How to concatenate a std::string and an int

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 C++11
result = name + std::to_string(age);

// 3. with FastFormat.Format
fastformat::fmt(result, "{0}{1}", name, age);

// 4. with FastFormat.Write
fastformat::write(result, name, age);

// 5. with the {fmt} library
result = fmt::format("{}{}", name, age);

// 6. with IOStreams
std::stringstream sstm;
sstm << name << age;
result = sstm.str();

// 7. with itoa
char numstr[21]; // enough to hold all numbers up to 64-bits
result = name + itoa(age, numstr, 10);

// 8. with sprintf
char numstr[21]; // enough to hold all numbers up to 64-bits
sprintf(numstr, "%d", age);
result = name + numstr;

// 9. 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);

// 10. with STLSoft's winstl::int_to_string()
result = name + winstl::int_to_string(age);

// 11. With Poco NumberFormatter
result = name + Poco::NumberFormatter().format(age);

1. is safe, but slow; requires Boost (header-only); most/all platforms
2. is safe, requires C++11 (to_string() is already included in #include <string>)
3. is safe, and fast; requires FastFormat, which must be compiled; most/all platforms
4. (ditto)
5. is safe, and fast; requires the {fmt} library, which can either be compiled or used in a header-only mode; most/all platforms
6. safe, slow, and verbose; requires #include <sstream> (from standard C++)
7. 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
8. is brittle (you must supply a large enough buffer), fast, and verbose; requires nothing (is standard C++); all platforms
9. is brittle (you must supply a large enough buffer), probably the fastest-possible conversion, verbose; requires STLSoft (header-only); most/all platforms
10. safe-ish (you don't use more than one int_to_string() call in a single statement), fast; requires STLSoft (header-only); Windows-only
11. is safe, but slow; requires Poco C++ ; most/all platforms

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.

#include <iostream>
#include <sstream>

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

#include <iostream>
#include <string>
#include <sstream>
using namespace std;
string itos(int i) // convert int to string
{
    stringstream s;
    s << i;
    return s.str();
}

Shamelessly stolen from http://www.research.att.com/~bs/bs_faq2.html.

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

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

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

#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 ]

This problem can be done in many ways. I will show it in two ways:

1. Convert the number to string using to_string(i).

2. Using string streams.

   Code:

   #include <string>
   #include <sstream>
   #include <bits/stdc++.h>
   #include <iostream>
   using namespace std;
   
   int main() {
       string name = "John";
       int age = 21;
   
       string answer1 = "";
       // Method 1). string s1 = to_string(age).
   
       string s1=to_string(age); // Know the integer get converted into string
       // where as we know that concatenation can easily be done using '+' in C++
   
       answer1 = name + s1;
   
       cout << answer1 << endl;
   
       // Method 2). Using string streams
   
       ostringstream s2;
   
       s2 << age;
   
       string s3 = s2.str(); // The str() function will convert a number into a string
   
       string answer2 = "";  // For concatenation of strings.
   
       answer2 = name + s3;
   
       cout << answer2 << endl;
   
       return 0;
   }
   

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

auto result = std::format("{}{}", name, age);

In the meantime you can use the {fmt} library, std::format is based on:

auto result = fmt::format("{}{}", name, age);

Disclaimer: I'm the author of the {fmt} library and C++20 std::format.

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.

As a one liner: name += std::to_string(age);

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"
}

• std::ostringstream

#include <sstream>

std::ostringstream s;
s << "John " << age;
std::string query(s.str());

• std::to_string (C++11)

std::string query("John " + std::to_string(age));

• boost::lexical_cast

#include <boost/lexical_cast.hpp>

std::string query("John " + boost::lexical_cast<std::string>(age));

Common Answer: itoa()

This is bad. itoa is non-standard, as pointed out here.

Here is an implementation of how to append an int to a string using the parsing and formatting facets from the IOStreams library.

#include <iostream>
#include <locale>
#include <string>

template <class Facet>
struct erasable_facet : Facet
{
    erasable_facet() : Facet(1) { }
    ~erasable_facet() { }
};

void append_int(std::string& s, int n)
{
    erasable_facet<std::num_put<char,
                                std::back_insert_iterator<std::string>>> facet;
    std::ios str(nullptr);

    facet.put(std::back_inserter(s), str,
                                     str.fill(), static_cast<unsigned long>(n));
}

int main()
{
    std::string str = "ID: ";
    int id = 123;

    append_int(str, id);

    std::cout << str; // ID: 123
}

You can concatenate int to string by using the given below simple trick, but note that this only works when integer is of single digit. Otherwise, add integer digit by digit to that string.

string name = "John";
int age = 5;
char temp = 5 + '0';
name = name + temp;
cout << name << endl;

Output:  John5

In C++ 20 you can have a variadic lambda that does concatenate arbitrary streamable types to a string in a few lines:

auto make_string=[os=std::ostringstream{}](auto&& ...p) mutable 
{ 
  (os << ... << std::forward<decltype(p)>(p) ); 
  return std::move(os).str();
};

int main() {
std::cout << make_string("Hello world: ",4,2, " is ", 42.0);
}

see https://godbolt.org/z/dEe9h75eb

using move(os).str() guarantees that the ostringstream object's stringbuffer is empty next time the lambda is called.