如何处理最后一个逗号,当用逗号分隔字符串时？

Question

可能重复:
请勿在最后一个数字后打印空格
使用逗号C++打印列表

#include <vector>
#include <iostream>
#include <sstream>
#include <boost/foreach.hpp>
using namespace std;

int main()
{
   vector<int> VecInts;

   VecInts.push_back(1);
   VecInts.push_back(2);
   VecInts.push_back(3);
   VecInts.push_back(4);
   VecInts.push_back(5);

   stringstream ss;
   BOOST_FOREACH(int i, VecInts)
   {
      ss << i << ",";
   }

   cout << ss.str();

   return 0;
}

这打印出来:1,2,3,4,5, 但是我想要:1,2,3,4,5

我怎样才能以优雅的方式实现这一目标？

我看到我对"优雅"的含义有些困惑:例如,我的循环中没有减慢"if-clause"的速度.想象一下向量中的100.000个条目!如果这就是你要提供的全部内容,我宁愿在完成循环后删除最后一个逗号.

Answer 1

这个怎么样:

#include <iostream>
#include <vector>
#include <algorithm>
#include <iterator>
#include <string>
#include <sstream>

int main()
{
   std::vector<int> v;

   v.push_back(1);
   v.push_back(2);
   v.push_back(3);
   v.push_back(4);
   v.push_back(5);

   std::ostringstream ss;
   if(!v.empty()) {
      std::copy(v.begin(), std::prev(v.end()), std::ostream_iterator<int>(ss, ", "));
      ss << v.back();
   }
   std::cout << ss.str() << "\n";
}

无需添加额外的变量,甚至不依赖于提升!实际上,除了"循环中没有额外的变量"要求之外,可以说甚至没有循环:)

Answer 2

在最后检测之前总是很棘手,检测第一个很容易.

bool first = true;
stringstream ss;
BOOST_FOREACH(int i, VecInts)
{
  if (!first) { ss << ","; }
  first = false;
  ss << i;
}

Answer 3

使用来自提升精神的业力 - 以快速而闻名.

#include <iostream>
#include <vector>
#include <boost/spirit/include/karma.hpp>

int main()
{
  std::vector<int> v;
  v.push_back(1);
  v.push_back(2);
  v.push_back(3);

  using namespace boost::spirit::karma;
  std::cout << format(int_ % ',', v) << std::endl;
}

Answer 4

尝试:

if (ss.tellp ())
{
   ss << ",";
}
ss << i;

或者,如果"if"让你担心:

char *comma = "";
BOOST_FOREACH(int i, VecInts)
{
   ss << comma << i;
   comma = ",";
}

Answer 5

就个人而言,我喜欢一种不会导致潜在内存分配的解决方案(因为字符串变得比需要的大).由于分支目标缓冲,循环体内的额外if应该是易处理的,但我会这样做:

#include <vector>
#include <iostream>

int main () {
    using std::cout;
    typedef std::vector<int>::iterator iterator;

    std::vector<int> ints;    
    ints.push_back(5);
    ints.push_back(1);
    ints.push_back(4);
    ints.push_back(2);
    ints.push_back(3);


    if (!ints.empty()) {
        iterator        it = ints.begin();
        const iterator end = ints.end();

        cout << *it;
        for (++it; it!=end; ++it) {
            cout << ", " << *it;
        }
        cout << std::endl;
    }
}

或者,BYORA(带上您自己的可重用算法):

// Follow the signature of std::getline. Allows us to stay completely
// type agnostic.
template <typename Stream, typename Iter, typename Infix>
inline Stream& infix (Stream &os, Iter from, Iter to, Infix infix_) {
    if (from == to) return os;
    os << *from;
    for (++from; from!=to; ++from) {
        os << infix_ << *from;
    }
    return os;
}

template <typename Stream, typename Iter>
inline Stream& comma_seperated (Stream &os, Iter from, Iter to) {
    return infix (os, from, to, ", ");
}

以便

...
comma_seperated(cout, ints.begin(), ints.end()) << std::endl;

infix(cout, ints.begin(), ints.end(), "-") << std::endl;
infix(cout, ints.begin(), ints.end(), "> <") << std::endl;
...

输出:

5, 1, 4, 2, 3
5-1-4-2-3
5> <1> <4> <2> <3

整洁的事情是它适用于每个输出流,任何具有前向迭代器的容器,任何中缀,以及任何中缀类型(有趣的是,例如当你使用宽字符串时).