将 vector<int> 转换为字符串

绝对不如 Python 优雅，但没有什么比 C++ 中的 Python 更优雅了。

您可以使用 stringstream ...

#include <sstream>
//...

std::stringstream ss;
for(size_t i = 0; i < v.size(); ++i)
{
  if(i != 0)
    ss << ",";
  ss << v[i];
}
std::string s = ss.str();

您也可以改用 std::for_each。

---

使用 std::for_each 和 lambda 你可以做一些有趣的事情。

#include <iostream>
#include <sstream>

int main()
{
     int  array[] = {1,2,3,4};
     std::for_each(std::begin(array), std::end(array),
                   [&std::cout, sep=' '](int x) mutable {
                       out << sep << x; sep=',';
                   });
}

请参阅 this question 了解我编写的一个小类。这不会打印尾随逗号。此外，如果我们假设 C++14 将继续为我们提供基于范围的算法等价物，如下所示：

namespace std {
   // I am assuming something like this in the C++14 standard
   // I have no idea if this is correct but it should be trivial to write if it  does not appear.
   template<typename C, typename I>
   void copy(C const& container, I outputIter) {copy(begin(container), end(container), outputIter);}
}
using POI = PrefexOutputIterator;   
int main()
{
     int  array[] = {1,2,3,4};
     std::copy(array, POI(std::cout, ","));
  // ",".join(map(str,array))               // closer
}

---

您可以使用 std::accumulate。考虑以下示例

if (v.empty() 
    return std::string();
std::string s = std::accumulate(v.begin()+1, v.end(), std::to_string(v[0]),
                     [](const std::string& a, int b){
                           return a + ',' + std::to_string(b);
                     });

---

另一种选择是使用 std::copy 和 ostream_iterator 类：

#include <iterator>  // ostream_iterator
#include <sstream>   // ostringstream
#include <algorithm> // copy

std::ostringstream stream;
std::copy(array.begin(), array.end(), std::ostream_iterator<>(stream));
std::string s=stream.str();
s.erase(s.length()-1);

也不如 Python 好。为此，我创建了一个 join 函数：

template <class T, class A>
T join(const A &begin, const A &end, const T &t)
{
  T result;
  for (A it=begin;
       it!=end;
       it++)
  {
    if (!result.empty())
      result.append(t);
    result.append(*it);
  }
  return result;
}

然后像这样使用它：

std::string s=join(array.begin(), array.end(), std::string(","));

你可能会问我为什么要传入迭代器。好吧，实际上我想反转数组，所以我这样使用它：

std::string s=join(array.rbegin(), array.rend(), std::string(","));

理想情况下，我想模板化到可以推断 char 类型并使用字符串流的程度，但我还想不通。

---

使用 Boost 和 C++11 可以这样实现：

auto array = {1,2,3,4};
join(array | transformed(tostr), ",");

嗯，差不多。这是完整的示例：

#include <array>
#include <iostream>

#include <boost/algorithm/string/join.hpp>
#include <boost/range/adaptor/transformed.hpp>

int main() {
    using boost::algorithm::join;
    using boost::adaptors::transformed;
    auto tostr = static_cast<std::string(*)(int)>(std::to_string);

    auto array = {1,2,3,4};
    std::cout << join(array | transformed(tostr), ",") << std::endl;

    return 0;
}

归功于 Praetorian。

您可以像这样处理任何值类型：

template<class Container>
std::string join(Container const & container, std::string delimiter) {
  using boost::algorithm::join;
  using boost::adaptors::transformed;
  using value_type = typename Container::value_type;

  auto tostr = static_cast<std::string(*)(value_type)>(std::to_string);
  return join(container | transformed(tostr), delimiter);
};

---

这只是试图解决 1800 INFORMATION's remark 给出的第二个解决方案缺乏通用性的谜语，而不是试图回答这个问题：

template <class Str, class It>
Str join(It begin, const It end, const Str &sep)
{
  typedef typename Str::value_type     char_type;
  typedef typename Str::traits_type    traits_type;
  typedef typename Str::allocator_type allocator_type;
  typedef std::basic_ostringstream<char_type,traits_type,allocator_type>
                                       ostringstream_type;
  ostringstream_type result;

  if(begin!=end)
    result << *begin++;
  while(begin!=end) {
    result << sep;
    result << *begin++;
  }
  return result.str();
}

