如何将std::vector的内容打印到屏幕上?


实现以下操作符<<的解决方案也很好:

template<container C, class T, String delim = ", ", String open = "[", String close = "]">
std::ostream & operator<<(std::ostream & o, const C<T> & x)
{
  // ... What can I write here?
}

以下是目前为止我所做的,没有单独的函数:

#include <iostream>
#include <fstream>
#include <string>
#include <cmath>
#include <vector>
#include <sstream>
#include <cstdio>
using namespace std;

int main()
{
    ifstream file("maze.txt");
    if (file) {
        vector<char> vec(istreambuf_iterator<char>(file), (istreambuf_iterator<char>()));
        vector<char> path;
        int x = 17;
        char entrance = vec.at(16);
        char firstsquare = vec.at(x);
        if (entrance == 'S') { 
            path.push_back(entrance); 
        }
        for (x = 17; isalpha(firstsquare); x++) {
            path.push_back(firstsquare);
        }
        for (int i = 0; i < path.size(); i++) {
            cout << path[i] << " ";
        }
        cout << endl;
        return 0;
    }
}

当前回答

这里是一个工作库,作为一个完整的工作程序,我刚刚把它组合在一起:

#include <set>
#include <vector>
#include <iostream>

#include <boost/utility/enable_if.hpp>

// Default delimiters
template <class C> struct Delims { static const char *delim[3]; };
template <class C> const char *Delims<C>::delim[3]={"[", ", ", "]"};
// Special delimiters for sets.                                                                                                             
template <typename T> struct Delims< std::set<T> > { static const char *delim[3]; };
template <typename T> const char *Delims< std::set<T> >::delim[3]={"{", ", ", "}"};

template <class C> struct IsContainer { enum { value = false }; };
template <typename T> struct IsContainer< std::vector<T> > { enum { value = true }; };
template <typename T> struct IsContainer< std::set<T>    > { enum { value = true }; };

template <class C>
typename boost::enable_if<IsContainer<C>, std::ostream&>::type
operator<<(std::ostream & o, const C & x)
{
  o << Delims<C>::delim[0];
  for (typename C::const_iterator i = x.begin(); i != x.end(); ++i)
    {
      if (i != x.begin()) o << Delims<C>::delim[1];
      o << *i;
    }
  o << Delims<C>::delim[2];
  return o;
}

template <typename T> struct IsChar { enum { value = false }; };
template <> struct IsChar<char> { enum { value = true }; };

template <typename T, int N>
typename boost::disable_if<IsChar<T>, std::ostream&>::type
operator<<(std::ostream & o, const T (&x)[N])
{
  o << "[";
  for (int i = 0; i != N; ++i)
    {
      if (i) o << ",";
      o << x[i];
    }
  o << "]";
  return o;
}

int main()
{
  std::vector<int> i;
  i.push_back(23);
  i.push_back(34);

  std::set<std::string> j;
  j.insert("hello");
  j.insert("world");

  double k[] = { 1.1, 2.2, M_PI, -1.0/123.0 };

  std::cout << i << "\n" << j << "\n" << k << "\n";
}

它目前只适用于vector和set,但通过扩展IsContainer专门化,可以使其适用于大多数容器。我没有过多地考虑这些代码是否最少,但我不能立即想到任何可以去掉的冗余代码。

编辑:只是为了好玩,我包含了一个处理数组的版本。我不得不排除字符数组,以避免进一步的歧义;使用wchar_t[]仍然会遇到麻烦。

其他回答

这已经被编辑了几次,我们决定调用包装集合的主类RangePrinter。

一旦您编写了一次性操作符<< overload,这将自动适用于任何集合,除非您将需要一个特殊的映射来打印对,并且可能希望在那里自定义分隔符。

您还可以在项目上使用特殊的“print”函数,而不是直接输出它,有点像STL算法允许您传入自定义谓词。对于map,您可以这样使用它,使用std::pair的自定义打印机。

您的“默认”打印机只会将其输出到流中。

好的,让我们用一台定制打印机。我将外层类改为RangePrinter。所以我们有两个迭代器和一些分隔符,但还没有定制如何打印实际的项。

struct DefaultPrinter
{
   template< typename T >
   std::ostream & operator()( std::ostream& os, const T& t ) const
   {
     return os << t;
   }

   // overload for std::pair
   template< typename K, typename V >
   std::ostream & operator()( std::ostream & os, std::pair<K,V> const& p)
   {
      return os << p.first << '=' << p.second;
   }
};

