CODEWITHSUNDEEP
c++stringvector
thirdDeveloper
thirdDeveloper

Reputation: 895

Convert vector<int> to delimited string

As I see here there is a fast and short way to Convert vector to string separated with a character in c#:

var result = string.Join(";", data); 
var result = string.Join(";", data.Select(x => x.ToString()).ToArray());

I wan to know is there a same way in c++ to do this?

Upvotes: 12

Views: 9326

Answers (2)

BenS1
BenS1

Reputation: 105

Expanding on the accepted answer a little, I took the code and made a generic template version that works not only with a vector of ints but works with any types that work with stringstream:

template <typename T>
std::string VectorToCSV(const std::vector<T>& vec)
{
    std::ostringstream out;
    if (!vec.empty())
    {
        std::copy(std::begin(vec), std::end(vec) - 1, std::ostream_iterator<T>(out, ";"));
        out << vec.back();
    }

    return out.str();
}

This works with Template Argument Deduction, so to call it like this:

// Vector of ints example
std::vector<int> vec_of_ints = { 1, 2, 3, 4};    
std::string ints_csv = VectorToCSV(vec_of_ints);

// Vector of strings example
std::vector<std::string> vec_of_strings = { "Hello", "World"};
std::string strings_csv = VectorToCSV(vec_of_strings);

Upvotes: 2

dyp
dyp

Reputation: 39151

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

int main()
{
    std::vector<int> data = {42, 1, 2, 3, 4, 5};

    std::ostringstream oss;
    std::copy(data.begin(), data.end(), std::ostream_iterator<int>(oss, ";"));

    std::string result( oss.str() );
    std::cout << result << "\n";
}

N.B. In C++11, you can use the more general form

using std::begin;
using std::end;
std::copy(begin(data), end(data), std::ostream_iterator<int>(oss, ";"));

Where the using-declarations are not required if ADL can be used (like in the example above).

Also possible, but maybe a bit less efficient:

std::string s;
for(auto const& e : v) s += std::to_string(e) + ";";

which can be written via std::accumulate in <algorithm> as:

std::string s = std::accumulate(begin(v), end(v), std::string{},
    [](std::string r, int p){ return std::move(r) + std::to_string(p) + ";"; });

(IIRC there was some method to eliminate the copying, maybe by taking the lambda-parameter by reference std::string& r.)

A version w/o the trailing semi-colon (thanks to Dietmar Kühl):

std::vector<int> data = {42, 1, 2, 3, 4, 5};

std::ostringstream out;
if (!v.empty())
{
    std::copy(v.begin(), v.end() - 1, std::ostream_iterator<int>(out, ";"));
    out << v.back();
}

std::string result( out.str() );
std::cout << result << "\n";

Upvotes: 24

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