在我的机器上工作 (TM)。

---

很多模板/想法。我的没有那么通用或高效，但我也遇到了同样的问题，想把它作为简短而甜蜜的东西加入其中。它以最短的行数获胜...... :)

std::stringstream joinedValues;
for (auto value: array)
{
    joinedValues << value << ",";
}
//Strip off the trailing comma
std::string result = joinedValues.str().substr(0,joinedValues.str().size()-1);

---

string s;
for (auto i : v)
    s += (s.empty() ? "" : ",") + to_string(i);

---

如果您想做 std::cout << join(myVector, ",") << std::endl;，您可以执行以下操作：

template <typename C, typename T> class MyJoiner
{
    C &c;
    T &s;
    MyJoiner(C &&container, T&& sep) : c(std::forward<C>(container)), s(std::forward<T>(sep)) {}
public:
    template<typename C, typename T> friend std::ostream& operator<<(std::ostream &o, MyJoiner<C, T> const &mj);
    template<typename C, typename T> friend MyJoiner<C, T> join(C &&container, T&& sep);
};

template<typename C, typename T> std::ostream& operator<<(std::ostream &o, MyJoiner<C, T> const &mj)
{
    auto i = mj.c.begin();
    if (i != mj.c.end())
    {
        o << *i++;
        while (i != mj.c.end())
        {
            o << mj.s << *i++;
        }
    }

    return o;
}

template<typename C, typename T> MyJoiner<C, T> join(C &&container, T&& sep)
{
    return MyJoiner<C, T>(std::forward<C>(container), std::forward<T>(sep));
}

请注意，此解决方案将连接直接连接到输出流中，而不是创建辅助缓冲区，并且适用于在 ostream 上具有 operator<< 的任何类型。

当您使用 vector<char*> 而不是 vector<string> 时，这也适用于 boost::algorithm::join() 失败的情况。

---

我喜欢1800的回答。但是，我会将第一次迭代移出循环，因为 if 语句的结果在第一次迭代后仅更改一次

template <class T, class A>
T join(const A &begin, const A &end, const T &t)
{
  T result;
  A it = begin;
  if (it != end) 
  {
   result.append(*it);
   ++it;
  }

  for( ;
       it!=end;
       ++it)
  {
    result.append(t);
    result.append(*it);
  }
  return result;
}

如果您愿意，这当然可以减少到更少的语句：

template <class T, class A>
T join(const A &begin, const A &end, const T &t)
{
  T result;
  A it = begin;
  if (it != end) 
   result.append(*it++);

  for( ; it!=end; ++it)
   result.append(t).append(*it);
  return result;
}

---

有一些有趣的尝试可以为这个问题提供一个优雅的解决方案。我有一个想法，使用模板流来有效地回答 OP 最初的困境。尽管这是一篇旧文章，但我希望偶然发现此问题的未来用户会发现我的解决方案是有益的。

首先，一些答案（包括接受的答案）不会促进可重用性。由于 C++ 没有提供一种在标准库中连接字符串的优雅方法（我已经看到），因此创建一个灵活且可重用的方法变得很重要。这是我的镜头：

// Replace with your namespace //
namespace my {
    // Templated join which can be used on any combination of streams, iterators and base types //
    template <typename TStream, typename TIter, typename TSeperator>
    TStream& join(TStream& stream, TIter begin, TIter end, TSeperator seperator) {
        // A flag which, when true, has next iteration prepend our seperator to the stream //
        bool sep = false;                       
        // Begin iterating through our list //
        for (TIter i = begin; i != end; ++i) {
            // If we need to prepend a seperator, do it //
            if (sep) stream << seperator;
            // Stream the next value held by our iterator //
            stream << *i;
            // Flag that next loops needs a seperator //
            sep = true;
        }
        // As a convenience, we return a reference to the passed stream //
        return stream;
    }
}

现在要使用它，您可以简单地执行以下操作：

// Load some data //
std::vector<int> params;
params.push_back(1);
params.push_back(2);
params.push_back(3);
params.push_back(4);

// Store and print our results to standard out //
std::stringstream param_stream;
std::cout << my::join(param_stream, params.begin(), params.end(), ",").str() << std::endl;

// A quick and dirty way to print directly to standard out //
my::join(std::cout, params.begin(), params.end(), ",") << std::endl;

