Combining a vector of strings

Question

I've been reading Accelerated C++ and I have to say it's an interesting book.

In chapter 6, I have to use a function from <algorithm> to concatenate from a vector<string> into a single string. I could use accumulate, but it doesn't help because string containers can only push_back characters.

int main () {
  using namespace std;
  string str = "Hello, world!";
  vector<string>  vec (10, str);
  // Concatenate here?

  return 0;
}

How do I join the strings together?

Valor

Please consider changing the accepted solution to something less inefficient. – Nystagmus

Answer 1

Assuming this is question 6.8, it doesn't say you have to use accumulate - it says use "a library algorithm". However, you can use accumulate:

#include <numeric>
    
int main () {
    std::string str = "Hello World!";
    std::vector<std::string> vec(10,str);
    std::string a = std::accumulate(vec.begin(), vec.end(), std::string(""));
    std::cout << a << std::endl;
}

All that accumulate does is set 'sum' to the third parameter, and then for all of the values 'val' from first parameter to second parameter, do:

sum = sum + val

it then returns 'sum'. Despite the fact that accumulate is declared in <numeric> it will work for anything that implements operator+()

---

Note: This solution, while elegant, is inefficient, as a new string will be allocated and populated for each element of vec.

Monometallism

Thanks, I tried using accumulate with the 3rd parameter as a.begin, didn't work, also I tried with back_inserter which failed too. Can you explain how this works? Thanks alot. – Valor

It takes each element in the vector from .begin() to .end() and accumulates them in to the third parameter, which is an empty std::string passed in as a temporary. std::acumulate()'s return value is the result of the accumulation, passed by value. – Avesta

Btw, this approach possibly scales very badly since there might be a lof of copying / reallicating involved. – Preposterous

std::reduce (C++17) - more faster than std::accumulate: std::string a = std::reduce(vec.begin(), vec.end(), std::string("")); – Ludly

Answer 2

How about std::copy?

std::ostringstream os;
std::copy( vec_strings.begin(), vec_string.end(), std::ostream_iterator<std::string>( os ) );
std::cout << os.str() << std::endl;

Brunel

Can you compare this solution to the accepted one? – Nystagmus

accepted answer is O(N^2): #15347623 – Lodicule

Answer 3

The following snippet compiles in Visual C++ 2012 and uses a lambda function:

int main () {
    string str = "Hello World!";
    vector<string>  vec (10,str);

    stringstream ss;
    for_each(vec.begin(), vec.end(), [&ss] (const string& s) { cat(ss, s); });

    cout << ss.str() << endl;
}

The accumulate example in the 1st answer is elegant, but as sellibitze pointed out, it reallocates with each concatenation and scales at O(N²). This for_each snippet scales at about O(N). I profiled both solutions with 100K strings; the accumulate example took 23.6 secs, but this for_each snippet took 0.054 sec.

Bibliology

It could be even faster to create an std::sring call reserve(final-size), then just use += which should be fast since the buffer is already of the right size. – Ofeliaofella

@pixelgrease, what is the cat function? how is it defined? thanks – Heptarchy

Answer 4

I am not sure about your question.Where lies the problem? Its just a matter of a loop.

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

int main () 
{
    std::string str = "Hello World!";
    std::vector<string>  vec (10,str);

    for(size_t i=0;i!=vec.size();++i)
        str=str+vec[i];
    std::cout<<str;
}

EDIT :

Use for_each() from <algorithm>

Try this:

#include<vector>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
string i;
void func(string &k)
{
  i+=k;
}
int main () {
    string str = "Hello World!";
    vector<string>  vec (10,str);

    for_each(vec.begin(),vec.end(),func);
    cout<<i;
    return 0;
  }

Pyrethrin

I have to use a function from algorithm header to do this. – Valor

Or use a proper functor With for_each instead of a free function. Saves you having a global variable – Adscription

Answer 5

This answer is an improved version of the one from @PrasoonSaurav by adding a helpful sep argument.

inline std::string joinStrings(std::vector<std::string> arr, std::string sep) {
    std::string out = arr[0];
    for(unsigned int i = 1; i < arr.size(); i++) {
        out += sep + arr[i];
    }
    return out;
}

// Overload function parameter to add default value for the separator
inline std::string joinStrings(std::vector<std::string> arr) {
    return joinStrings(arr, std::string(", "));
}

Pence