How to construct a std::string from a std::vector<string>?
C++StlString ConcatenationStringstreamStdstringC++ Problem Overview
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?
C++ Solutions
Solution 1 - C++
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{});
Solution 2 - C++
You could use the std::accumulate()
standard function from the <numeric>
header (it works because an overload of operator +
is defined for string
s 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!"
}
Solution 3 - C++
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
Solution 4 - C++
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.
Solution 5 - C++
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);
Solution 6 - C++
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"})
Solution 7 - C++
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();
}
Solution 8 - C++
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