请注意，流的使用如何使该解决方案非常灵活，因为我们可以将结果存储在字符串流中以便以后回收，或者我们可以直接写入标准输出、文件，甚至写入实现为流的网络连接。打印的类型必须是可迭代的并且与源流兼容。 STL 提供了多种兼容多种类型的流。所以你真的可以带着这个去城里。在我的脑海中，您的向量可以是 int、float、double、string、unsigned int、SomeObject* 等。

---

为什么这里的答案如此复杂

string vec2str( vector<int> v){
        string s="";
        for (auto e: v){
            s+=to_string(e);
            s+=',';
        }
        s.pop_back();
        return s;
   }

---

我创建了一个帮助头文件来添加扩展连接支持。

只需将下面的代码添加到您的通用头文件中，并在需要时包含它。

用法示例：

/* An example for a mapping function. */
ostream&
map_numbers(ostream& os, const void* payload, generic_primitive data)
{
    static string names[] = {"Zero", "One", "Two", "Three", "Four"};
    os << names[data.as_int];
    const string* post = reinterpret_cast<const string*>(payload);
    if (post) {
        os << " " << *post;
    }
    return os;
}

int main() {
    int arr[] = {0,1,2,3,4};
    vector<int> vec(arr, arr + 5);
    cout << vec << endl; /* Outputs: '0 1 2 3 4' */
    cout << join(vec.begin(), vec.end()) << endl; /* Outputs: '0 1 2 3 4' */
    cout << join(vec.begin(), vec.begin() + 2) << endl; /* Outputs: '0 1 2' */
    cout << join(vec.begin(), vec.end(), ", ") << endl; /* Outputs: '0, 1, 2, 3, 4' */
    cout << join(vec.begin(), vec.end(), ", ", map_numbers) << endl; /* Outputs: 'Zero, One, Two, Three, Four' */
    string post = "Mississippi";
    cout << join(vec.begin() + 1, vec.end(), ", ", map_numbers, &post) << endl; /* Outputs: 'One Mississippi, Two mississippi, Three mississippi, Four mississippi' */
    return 0;
}

幕后代码：

#include <iostream>
#include <vector>
#include <list>
#include <set>
#include <unordered_set>
using namespace std;

#define GENERIC_PRIMITIVE_CLASS_BUILDER(T) generic_primitive(const T& v) { value.as_##T = v; }
#define GENERIC_PRIMITIVE_TYPE_BUILDER(T) T as_##T;

typedef void* ptr;

/** A union that could contain a primitive or void*,
 *    used for generic function pointers.
 * TODO: add more primitive types as needed.
 */
struct generic_primitive {
    GENERIC_PRIMITIVE_CLASS_BUILDER(int);
    GENERIC_PRIMITIVE_CLASS_BUILDER(ptr);
    union {
        GENERIC_PRIMITIVE_TYPE_BUILDER(int);
        GENERIC_PRIMITIVE_TYPE_BUILDER(ptr);
    };
};

typedef ostream& (*mapping_funct_t)(ostream&, const void*, generic_primitive);
template<typename T>
class Join {
public:
    Join(const T& begin, const T& end,
            const string& separator = " ",
            mapping_funct_t mapping = 0,
            const void* payload = 0):
            m_begin(begin),
            m_end(end),
            m_separator(separator),
            m_mapping(mapping),
            m_payload(payload) {}

    ostream&
    apply(ostream& os) const
    {
        T begin = m_begin;
        T end = m_end;
        if (begin != end)
            if (m_mapping) {
                m_mapping(os, m_payload, *begin++);
            } else {
                os << *begin++;
            }
        while (begin != end) {
            os << m_separator;
            if (m_mapping) {
                m_mapping(os, m_payload, *begin++);
            } else {
                os << *begin++;
            }
        }
        return os;
    }
private:
    const T& m_begin;
    const T& m_end;
    const string m_separator;
    const mapping_funct_t m_mapping;
    const void* m_payload;
};

template <typename T>
Join<T>
join(const T& begin, const T& end,
     const string& separator = " ",
     ostream& (*mapping)(ostream&, const void*, generic_primitive) = 0,
     const void* payload = 0)
{
    return Join<T>(begin, end, separator, mapping, payload);
}

template<typename T>
ostream&
operator<<(ostream& os, const vector<T>& vec) {
    return join(vec.begin(), vec.end()).apply(os);
}

template<typename T>
ostream&
operator<<(ostream& os, const list<T>& lst) {
    return join(lst.begin(), lst.end()).apply(os);
}

template<typename T>
ostream&
operator<<(ostream& os, const set<T>& s) {
    return join(s.begin(), s.end()).apply(os);
}

template<typename T>
ostream&
operator<<(ostream& os, const Join<T>& vec) {
    return vec.apply(os);
}

