Question

只是测试/学习在C ++中编写管道，使|运算符重载，以下程序无法编译：

忽略二进制表达式候选模板的无效操作数： ... 无法匹配＆＃39; bitset＆＃39;反对＆＃39; vector＆＃39;

似乎编译器正在尝试使用标准|定义。该代码使用显式调用和显式类型参数设置。

// g++ -std=c++11 Pipeline.cpp 
#include <iostream>
#include <vector>

// ..............................................................
// ..............................................................
std::string dup (std::string s) {
  return s + s;
}

// ..............................................................
// ..............................................................
template <typename TI, typename TO>
std::vector<TO> operator | (const std::vector<TI> & in, std::function<TO(TI)> f) {
  std::vector<TO> out;
  for (auto i : in) {
    out.push_back ( f(i) );
  }
  return out;
}

// ..............................................................
// ..............................................................
int main () {
  std::cout << " hello " << std::endl;

  std::vector<std::string> vs = {"one", "two", "three", "four", "five"};

  auto res = vs | dup;
  // OK: vector<string> res = operator|<string,string> (vs, dup);

  for (auto s : res) { std::cout << s << std::endl; }

} // ()

完整的错误消息：

Pipeline.cpp:29:17: error: invalid operands to binary expression
      ('std::vector<std::string>' and 'std::string (*)(std::string)')
  auto res = vs | dup;
             ~~ ^ ~~~
/Library/Developer/CommandLineTools/usr/bin/../include/c++/v1/bitset:1045:1: note: 
      candidate template ignored: could not match 'bitset' against 'vector'
operator|(const bitset<_Size>& __x, const bitset<_Size>& __y) _NOEXCEPT
^
Pipeline.cpp:13:17: note: candidate template ignored: could not match
      'function<type-parameter-0-1 (type-parameter-0-0)>' against
      'std::__1::basic_string<char> (*)(std::__1::basic_string<char>)'
std::vector<TO> operator | (const std::vector<TI> & in, std::function<TO...
                ^
1 error generated.

Answer 1

模板参数推导不会查看隐式转换。函数指针不是std::function，因此编译器无法推断出用operator|实例化的模板参数。通常，std::function<...>，其中...包含模板参数，不是接受任意函数或函数对象的正确方法。

相反，接受仿函数的任何类型：

template <typename TI, typename F>
/* something */ operator | (const std::vector<TI> & in, F f)

然后找出返回类型。

返回类型是std::vector，其值类型是通过在f元素上调用in返回的类型 - 换句话说，decltype(f(in[0]))。但是，f可以返回引用类型，并且您确实希望在这些情况下向量的值类型是引用的类型。使用std::decay（或std::remove_reference）去除任何参考资料：

template <typename TI, typename F>
auto operator | (const std::vector<TI> & in, F f) -> std::vector<typename std::decay<decltype(f(in[0]))>::type> {
    std::vector<typename std::decay<decltype(f(in[0]))>::type> out;
    /*...*/
}

Demo

Answer 2

T.C。

的类似解决方案

template <typename CONT, typename F>  // types: container , function
auto operator | (const CONT & a, F f) {
  // find out input type and output type
  using TI = typename CONT::value_type;
  TI aux;
  using TO = decltype(f(aux));
  // do the task
  vector<TO> res;
  for (auto & i : a) {
    res.push_back( f(i) );
  }
  return res;
}

修改完成＆＃34;实验＆＃34;码。将运营商更改为因此，==：转换，!=过滤器和>>=缩减/访问避免在像vs == dup != lengthy != contains == length >>= add;这样的表达式中需要括号（我们需要变换和文件管理器具有相同的优先级，并且reduce更少）。不知道在使用它们的其他表达方式中是否会产生附带问题。

// g++ -std=c++1y Pipeline.cpp #include <iostream> #include <vector> #include <algorithm> using namespace std; // .............................................................. // transform // .............................................................. template <typename CONT, typename F> auto operator == (const CONT & a, F f) { using IT = typename CONT::value_type; IT aux; using OT = decltype(f(aux)); vector<OT> res; for (auto & i : a) { res.push_back( f(i) ); } return res; } // .............................................................. // filter // .............................................................. template <typename CONT, typename F> auto operator != (const CONT & a, F f) { using IT = typename CONT::value_type; IT aux; vector<IT> res; for (auto & i : a) { if (f(i)) res.push_back( i ); } return res; } // .............................................................. // reduce / visit // .............................................................. template <typename CONT, typename F> auto operator >>= (const CONT & a, F f) { for (auto & i : a) { f(i); } return f; } // .............................................................. // .............................................................. int main () { vector<string> vs = {"one", "two", "three", "four", "five"}; int sum = 0; auto dup = [] (const string &s) -> string { return s+s; }; auto lengthy = [] (const string &s) -> bool { return s.length()>6; }; auto contains = [] (const string &s) -> bool { return s.find('f') != string::npos; }; auto length = [] (const string &s) { return s.length(); }; auto add = [&sum] (int i) { sum += i;}; // == transform // != filter // >>= reduce / visit vs == dup != lengthy != contains == length >>= add; cout << sum << endl; auto res1 = vs == dup != lengthy != contains; for (auto & s : res1) { cout << s << endl; } } // ()

重载运算符中的类型不匹配（写入管道）

2 个答案: