C++ Cookbook

4.8. Joining a Sequence of Strings

Problem

Given a sequence of strings, such as output from Example 4-10, you want to join them together into a single, long string, perhaps with a delimiter.

Solution

Loop through the sequence and append each string to the output string. You can handle any standard sequence as input; Example 4-13 uses a vector of strings.

Example 4-13. Join a sequence of strings

#include <string>
#include <vector>
#include <iostream>

using namespace std;

void join(const vector<string>& v, char c, string& s) {

   s.clear();

   for (vector<string>::const_iterator p = v.begin();
        p != v.end(); ++p) {
      s += *p;
      if (p != v.end() - 1)
        s += c;
   }
}

int main() {

   vector<string> v;
   vector<string> v2;
   string s;

   v.push_back(string("fee"));
   v.push_back(string("fi"));
   v.push_back(string("foe"));
   v.push_back(string("fum"));

   join(v, '/', s);

   cout << s << '\n';
}

Discussion

Example 4-13 has one technique that is slightly different from previous examples. Look at this line:

for (vector<string>::const_iterator p = v.begin();

The previous string examples simply used iterators, without the “const” part, but you can’t get away with that here because v is declared as a reference to a const object. If you have a const container object, you can only use a const_iterator to access its elements. This is because a plain iterator allows writes to the object it refers to, which, of course, you can’t do if your container object is const.

I declared v const for two reasons. First, I know I’m not going to be modifying ...