---

这是一个通用的 C++11 解决方案，可以让你做

int main() {
    vector<int> v {1,2,3};
    cout << join(v, ", ") << endl;
    string s = join(v, '+').str();
}

代码是：

template<typename Iterable, typename Sep>
class Joiner {
    const Iterable& i_;
    const Sep& s_;
public:
    Joiner(const Iterable& i, const Sep& s) : i_(i), s_(s) {}
    std::string str() const {std::stringstream ss; ss << *this; return ss.str();}
    template<typename I, typename S> friend std::ostream& operator<< (std::ostream& os, const Joiner<I,S>& j);
};

template<typename I, typename S>
std::ostream& operator<< (std::ostream& os, const Joiner<I,S>& j) {
    auto elem = j.i_.begin();
    if (elem != j.i_.end()) {
        os << *elem;
        ++elem;
        while (elem != j.i_.end()) {
            os << j.s_ << *elem;
            ++elem;
        }
    }
    return os;
}

template<typename I, typename S>
inline Joiner<I,S> join(const I& i, const S& s) {return Joiner<I,S>(i, s);}

---

以下是将 vector 中的元素转换为 string 的简单实用方法：

std::string join(const std::vector<int>& numbers, const std::string& delimiter = ",") {
    std::ostringstream result;
    for (const auto number : numbers) {
        if (result.tellp() > 0) { // not first round
            result << delimiter;
        }
        result << number;
    }
    return result.str();
}

对于 ostringstream，您需要 #include <sstream>。

---

扩展 @sbi 在不限于 std::vector<int> 或特定返回字符串类型的 generic solution 的尝试。下面给出的代码可以这样使用：

std::vector<int> vec{ 1, 2, 3 };

// Call modern range-based overload.
auto str     = join( vec,  "," );
auto wideStr = join( vec, L"," );

// Call old-school iterator-based overload.
auto str     = join( vec.begin(), vec.end(),  "," );
auto wideStr = join( vec.begin(), vec.end(), L"," );

在原始代码中，如果分隔符是字符串文字（如上面的示例中所示），则模板参数推导无法生成正确的返回字符串类型。在这种情况下，函数体中的 Str::value_type 之类的 typedef 是不正确的。该代码假定 Str 始终是类似 std::basic_string 的类型，因此对于字符串文字显然失败。

为了解决这个问题，以下代码尝试仅从分隔符参数中推断出 character 类型，并使用它来生成默认的返回字符串类型。这是使用 boost::range_value 实现的，它从给定的 range 类型中提取元素类型。

#include <string>
#include <sstream>
#include <boost/range.hpp>

template< class Sep, class Str = std::basic_string< typename boost::range_value< Sep >::type >, class InputIt >
Str join( InputIt first, const InputIt last, const Sep& sep )
{
    using char_type          = typename Str::value_type;
    using traits_type        = typename Str::traits_type;
    using allocator_type     = typename Str::allocator_type;
    using ostringstream_type = std::basic_ostringstream< char_type, traits_type, allocator_type >;

    ostringstream_type result;

    if( first != last )
    {
        result << *first++;
    }
    while( first != last ) 
    {
        result << sep << *first++;
    }
    return result.str();
}

现在我们可以很容易地提供一个基于范围的重载，它简单地转发到基于迭代器的重载：

template <class Sep, class Str = std::basic_string< typename boost::range_value<Sep>::type >, class InputRange>
Str join( const InputRange &input, const Sep &sep )
{
    // Include the standard begin() and end() in the overload set for ADL. This makes the 
    // function work for standard types (including arrays), aswell as any custom types 
    // that have begin() and end() member functions or overloads of the standalone functions.
    using std::begin; using std::end;

    // Call iterator-based overload.
    return join( begin(input), end(input), sep );
}

