Easiest way to convert int to string in C++

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What is the easiest way to convert from int to equivalent string in C++. I am aware of two methods. Is there any easier way?


int a = 10;
char *intStr = itoa(a);
string str = string(intStr);


int a = 10;
stringstream ss;
ss << a;
string str = ss.str();


answered 7 years ago Mehrdad #1

Not that I know of, in pure C++. But a little modification of what you mentioned

string s = string(itoa(a));

should work, and it's pretty short.

answered 7 years ago DevSolar #2

Picking up a discussion with @v.oddou a couple of years later, C++17 has finally delivered a way to do the originally macro-based type-agnostic solution (preserved below) without going through macro uglyness.

template < typename... Args >
std::string sstr( Args &&... args )
    std::ostringstream sstr;
    ( sstr << std::dec << ... << args );
    return sstr.str();


int i = 42;
std::string s = sstr( "i is: ", i );
puts( sstr( i ).c_str() );

Foo x( 42 );
throw std::runtime_error( sstr( "Foo is '", x, "', i is ", i ) );

Original answer:

Since "converting ... to string" is a recurring problem, I always define the SSTR() macro in a central header of my C++ sources:

#include <sstream>

#define SSTR( x ) static_cast< std::ostringstream & >( \
        ( std::ostringstream() << std::dec << x ) ).str()

Usage is as easy as could be:

int i = 42;
std::string s = SSTR( "i is: " << i );
puts( SSTR( i ).c_str() );

Foo x( 42 );
throw std::runtime_error( SSTR( "Foo is '" << x << "', i is " << i ) );

The above is C++98 compatible (if you cannot use C++11 std::to_string), and does not need any third-party includes (if you cannot use Boost lexical_cast<>); both these other solutions have a better performance though.

answered 7 years ago Jerry Coffin #3

Probably the most common easy way wraps essentially your second choice into a template named lexical_cast, such as the one in Boost, so your code looks like this:

int a = 10;
string s = lexical_cast<string>(a);

One nicety of this is that it supports other casts as well (e.g., in the opposite direction works just as well).

Also note that although Boost lexical_cast started out as just writing to a stringstream, then extracting back out of the stream, it now has a couple of additions. First of all, specializations for quite a few types have been added, so for many common types, it's substantially faster than using a stringstream. Second, it now checks the result, so (for example) if you convert from a string to an int, it can throw an exception if the string contains something that couldn't be converted to an int (e.g., 1234 would succeed, but 123abc would throw).

As of C++11, there's a std::to_string function overloaded for integer types, so you can use code like:

int a = 20;
std::string s = to_string(a);

The standard defines these as being equivalent to doing the conversion with sprintf (using the conversion specifier that matches the supplied type of object, such as %d for int), into a buffer of sufficient size, then creating an std::string of the contents of that buffer.

answered 7 years ago Throwback1986 #4

sprintf() is pretty good for format conversion. You can then assign the resulting C string to the C++ string as you did in 1.

answered 7 years ago Kevin #5

If you have Boost installed (which you should):

#include <boost/lexical_cast.hpp>

int num = 4;
std::string str = boost::lexical_cast<std::string>(num);

answered 7 years ago Matthieu M. #6

C++11 introduces std::stoi (and variants for each numeric type) and std::to_string, the counterparts of the C atoi and itoa but expressed in term of std::string.

#include <string> 

std::string s = std::to_string(42);

is therefore the shortest way I can think of. You can even omit naming the type, using the auto keyword:

auto s = std::to_string(42);

Note: see [string.conversions] (21.5 in n3242)

answered 6 years ago Rasoul #7

I usually use the following method:

#include <sstream>

template <typename T>
  std::string NumberToString ( T Number )
     std::ostringstream ss;
     ss << Number;
     return ss.str();

described in details here.

answered 5 years ago user2047065 #8

char * bufSecs = new char[32];
char * bufMs = new char[32];

answered 5 years ago Gilad Darmon #9

namespace std
    inline string to_string(int _Val)
    {   // convert long long to string
        char _Buf[2 * _MAX_INT_DIG];
        snprintf(_Buf, "%d", _Val);
        return (string(_Buf));

you can now use to_string(5)

answered 5 years ago user2287915 #10

Wouldn't it be easier using stringstreams?