// some prototypes
template< typename FwdIter, typename Printer > class RangePrinter;

template< typename FwdIter, typename Printer > 
  std::ostream & operator<<( std::ostream &, 
        RangePrinter<FwdIter, Printer> const& );

template< typename FwdIter, typename Printer=DefaultPrinter >
class RangePrinter
{
    FwdIter begin;
    FwdIter end;
    std::string delim;
    std::string open;
    std::string close;
    Printer printer;

    friend std::ostream& operator<< <>( std::ostream&, 
         RangePrinter<FwdIter,Printer> const& );

public:
    RangePrinter( FwdIter b, FwdIter e, Printer p,
         std::string const& d, std::string const & o, std::string const& c )
      : begin( b ), end( e ), printer( p ), open( o ), close( c )
    {
    } 

     // with no "printer" variable
    RangePrinter( FwdIter b, FwdIter e,
         std::string const& d, std::string const & o, std::string const& c )
      : begin( b ), end( e ), open( o ), close( c )
    {
    } 

};


template<typename FwdIter, typename Printer>
std::ostream& operator<<( std::ostream& os, 
          RangePrinter<FwdIter, Printer> const& range )
{
    const Printer & printer = range.printer;

    os << range.open;
    FwdIter begin = range.begin, end = range.end;

    // print the first item
    if (begin == end) 
    { 
      return os << range.close; 
    }

    printer( os, *begin );

    // print the rest with delim as a prefix
    for( ++begin; begin != end; ++begin )
    {
       os << range.delim;
       printer( os, *begin );
    }
    return os << range.close;
}

现在,默认情况下,只要键和值类型都是可打印的,并且当它们都不是可打印的(就像任何其他类型一样),或者如果您不想要“=”作为分隔符,它就可以用于映射,并且您可以使用自己的特殊项目打印机。

现在,我将自由函数移到末尾来创建这些:

free-function(迭代器版本)看起来像这样,你甚至可以有默认值:

template<typename Collection>
RangePrinter<typename Collection::const_iterator> rangePrinter
    ( const Collection& coll, const char * delim=",", 
       const char * open="[", const char * close="]")
{
   return RangePrinter< typename Collection::const_iterator >
     ( coll.begin(), coll.end(), delim, open, close );
}

然后你可以将它用于std::set by

 std::cout << outputFormatter( mySet );

您还可以编写带有自定义打印机和带有两个迭代器的自由函数版本。在任何情况下,它们都将为您解析模板参数,并且您将能够将它们作为临时参数传递。

这个答案是基于Zorawar的回答,但我不能在那里留下评论。

你可以使用cbegin和cend来使auto (c++ 11)/typedef版本为const

for (auto i = path.cbegin(); i != path.cend(); ++i)
    std::cout << *i << ' ';

问题可能在前面的循环中:

(x = 17; isalpha(firstsquare); x++)

这个循环将根本不运行(如果firstsquare是非字母),或者将永远运行(如果它是字母)。原因是firstsquare不会随着x的增加而改变。

模板收集:

应用std::cout <<和std::to_string

std::vector、std::array和std::tuple

由于在cpp中打印一个向量被证明是惊人的工作量(至少与这个任务的基本程度相比),并且作为再次跨越相同问题的一个步骤,当使用其他容器时,这里有一个更通用的解决方案…

模板收集内容

这个模板集合处理3种容器类型: Std::vector, Std::array和Std::tuple。 它为这些对象定义了std::to_string(),并可以通过std::cout << container;直接将它们打印出来。

此外,它还为std::string << container定义了<<运算符。 这样就可以以紧凑的方式构造包含这些容器类型的字符串。

From

std::string s1 = "s1: " + std::to_string(arr) + "; " + std::to_string(vec) + "; " + std::to_string(tup);

我们会讲到

std::string s2 = STR() << "s2: " << arr << "; " << vec << "; " << tup;

Code

您可以交互地测试这段代码:这里。

#include <iostream>
#include <string>
#include <tuple>
#include <vector>
#include <array>

namespace std
{   
    // declations: needed for std::to_string(std::vector<std::tuple<int, float>>)
    std::string to_string(std::string str);
    std::string to_string(const char *str);
    template<typename T, size_t N>
    std::string to_string(std::array<T, N> const& arr);
    template<typename T>
    std::string to_string(std::vector<T> const& vec);
    template<typename... Args>
    std::string to_string(const std::tuple<Args...>& tup);
    
    std::string to_string(std::string str)
    {
        return std::string(str);
    }
    std::string to_string(const char *str)
    {
        return std::string(str);
    }

    template<typename T, size_t N>
    std::string to_string(std::array<T, N> const& arr)
    {
        std::string s="{";
        for (std::size_t t = 0; t != N; ++t)
            s += std::to_string(arr[t]) + (t+1 < N ? ", ":"");
        return s + "}";
    }

    template<typename T>
    std::string to_string(std::vector<T> const& vec)
    {
        std::string s="[";
        for (std::size_t t = 0; t != vec.size(); ++t)
            s += std::to_string(vec[t]) + (t+1 < vec.size() ? ", ":"");
        return s + "]";
    }
    