Live Demo at Coliru

---

正如@capone 所做的那样，

std::string join(const std::vector<std::string> &str_list , 
                 const std::string &delim=" ")
{
    if(str_list.size() == 0) return "" ;
    return std::accumulate( str_list.cbegin() + 1, 
                            str_list.cend(), 
                            str_list.at(0) , 
                            [&delim](const std::string &a , const std::string &b)
                            { 
                                return a + delim + b ;
                            }  ) ; 
}

template <typename ST , typename TT>
std::vector<TT> map(TT (*op)(ST) , const vector<ST> &ori_vec)
{
    vector<TT> rst ;
    std::transform(ori_vec.cbegin() ,
                  ori_vec.cend() , back_inserter(rst) , 
                  [&op](const ST& val){ return op(val)  ;} ) ;
    return rst ;
}

然后我们可以像下面这样调用：

int main(int argc , char *argv[])
{
    vector<int> int_vec = {1,2,3,4} ;
    vector<string> str_vec = map<int,string>(to_string, int_vec) ;
    cout << join(str_vec) << endl ;
    return 0 ;
}

就像蟒蛇：

>>> " ".join( map(str, [1,2,3,4]) )

---

我用这样的东西

namespace std
{

// for strings join
string to_string( string value )
{
    return value;
}

} // namespace std

namespace // anonymous
{

template< typename T >
std::string join( const std::vector<T>& values, char delimiter )
{
    std::string result;
    for( typename std::vector<T>::size_type idx = 0; idx < values.size(); ++idx )
    {
        if( idx != 0 )
            result += delimiter;
        result += std::to_string( values[idx] );
    }
    return result;
}

} // namespace anonymous

---

我从@sbi 的答案开始，但大多数时候最终将结果字符串传输到流中，因此创建了下面的解决方案，可以将其传输到流中，而无需在内存中创建完整字符串的开销。

它的用法如下：

#include "string_join.h"
#include <iostream>
#include <vector>

int main()
{
  std::vector<int> v = { 1, 2, 3, 4 };
  // String version
  std::string str = join(v, std::string(", "));
  std::cout << str << std::endl;
  // Directly piped to stream version
  std::cout << join(v, std::string(", ")) << std::endl;
}

其中 string_join.h 是：

#pragma once

#include <iterator>
#include <sstream>

template<typename Str, typename It>
class joined_strings
{
  private:
    const It begin, end;
    Str sep;

  public:
    typedef typename Str::value_type char_type;
    typedef typename Str::traits_type traits_type;
    typedef typename Str::allocator_type allocator_type;

  private:
    typedef std::basic_ostringstream<char_type, traits_type, allocator_type>
      ostringstream_type;

  public:
    joined_strings(It begin, const It end, const Str &sep)
      : begin(begin), end(end), sep(sep)
    {
    }

    operator Str() const
    {
      ostringstream_type result;
      result << *this;
      return result.str();
    }

    template<typename ostream_type>
    friend ostream_type& operator<<(
      ostream_type &ostr, const joined_strings<Str, It> &joined)
    {
      It it = joined.begin;
      if(it!=joined.end)
        ostr << *it;
      for(++it; it!=joined.end; ++it)
        ostr << joined.sep << *it;
      return ostr;
    }
};

template<typename Str, typename It>
inline joined_strings<Str, It> join(It begin, const It end, const Str &sep)
{
  return joined_strings<Str, It>(begin, end, sep);
}

template<typename Str, typename Container>
inline joined_strings<Str, typename Container::const_iterator> join(
  Container container, const Str &sep)
{
  return join(container.cbegin(), container.cend(), sep);
}

---

我写了以下代码。它基于 C# string.join。它适用于 std::string 和 std::wstring 以及许多类型的向量。 （评论中的例子）

像这样称呼它：

 std::vector<int> vVectorOfIds = {1, 2, 3, 4, 5};

 std::wstring wstrStringForSQLIn = Join(vVectorOfIds, L',');

代码：

// Generic Join template (mimics string.Join() from C#)
// Written by RandomGuy (stackoverflow) 09-01-2017
// Based on Brian R. Bondy anwser here:
// http://stackoverflow.com/questions/1430757/c-vector-to-string
// Works with char, wchar_t, std::string and std::wstring delimiters
// Also works with a different types of vectors like ints, floats, longs
template<typename T, typename D>
auto Join(const std::vector<T> &vToMerge, const D &delimiter)
{
    // We use std::conditional to get the correct type for the stringstream (char or wchar_t)
    // stringstream = basic_stringstream<char>, wstringstream = basic_stringstream<wchar_t>
    using strType =
        std::conditional<
        std::is_same<D, std::string>::value,
        char,
            std::conditional<
            std::is_same<D, char>::value,
            char,
            wchar_t
            >::type
        >::type;

    std::basic_stringstream<strType> ss;

    for (size_t i = 0; i < vToMerge.size(); ++i)
    {
        if (i != 0)
            ss << delimiter;
        ss << vToMerge[i];
    }
    return ss.str();
}

