C++ concatenate string and int

Question

I thought this would be really simple but it's presenting some difficulties. If I have

string name = "John"; int age = 21;

How do I combine them to get a single string "John21"?

Answer 1

Common Answer: itoa()

This is bad. itoa is non-standard, as pointed out in http://stackoverflow.com/questions/190229/where-is-the-itoa-function-in-linux

Answer 2

std::ostringstream o;
o << name << age;
std::cout << o.str();

Answer 3

If you have Boost, you can convert the integer to a string using boost::lexical_cast<std::string>(age).

Another way is to use stringstreams:

std::stringstream ss;
ss << age;
std::cout << name << ss.str() << std::endl;

A third approach would be to use sprintf or snprintf from the C library.

char buffer[128];
snprintf(buffer, sizeof(buffer), "%s%d", name.c_str(), age);
std::cout << buffer << std::endl;

Other posters suggested using itoa. This is NOT a standard function, so your code will not be portable if you use it. There are compilers that don't support it.

Answer 4

#include <string>
#include <sstream>
using namespace std;
string concatenate(std::string const& name, int i)
{
    stringstream s;
    s << name << i;
    return s.str();
}

Answer 5

It seems to me that the simplest answer is to use the sprintf function:

sprintf(outString,"%s%d",name,age);

Answer 6

Herb Sutter has a good article on this subject: "The String Formatters of Manor Farm". He covers Boost::lexical_cast, std::stringstream, std::strstream (which is deprecated), and sprintf vs. snprintf.

Answer 7

If you are using MFC, you can use a CString

CString nameAge = "";
nameAge.Format("%s%d", "John", 21);

Managed C++ also has a string formatter:

Answer 8

#include <sstream>

template <class T>
inline std::string to_string (const T& t)
{
   std::stringstream ss;
   ss << t;
   return ss.str();
}

Then your usage would look something like this

   std::string szName = "John";
   int numAge = 23;
   szName += to_string<int>(numAge);
   cout << szName << endl;

Googled [and tested :p ]

Answer 9

The std::ostringstream is a good method, but sometimes this additional trick might get handy transforming the formatting to a one-liner:

#include <sstream>
#define MAKE_STRING(tokens) /****************/ \
    static_cast<std::ostringstream&>(          \
        std::ostringstream().flush() << tokens \
    ).str()                                    \
    /**/

Now you can format strings like this:

int main() {
    int i = 123;
    std::string message = MAKE_STRING("i = " << i);
    std::cout << message << std::endl; // prints: "i = 123"
}

Answer 10

EDITED: My former version was incorrectly written. Here's a quick one using the sprintf_s functionality.

#include <iostream>
#include <string>
using namespace std;
int main()
{
   char name[] = "John";
   int age = 21;
   char nameAndAge[50];
   sprintf_s(nameAndAge,"%s%d",name,age);
   cout << nameAndAge << endl;
   return 0;
}

Answer 11

I don't have karma enough to comment (let alone edit), but Jay's post (currently the top-voted one at 27) contains an error. This code:

std::stringstream ss;
ss << age;
std::cout << name << ss.str() << std::endl;

Does not solve the stated problem of creating a string consisting of a concatenated string and integer. I think Jay meant something more like this:

std::stringstream ss;
ss << name;
ss << age;
std::cout << "built string: " << ss.str() << std::endl;

The final line is just to print the result, and shows how to access the final concatenated string.

Answer 12

If using sstream for completeness refer to http://stackoverflow.com/questions/20731/in-c-how-do-you-clear-a-stringstream-variable. Continued use of the stringstream variable causes grief, e.g.

char loop = 'n'; //y to continue looping
int count = 1; 
stringstream out;
string prefix;
do
{
    out << count;
    prefix = out.str();
    //out.str("");//uncomment to clear the stream
    cout << prefix << endl; 
    cout << "Continue (y/n): ";
    cin >> loop;
    cout << endl;
    count++;
}while(tolower(loop) == 'y');

Answer 13

In alphabetical order:

std::string name = "John"; int age = 21;
std::string result;

// 1. with Boost
result = name + boost::lexical_cast<std::string>(age).

// 2. with FastFormat.Format
fastformat::fmt(result, "{0}{1}", name, age);

// 3. with FastFormat.Write
fastformat::write(result, name, age);

// 4. with IOStreams
std::stringstream sstm;
sstm << name << age;
result = sstm.str();

// 5. with itoa
char numstr[21]; // enough to hold all numbers up to 64-bits
result = name + itoa(age, numstr, 10);

// 6. with sprintf
char numstr[21]; // enough to hold all numbers up to 64-bits
sprintf(numstr, "%d", age);
result = name + numstr;

// 7. with STLSoft's integer_to_string
char numstr[21]; // enough to hold all numbers up to 64-bits
result = name + stlsoft::integer_to_string(numstr, 21, age);

// 8. with STLSoft's winstl::int_to_string()
result = name + winstl::int_to_string(age);

1. is safe, but slow; requires Boost (header-only); most/all platforms
2. is safe, and fast; requires FastFormat, which must be compiled; most/all platforms
3. is safe, and fast; requires FastFormat, which must be compiled; most/all platforms
4. safe, slow, and verbose; requires nothing (is standard C++)
5. is brittle (you must supply a large enough buffer), fast, and verbose; itoa() is a non-standard extension, and not guaranteed to be available for all platforms
6. is brittle (you must supply a large enough buffer), fast, and verbose; requires nothing (is standard C++); all platforms
7. is brittle (you must supply a large enough buffer), probably the fastest-possible conversion, verbose; requires STLSoft (header-only); most/all platforms
8. safe-ish (you not use more than one int_to_string() call in a single statement), fast; requires STLSoft (header-only); Windows-only

Answer 14

There are more options possible to use to concatenate integer (or other numerric object) with string. It is Boost.Format

#include <boost/format.hpp>
#include <string>
int main()
{
    using boost::format;

    int age = 22;
    std::string str_age = str(format("age is %1%") % age);
}

and Karma from Boost.Spirit (v2)

#include <boost/spirit/include/karma.hpp>
#include <iterator>
#include <string>
int main()
{
    using namespace boost::spirit;

    int age = 22;
    std::string str_age("age is ");
    std::back_insert_iterator<std::string> sink(str_age);
    karma::generate(sink, int_, age);

    return 0;
}

Boost.Spirit Karma claims to be one of the fastest option for integer to string conversion.