How do I concatenate multiple C++ strings on one line?

Question

C# has a syntax feature where you can concatenate many data types together on 1 line.

string s = new String();
s += "Hello world, " + myInt + niceToSeeYouString;
s += someChar1 + interestingDecimal + someChar2;

What would be the equivalent in C++? As far as I can see, you'd have to do it all on separate lines as it doesn't support multiple strings/variables with the + operator. This is OK, but doesn't look as neat.

string s;
s += "Hello world, " + "nice to see you, " + "or not.";

The above code produces an error.

Answer 1

#include <sstream>
#include <string>

std::stringstream ss;
ss << "Hello, world, " << myInt << niceToSeeYouString;
std::string s = ss.str();

Take a look at this Guru Of The Week article from Herb Sutter: The String Formatters of Manor Farm

Answer 2

In 5 years nobody has mentioned .append?

#include <string>

std::string s;
s.append("Hello world, ");
s.append("nice to see you, ");
s.append("or not.");

Answer 3

s += "Hello world, " + "nice to see you, " + "or not.";

Those character array literals are not C++ std::strings - you need to convert them:

s += string("Hello world, ") + string("nice to see you, ") + string("or not.");

To convert ints (or any other streamable type) you can use a boost lexical_cast or provide your own function:

template <typename T>
string Str( const T & t ) {
   ostringstream os;
   os << t;
   return os.str();
}

You can now say things like:

string s = string("The meaning is ") + Str( 42 );

Answer 4

Your code can be written as¹,

s = "Hello world," "nice to see you," "or not."

...but I doubt that's what you're looking for. In your case, you are probably looking for streams:

std::stringstream ss;
ss << "Hello world, " << 42 << "nice to see you.";
std::string s = ss.str();

---

¹ "can be written as" : This only works for string literals. The concatenation is done by the compiler.

Answer 5

Using C++14 user defined literals and std::to_string the code becomes easier.

using namespace std::literals::string_literals;
std::string str;
str += "Hello World, "s + "nice to see you, "s + "or not"s;
str += "Hello World, "s + std::to_string(my_int) + other_string;

Note that concatenating string literals can be done at compile time. Just remove the +.

str += "Hello World, " "nice to see you, " "or not";

Answer 6

In C++20 you'll be able to do:

auto s = std::format("{}{}{}", "Hello world, ", myInt, niceToSeeYouString);

Until then you could do the same with the {fmt} library:

auto s = fmt::format("{}{}{}", "Hello world, ", myInt, niceToSeeYouString);

Disclaimer: I'm the author of {fmt}.

Answer 7

To offer a solution that is more one-line-ish: A function concat can be implemented to reduce the "classic" stringstream based solution to a single statement. It is based on variadic templates and perfect forwarding.

---

Usage:

std::string s = concat(someObject, " Hello, ", 42, " I concatenate", anyStreamableType);

---

Implementation:

void addToStream(std::ostringstream&)
{
}

template<typename T, typename... Args>
void addToStream(std::ostringstream& a_stream, T&& a_value, Args&&... a_args)
{
    a_stream << std::forward<T>(a_value);
    addToStream(a_stream, std::forward<Args>(a_args)...);
}

template<typename... Args>
std::string concat(Args&&... a_args)
{
    std::ostringstream s;
    addToStream(s, std::forward<Args>(a_args)...);
    return s.str();
}

Answer 8

boost::format

or std::stringstream

std::stringstream msg;
msg << "Hello world, " << myInt  << niceToSeeYouString;
msg.str(); // returns std::string object

Answer 9

The actual problem was that concatenating string literals with + fails in C++:

string s;
s += "Hello world, " + "nice to see you, " + "or not.";
The above code produces an error.