---

这是将整数向量转换为字符串的简单方法。

#include <bits/stdc++.h>
using namespace std;
int main()
{
    vector<int> A = {1, 2, 3, 4};
    string s = "";
    for (int i = 0; i < A.size(); i++)
    {
        s = s + to_string(A[i]) + ",";
    }
    s = s.substr(0, s.length() - 1); //Remove last character
    cout << s;
}

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使用模板函数加入

我使用 template function 加入 vector 项，并通过仅迭代 vector 中的第一个到倒数第二个项目，然后加入 for 之后的最后一个项目，删除了不必要的 if 语句环形。这也避免了需要额外的代码来删除连接字符串末尾的额外分隔符。因此，没有 if 语句会减慢迭代速度，也没有需要整理的多余分隔符。

这会产生一个优雅的函数调用来加入 string、integer 或 double 等的 vector。

我写了两个版本：一个返回一个字符串；另一个返回一个字符串。另一个直接写入流。

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

// Return a string of joined vector items.
template<typename T>
string join(const vector<T>& v, const string& sep)
{
    ostringstream oss;
    const auto LAST = v.end() - 1;
    // Iterate through the first to penultimate items appending the separator.
    for (typename vector<T>::const_iterator p = v.begin(); p != LAST; ++p)
    {
        oss << *p << sep;
    }
    // Join the last item without a separator.
    oss << *LAST;
    return oss.str();
}

// Write joined vector items directly to a stream.
template<typename T>
void join(const vector<T>& v, const string& sep, ostream& os)
{
    const auto LAST = v.end() - 1;
    // Iterate through the first to penultimate items appending the separator.
    for (typename vector<T>::const_iterator p = v.begin(); p != LAST; ++p)
    {
        os << *p << sep;
    }
    // Join the last item without a separator.
    os << *LAST;
}

int main()
{
    vector<string> strings
    {
        "Joined",
        "from",
        "beginning",
        "to",
        "end"
    };
    vector<int> integers{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
    vector<double> doubles{ 1.2, 3.4, 5.6, 7.8, 9.0 };

    cout << join(strings, "... ") << endl << endl;
    cout << join(integers, ", ") << endl << endl;
    cout << join(doubles, "; ") << endl << endl;

    join(strings, "... ", cout);
    cout << endl << endl;
    join(integers, ",  ", cout);
    cout << endl << endl;
    join(doubles, ";  ", cout);
    cout << endl << endl;

    return 0;
}

输出

Joined... from... beginning... to... end

1, 2, 3, 4, 5, 6, 7, 8, 9, 10

1.2; 3.4; 5.6; 7.8; 9

Joined... from... beginning... to... end

1, 2, 3, 4, 5, 6, 7, 8, 9, 10

1.2; 3.4; 5.6; 7.8; 9

---

#include <iostream>
#include <vector>

int main()
{
    std::vector<int> v{{1,2,3,4}};
    std::string str;

    // ----->
    if (! v.empty())
    {
        str = std::to_string(*v.begin());
        for (auto it = std::next(v.begin()); it != v.end(); ++it)
            str.append("," + std::to_string(*it));
    }
    // <-----
    
    std::cout << str << "\n";
}

---

使用数字库 (#include <numeric>) 中的 std::accumulate 解决此问题的另一种方法：

std::vector<int> v{1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
auto comma_fold = [](std::string a, int b) { return std::move(a) + ',' + std::to_string(b); };
std::string s = std::accumulate(std::next(v.begin()), v.end(),
                                std::to_string(v[0]), // start with first element
                                comma_fold);
std::cout << s << std::endl; // 1,2,3,4,5,6,7,8,9,10

---

简单的解决方案/破解......不优雅但它有效

const auto vecToString = [](std::vector<int> input_vector)
{
    std::string holder = "";

    for (auto s : input_vector){
        holder += std::to_string(s);
        if(input_vector.back() != s){
            holder += ", ";
        }
    }

    return holder;
};

---