    // to_string(tuple)
    // https://en.cppreference.com/w/cpp/utility/tuple/operator%3D
    template<class Tuple, std::size_t N>
    struct TupleString
    {
        static std::string str(const Tuple& tup)
        {
            std::string out;
            out += TupleString<Tuple, N-1>::str(tup);
            out += ", ";
            out += std::to_string(std::get<N-1>(tup));
            return out;
        }
    };
    template<class Tuple>
    struct TupleString<Tuple, 1>
    {
        static std::string str(const Tuple& tup)
        {
            std::string out;
            out += std::to_string(std::get<0>(tup));
            return out;
        }
    };
    template<typename... Args>
    std::string to_string(const std::tuple<Args...>& tup)
    {
        std::string out = "(";
        out += TupleString<decltype(tup), sizeof...(Args)>::str(tup);
        out += ")";
        return out;
    }
} // namespace std


/**
 * cout: cout << continer
 */
template <typename T, std::size_t N> // cout << array
std::ostream& operator <<(std::ostream &out, std::array<T, N> &con)
{
    out <<  std::to_string(con);
    return out;
}
template <typename T, typename A> // cout << vector
std::ostream& operator <<(std::ostream &out, std::vector<T, A> &con)
{
    out <<  std::to_string(con);
    return out;
}
template<typename... Args> // cout << tuple
std::ostream& operator <<(std::ostream &out, std::tuple<Args...> &con)
{
    out <<  std::to_string(con);
    return out;
}

/**
 * Concatenate: string << continer
 */
template <class C>
std::string operator <<(std::string str, C &con)
{
    std::string out = str;
    out += std::to_string(con);
    return out;
}
#define STR() std::string("")

int main()
{
    std::array<int, 3> arr {1, 2, 3};
    std::string sArr = std::to_string(arr);
    std::cout << "std::array" << std::endl;
    std::cout << "\ttest to_string: " << sArr << std::endl;
    std::cout << "\ttest cout <<: " << arr << std::endl;
    std::cout << "\ttest string <<: " << (std::string() << arr) << std::endl;
    
    std::vector<std::string> vec {"a", "b"};
    std::string sVec = std::to_string(vec);
    std::cout << "std::vector" << std::endl;
    std::cout << "\ttest to_string: " << sVec << std::endl;
    std::cout << "\ttest cout <<: " << vec << std::endl;
    std::cout << "\ttest string <<: " << (std::string() << vec) << std::endl;
    
    std::tuple<int, std::string> tup = std::make_tuple(5, "five");
    std::string sTup = std::to_string(tup);
    std::cout << "std::tuple" << std::endl;
    std::cout << "\ttest to_string: " << sTup << std::endl;
    std::cout << "\ttest cout <<: " << tup << std::endl;
    std::cout << "\ttest string <<: " << (std::string() << tup) << std::endl;
    
    std::vector<std::tuple<int, float>> vt {std::make_tuple(1, .1), std::make_tuple(2, .2)};
    std::string sVt = std::to_string(vt);
    std::cout << "std::vector<std::tuple>" << std::endl;
    std::cout << "\ttest to_string: " << sVt << std::endl;
    std::cout << "\ttest cout <<: " << vt << std::endl;
    std::cout << "\ttest string <<: " << (std::string() << vt) << std::endl;
    
    std::cout << std::endl;
    
    std::string s1 = "s1: " + std::to_string(arr) + "; " + std::to_string(vec) + "; " + std::to_string(tup);
    std::cout << s1 << std::endl;
    
    std::string s2 = STR() << "s2: " << arr << "; " << vec << "; " << tup;
    std::cout << s2 << std::endl;

    return 0;
}

输出

std::array
    test to_string: {1, 2, 3}
    test cout <<: {1, 2, 3}
    test string <<: {1, 2, 3}
std::vector
    test to_string: [a, b]
    test cout <<: [a, b]
    test string <<: [a, b]
std::tuple
    test to_string: (5, five)
    test cout <<: (5, five)
    test string <<: (5, five)
std::vector<std::tuple>
    test to_string: [(1, 0.100000), (2, 0.200000)]
    test cout <<: [(1, 0.100000), (2, 0.200000)]
    test string <<: [(1, 0.100000), (2, 0.200000)]

s1: {1, 2, 3}; [a, b]; (5, five)
s2: {1, 2, 3}; [a, b]; (5, five)
template <typename T>
std::ostream& operator<<( std::ostream& ostrm, const std::vector<T>& vec ){
    ostrm << "[";
    for( int j = 0, n = vec.size(); j < n; ++j ){
        ostrm << " " << vec[ j ] << " ,"[ j < n - 1 ];
    }
    return ostrm << "]";
}

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