42.0f ;
    }
};

这是驱动程序:

int main ()
{
    float out = 0 ;
    StringToInt ("42") >> IntSquare () >> DivideBy42F () >> out ;
    std::cout << out << "\n" ;

    return 0 ;
}

Ideone抱怨模板扣除并且无法找到正确的operator>>候选函数:

prog.cpp: In function ‘int main()’:
prog.cpp:75:21: error: no match for ‘operator>>’ (operand types are ‘StringToInt’ and ‘IntSquare’)
  StringToInt ("42") >> IntSquare () >> DivideBy42F () >> out ;
                     ^
prog.cpp:75:21: note: candidates are:
prog.cpp:23:6: note: Pipe<Out, Out2>& operator>>(const Pipe<In, Out>&, Pipe<Out, Out2>&) [with In = std::basic_string<char>; Out = int; Out2 = int]
 auto operator>> (const Pipe <In, Out> &lhs, Pipe <Out, Out2> &rhs) -> Pipe <Out, Out2>&
      ^
prog.cpp:23:6: note:   no known conversion for argument 2 from ‘IntSquare’ to ‘Pipe<int, int>&’
prog.cpp:30:6: note: template<class In, class Out> Out& operator>>(const Pipe<In, Out>&, Out&)
 auto operator>> (const Pipe <In, Out> &lhs, Out &rhs) -> Out&
      ^
prog.cpp:30:6: note:   template argument deduction/substitution failed:
prog.cpp:75:35: note:   deduced conflicting types for parameter ‘Out’ (‘int’ and ‘IntSquare’)
  StringToInt ("42") >> IntSquare () >> DivideBy42F () >> out ;

哪个编译器正确？如果Ideone是正确的,这个代码是否有任何简单的解决方法？

Answer 1

Ideone（实际上是 GCC）在这里是正确的。在 Visual Studio 中，它会因为一个臭名昭著的扩展而进行编译，该扩展允许临时变量绑定到非常量左值引用（标准禁止这样做）。

我看到了在标准 C++ 中解决这个问题的几种可能方法：

一，不要在管道阶段使用临时变量：

int main ()
{
    float out = 0 ;
    StringToInt stage1("42");
    IntSquare stage2;
    DivideBy24F stage3;
    stage1 >> stage2 >> stage3 >> out ;
    std::cout << out << "\n" ;

    return 0 ;
}

二、创建一个“stay”函数（与 相反std::move），并使用它：

template <class T>
T& stay(T &&x) { return x; }

int main ()
{
    float out = 0 ;
    stay(StringToInt ("42")) >> stay(IntSquare ()) >> stay(DivideBy42F ()) >> out ;
    std::cout << out << "\n" ;

    return 0 ;
}

operator >>三、提供获取右值引用的重载：

template <class In, class Out, class Out2>
auto operator>> (const Pipe <In, Out> &&lhs, Pipe <Out, Out2> &&rhs) -> Pipe <Out, Out2>&
{
    return lhs >> rhs;  // Notice that lhs and rhs are lvalues!
}

当然，理想情况下您还可以提供混合&, &&和重载。&&, &