In C++ (also in C), you concatenate string literals by just placing them right next to each other:

string s0 = "Hello world, " "nice to see you, " "or not.";
string s1 = "Hello world, " /*same*/ "nice to see you, " /*result*/ "or not.";
string s2 = 
    "Hello world, " /*line breaks in source code as well as*/ 
    "nice to see you, " /*comments don't matter*/ 
    "or not.";

This makes sense, if you generate code in macros:

#define TRACE(arg) cout << #arg ":" << (arg) << endl;

...a simple macro that can be used like this

int a = 5;
TRACE(a)
a += 7;
TRACE(a)
TRACE(a+7)
TRACE(17*11)

(live demo ...)

or, if you insist in using the + for string literals (as already suggested by underscore_d):

string s = string("Hello world, ")+"nice to see you, "+"or not.";

Another solution combines a string and a const char* for each concatenation step

string s;
s += "Hello world, "
s += "nice to see you, "
s += "or not.";

Answer 10

auto s = string("one").append("two").append("three")

Answer 11

You would have to define operator+() for every data type you would want to concenate to the string, yet since operator<< is defined for most types, you should use std::stringstream.

Damn, beat by 50 seconds...

Answer 12

If you write out the +=, it looks almost the same as C#

string s("Some initial data. "); int i = 5;
s = s + "Hello world, " + "nice to see you, " + to_string(i) + "\n";

Answer 13

As others said, the main problem with the OP code is that the operator + does not concatenate const char *; it works with std::string, though.

Here's another solution that uses C++11 lambdas and for_each and allows to provide a separator to separate the strings:

#include <vector>
#include <algorithm>
#include <iterator>
#include <sstream>

string join(const string& separator,
            const vector<string>& strings)
{
    if (strings.empty())
        return "";

    if (strings.size() == 1)
        return strings[0];

    stringstream ss;
    ss << strings[0];

    auto aggregate = [&ss, &separator](const string& s) { ss << separator << s; };
    for_each(begin(strings) + 1, end(strings), aggregate);

    return ss.str();
}

Usage:

std::vector<std::string> strings { "a", "b", "c" };
std::string joinedStrings = join(", ", strings);

It seems to scale well (linearly), at least after a quick test on my computer; here's a quick test I've written:

#include <vector>
#include <algorithm>
#include <iostream>
#include <iterator>
#include <sstream>
#include <chrono>

using namespace std;

string join(const string& separator,
            const vector<string>& strings)
{
    if (strings.empty())
        return "";

    if (strings.size() == 1)
        return strings[0];

    stringstream ss;
    ss << strings[0];

    auto aggregate = [&ss, &separator](const string& s) { ss << separator << s; };
    for_each(begin(strings) + 1, end(strings), aggregate);

    return ss.str();
}

int main()
{
    const int reps = 1000;
    const string sep = ", ";
    auto generator = [](){return "abcde";};

    vector<string> strings10(10);
    generate(begin(strings10), end(strings10), generator);

    vector<string> strings100(100);
    generate(begin(strings100), end(strings100), generator);

    vector<string> strings1000(1000);
    generate(begin(strings1000), end(strings1000), generator);

    vector<string> strings10000(10000);
    generate(begin(strings10000), end(strings10000), generator);

    auto t1 = chrono::system_clock::now();
    for(int i = 0; i<reps; ++i)
    {
        join(sep, strings10);
    }

    auto t2 = chrono::system_clock::now();
    for(int i = 0; i<reps; ++i)
    {
        join(sep, strings100);
    }

    auto t3 = chrono::system_clock::now();
    for(int i = 0; i<reps; ++i)
    {
        join(sep, strings1000);
    }

    auto t4 = chrono::system_clock::now();
    for(int i = 0; i<reps; ++i)
    {
        join(sep, strings10000);
    }

    auto t5 = chrono::system_clock::now();

    auto d1 = chrono::duration_cast<chrono::milliseconds>(t2 - t1);
    auto d2 = chrono::duration_cast<chrono::milliseconds>(t3 - t2);
    auto d3 = chrono::duration_cast<chrono::milliseconds>(t4 - t3);
    auto d4 = chrono::duration_cast<chrono::milliseconds>(t5 - t4);

    cout << "join(10)   : " << d1.count() << endl;
    cout << "join(100)  : " << d2.count() << endl;
    cout << "join(1000) : " << d3.count() << endl;
    cout << "join(10000): " << d4.count() << endl;
}

