CODEWITHSUNDEEP
c++stlstring-concatenationstringstreamstdstring
WilliamKF
WilliamKF

Reputation: 43099

How to construct a std::string from a std::vector<string>?

I'd like to build a std::string from a std::vector<std::string>.

I could use std::stringsteam, but imagine there is a shorter way:

std::string string_from_vector(const std::vector<std::string> &pieces) {
  std::stringstream ss;

  for(std::vector<std::string>::const_iterator itr = pieces.begin();
      itr != pieces.end();
      ++itr) {
    ss << *itr;
  }

  return ss.str();
}

How else might I do this?

Upvotes: 65

Views: 112267

Answers (8)

Oktalist
Oktalist

Reputation: 14714

C++03

std::string s;
for (std::vector<std::string>::const_iterator i = v.begin(); i != v.end(); ++i)
    s += *i;
return s;

C++11 (the MSVC 2010 subset)

std::string s;
std::for_each(v.begin(), v.end(), [&](const std::string &piece){ s += piece; });
return s;

C++11

std::string s;
for (const auto &piece : v) s += piece;
return s;

Don't use std::accumulate for string concatenation, it is a classic Schlemiel the Painter's algorithm, even worse than the usual example using strcat in C. Without C++11 move semantics, it incurs two unnecessary copies of the accumulator for each element of the vector. Even with move semantics, it still incurs one unnecessary copy of the accumulator for each element.

The three examples above are O(n).

std::accumulate is O(n²) for strings.

You could make std::accumulate O(n) for strings by supplying a custom functor:

std::string s = std::accumulate(v.begin(), v.end(), std::string{},
    [](std::string &s, const std::string &piece) -> decltype(auto) { return s += piece; });

Note that s must be a reference to non-const, the lambda return type must be a reference (hence decltype(auto)), and the body must use += not +.

C++20

In the current draft of what is expected to become C++20, the definition of std::accumulate has been altered to use std::move when appending to the accumulator, so from C++20 onwards, accumulate will be O(n) for strings, and can be used as a one-liner:

std::string s = std::accumulate(v.begin(), v.end(), std::string{});

Upvotes: 124

Ali
Ali

Reputation: 58431

My personal choice would be the range-based for loop, as in Oktalist's answer.

Boost also offers a nice solution:

#include <boost/algorithm/string/join.hpp>
#include <iostream>
#include <vector>

int main() {

    std::vector<std::string> v{"first", "second"};

    std::string joined = boost::algorithm::join(v, ", ");

    std::cout << joined << std::endl;
}

This prints:

first, second

Upvotes: 13

user9494810
user9494810

Reputation: 43

If requires no trailing spaces, use accumulate defined in <numeric> with custom join lambda.

#include <iostream>
#include <numeric>
#include <vector>

using namespace std;


int main() {
    vector<string> v;
    string s;

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

    s = accumulate(begin(v), end(v), string(),
                   [](string lhs, const string &rhs) { return lhs.empty() ? rhs : lhs + ' ' + rhs; }
    );
    cout << s << endl;
    return 0;
}

Output:

fee fi foe fum

Upvotes: 4

NoSenseEtAl
NoSenseEtAl

Reputation: 30006

Google Abseil has function absl::StrJoin that does what you need.

Example from their header file. Notice that separator can be also ""

//   std::vector<std::string> v = {"foo", "bar", "baz"};
//   std::string s = absl::StrJoin(v, "-");
//   EXPECT_EQ("foo-bar-baz", s);

Upvotes: 5

Andy Prowl
Andy Prowl

Reputation: 126422

You could use the std::accumulate() standard function from the <numeric> header (it works because an overload of operator + is defined for strings which returns the concatenation of its two arguments):

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

int main()
{
    std::vector<std::string> v{"Hello, ", " Cruel ", "World!"};
    std::string s;
    s = accumulate(begin(v), end(v), s);
    std::cout << s; // Will print "Hello, Cruel World!"
}

Alternatively, you could use a more efficient, small for cycle:

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

int main()
{
    std::vector<std::string> v{"Hello, ", "Cruel ", "World!"};
    std::string result;
    for (auto const& s : v) { result += s; }
    std::cout << result; // Will print "Hello, Cruel World!"
}

Upvotes: 40

Benjamin K.
Benjamin K.

Reputation: 1105

A little late to the party, but I liked the fact that we can use initializer lists:

std::string join(std::initializer_list<std::string> i)
{
  std::vector<std::string> v(i);
  std::string res;
  for (const auto &s: v) res += s;
  return res;   
}

Then you can simply invoke (Python style):

join({"Hello", "World", "1"})

Upvotes: 3

Galimov Albert
Galimov Albert

Reputation: 7357

With c++11 the stringstream way is not too scary:

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

int main()
{
    std::vector<std::string> v{"Hello, ", " Cruel ", "World!"};
   std::stringstream s;
   std::for_each(begin(v), end(v), [&s](const std::string &elem) { s << elem; } );
   std::cout << s.str();
}

Upvotes: 1

bstamour
bstamour

Reputation: 7766

Why not just use operator + to add them together?

std::string string_from_vector(const std::vector<std::string> &pieces) {
   return std::accumulate(pieces.begin(), pieces.end(), std::string(""));
}

std::accumulate uses std::plus under the hood by default, and adding two strings is concatenation in C++, as the operator + is overloaded for std::string.

Upvotes: 8

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