Results (milliseconds):

join(10)   : 2
join(100)  : 10
join(1000) : 91
join(10000): 898

Answer 14

Maybe you like my "Streamer" solution to really do it in one line:

#include <iostream>
#include <sstream>
using namespace std;

class Streamer // class for one line string generation
{
public:

    Streamer& clear() // clear content
    {
        ss.str(""); // set to empty string
        ss.clear(); // clear error flags
        return *this;
    }

    template <typename T>
    friend Streamer& operator<<(Streamer& streamer,T str); // add to streamer

    string str() // get current string
    { return ss.str();}

private:
    stringstream ss;
};

template <typename T>
Streamer& operator<<(Streamer& streamer,T str)
{ streamer.ss<<str;return streamer;}

Streamer streamer; // make this a global variable


class MyTestClass // just a test class
{
public:
    MyTestClass() : data(0.12345){}
    friend ostream& operator<<(ostream& os,const MyTestClass& myClass);
private:
    double data;
};

ostream& operator<<(ostream& os,const MyTestClass& myClass) // print test class
{ return os<<myClass.data;}


int main()
{
    int i=0;
    string s1=(streamer.clear()<<"foo"<<"bar"<<"test").str();                      // test strings
    string s2=(streamer.clear()<<"i:"<<i++<<" "<<i++<<" "<<i++<<" "<<0.666).str(); // test numbers
    string s3=(streamer.clear()<<"test class:"<<MyTestClass()).str();              // test with test class
    cout<<"s1: '"<<s1<<"'"<<endl;
    cout<<"s2: '"<<s2<<"'"<<endl;
    cout<<"s3: '"<<s3<<"'"<<endl;
}

Answer 15

Here's the one-liner solution:

#include <iostream>
#include <string>

int main() {
  std::string s = std::string("Hi") + " there" + " friends";
  std::cout << s << std::endl;

  std::string r = std::string("Magic number: ") + std::to_string(13) + "!";
  std::cout << r << std::endl;

  return 0;
}

Although it's a tiny bit ugly, I think it's about as clean as you cat get in C++.

We are casting the first argument to a std::string and then using the (left to right) evaluation order of operator+ to ensure that its left operand is always a std::string. In this manner, we concatenate the std::string on the left with the const char * operand on the right and return another std::string, cascading the effect.

Note: there are a few options for the right operand, including const char *, std::string, and char.

It's up to you to decide whether the magic number is 13 or 6227020800.

Answer 16

You may use this header for this regard: https://github.com/theypsilon/concat

using namespace concat;

assert(concat(1,2,3,4,5) == "12345");

Under the hood you will be using a std::ostringstream.

Answer 17

If you are willing to use c++11 you can utilize user-defined string literals and define two function templates that overload the plus operator for a std::string object and any other object. The only pitfall is not to overload the plus operators of std::string, otherwise the compiler doesn't know which operator to use. You can do this by using the template std::enable_if from type_traits. After that strings behave just like in Java or C#. See my example implementation for details.

Main code

#include <iostream>
#include "c_sharp_strings.hpp"

using namespace std;

int main()
{
    int i = 0;
    float f = 0.4;
    double d = 1.3e-2;
    string s;
    s += "Hello world, "_ + "nice to see you. "_ + i
            + " "_ + 47 + " "_ + f + ',' + d;
    cout << s << endl;
    return 0;
}

File c_sharp_strings.hpp

Include this header file in all all places where you want to have these strings.

#ifndef C_SHARP_STRING_H_INCLUDED
#define C_SHARP_STRING_H_INCLUDED

#include <type_traits>
#include <string>

inline std::string operator "" _(const char a[], long unsigned int i)
{
    return std::string(a);
}

template<typename T> inline
typename std::enable_if<!std::is_same<std::string, T>::value &&
                        !std::is_same<char, T>::value &&
                        !std::is_same<const char*, T>::value, std::string>::type
operator+ (std::string s, T i)
{
    return s + std::to_string(i);
}

template<typename T> inline
typename std::enable_if<!std::is_same<std::string, T>::value &&
                        !std::is_same<char, T>::value &&
                        !std::is_same<const char*, T>::value, std::string>::type
operator+ (T i, std::string s)
{
    return std::to_string(i) + s;
}

#endif // C_SHARP_STRING_H_INCLUDED

Answer 18

Something like this works for me

namespace detail {
    void concat_impl(std::ostream&) { /* do nothing */ }

    template<typename T, typename ...Args>
    void concat_impl(std::ostream& os, const T& t, Args&&... args)
    {
        os << t;
        concat_impl(os, std::forward<Args>(args)...);
    }
} /* namespace detail */

template<typename ...Args>
std::string concat(Args&&... args)
{
    std::ostringstream os;
    detail::concat_impl(os, std::forward<Args>(args)...);
    return os.str();
}
// ...
std::string s{"Hello World, "};
s = concat(s, myInt, niceToSeeYouString, myChar, myFoo);

Answer 19

Based on above solutions I made a class var_string for my project to make life easy. Examples:

var_string x("abc %d %s", 123, "def");
std::string y = (std::string)x;
const char *z = x.c_str();

The class itself:

#include <stdlib.h>
#include <stdarg.h>

class var_string
{
public:
    var_string(const char *cmd, ...)
    {
        va_list args;
        va_start(args, cmd);
        vsnprintf(buffer, sizeof(buffer) - 1, cmd, args);
    }

    ~var_string() {}

    operator std::string()
    {
        return std::string(buffer);
    }

    operator char*()
    {
        return buffer;
    }

    const char *c_str()
    {
        return buffer;
    }

    int system()
    {
        return ::system(buffer);
    }
private:
    char buffer[4096];
};

Still wondering if there will be something better in C++ ?

Answer 20

In c11:

void printMessage(std::string&& message) {
    std::cout << message << std::endl;
    return message;
}

this allow you to create function call like this:

printMessage("message number : " + std::to_string(id));

will print : message number : 10

Answer 21

you can also "extend" the string class and choose the operator you prefer ( <<, &, |, etc ...)

Here is the code using operator<< to show there is no conflict with streams

note: if you uncomment s1.reserve(30), there is only 3 new() operator requests (1 for s1, 1 for s2, 1 for reserve ; you can't reserve at constructor time unfortunately); without reserve, s1 has to request more memory as it grows, so it depends on your compiler implementation grow factor (mine seems to be 1.5, 5 new() calls in this example)

namespace perso {
class string:public std::string {
public:
    string(): std::string(){}

    template<typename T>
    string(const T v): std::string(v) {}

    template<typename T>
    string& operator<<(const T s){
        *this+=s;
        return *this;
    }
};
}

using namespace std;

int main()
{
    using string = perso::string;
    string s1, s2="she";
    //s1.reserve(30);
    s1 << "no " << "sunshine when " << s2 << '\'' << 's' << " gone";
    cout << "Aint't "<< s1 << " ..." <<  endl;

    return 0;
}

Answer 22

Stringstream with a simple preproccessor macro using a lambda function seems nice:

#include <sstream>
#define make_string(args) []{std::stringstream ss; ss << args; return ss;}() 

and then

auto str = make_string("hello" << " there" << 10 << '$');

Answer 23

This works for me:

#include <iostream>

using namespace std;

#define CONCAT2(a,b)     string(a)+string(b)
#define CONCAT3(a,b,c)   string(a)+string(b)+string(c)
#define CONCAT4(a,b,c,d) string(a)+string(b)+string(c)+string(d)

#define HOMEDIR "c:\\example"

int main()
{

    const char* filename = "myfile";

    string path = CONCAT4(HOMEDIR,"\\",filename,".txt");

    cout << path;
    return 0;
}

Output:

c:\example\myfile.txt

Answer 24

Have you tried to avoid the +=? instead use var = var + ... it worked for me.

#include <iostream.h> // for string

string myName = "";
int _age = 30;
myName = myName + "Vincent" + "Thorpe" + 30 + " " + 2019;