#include <sstream>

int x=42;            //The integer
string str;          //The string
ostringstream temp;  //temp as in temporary
str=temp.str();      //str is temp as string

Or make a function:

#include <sstream>

string IntToString (int a)
    ostringstream temp;
    return temp.str();

answered 5 years ago Alex #11

First include:

#include <string>
#include <sstream>

Second add the method:

template <typename T>
string NumberToString(T pNumber)
 ostringstream oOStrStream;
 oOStrStream << pNumber;
 return oOStrStream.str();

Use the method like this:



int x = 69;
string vStr = NumberToString(x) + " Hello word!."

answered 4 years ago J_C #12

I use:

int myint = 0;
long double myLD = 0.0;

string myint_str = static_cast<ostringstream*>( &(ostringstream() << myint) )->str();
string myLD_str = static_cast<ostringstream*>( &(ostringstream() << myLD) )->str();

It works on my windows and linux g++ compilers.

answered 4 years ago Tur1ng #13

Using CString:

int a = 10;
CString strA;
strA.Format("%d", a);

answered 4 years ago vitaut #14

You can use std::to_string available in C++11 as suggested by Matthieu M.:


Or, if performance is critical (for example, if you do lots of conversions), you can use fmt::FormatInt from the C++ Format library to convert an integer to std::string:


Or a C string:

fmt::FormatInt f(42);

The latter doesn't do any dynamic memory allocations and is more than 10 times faster than std::to_string on Boost Karma benchmarks. See Fast integer to string conversion in C++ for more details.

Unlike std::to_string, fmt::FormatInt doesn't require C++11 and works with any C++ compiler.

Disclaimer: I'm the author of the C++ Format library.

answered 3 years ago Abdullah Abdelmonem #15

string number_to_string(int x){
    if(!x) return "0";
        string s,s2;
            s.push_back(x%10 + '0');
    return s;

answered 2 years ago maverick9888 #16


#define convertToString(x) #x

int main()
    convertToString(42); // Returns const char* equivalent of 42

answered 2 years ago AdmiralSmith #17

You use a counter type of algorithm to convert to a string. I got this technique from programming Commodore 64 computers. It is also good for game programming.

  • You take the integer and take each digit that is weighted by powers of 10. So assume the integer is 950.

    • If the integer equals or is greater than 100,000 then subtract 100,000 and increase the counter in the string at ["000000"];
      keep doing it until no more numbers in position 100,000. Drop another power of ten

    • If the integer equals or is greater than 10,000 then subtract 10,000 and increase the counter in the string at ["000000"] + 1 position;
      keep doing it until no more numbers in position 10,000.

  • Drop another power of ten

  • Repeat pattern

I know 950 is too small to use as an example but I hope you get the idea.

answered 2 years ago Alek #18

If you need fast conversion of an integer with a fixed number of digits to char* left-padded with '0', this is a convenient example:

int n = 27;
char s[8];

If you are converting a two-digit number:

*(int32_t*)s = 0x3030 | (n/10) | (n%10) << 8;

If you are converting a three-digit number:

*(int32_t*)s = 0x303030 | (n/100) | (n/10%10) << 8 | (n%10) << 16;

If you are converting a four-digit number:

*(int64_t*)s = 0x30303030 | (n/1000) | (n/100%10)<<8 | (n/10%10)<<16 | (n%10)<<24;

And so on up to seven-digit numbers :)

answered 1 year ago Josh Detwiler #19

For C++98, there's a few options:


Boost is not a part of the C++ library, but contains many useful library extensions.

The lexical_cast function template offers a convenient and consistent form for supporting common conversions to and from arbitrary types when they are represented as text.
-- Boost's Documentation

#include "boost/lexical_cast.hpp"
#include <string>

int main() {
    int x = 5;
    std::string x_str = boost::lexical_cast<std::string>(x);
    return 0;

As for runtime, the lexical_cast operation takes about 80 microseconds (on my machine) on the first conversion, and then speeds up considerably afterwards if done redundantly.


This function is not defined in ANSI-C and is not part of C++, but is supported by some compilers.
-- cplusplus.com

This means that gcc/g++ cannot compile code using itoa.

#include <stdlib.h>

int main() {
    int x = 5;
    char * x_str = new char[2];
    x_str = itoa(x, x_str, 10); // base 10
    return 0;

No runtime to report. I don't have Visual Studio installed, which is reportedly able to compile itoa.


sprintf is a C standard library function that works on C strings, and is a perfectly valid alternative.

Composes a string with the same text that would be printed if format was used on printf, but instead of being printed, the content is stored as a C string in the buffer pointed by str.
-- cplusplus.com

#include <stdio.h>

int main() {
    int x = 5;
    char * x_str = new char[2];
    int chars_written = sprintf(x_str, "%d", x);
    return 0;

The stdio.h header may not be necessary. As for runtime, the sprintf operation takes about 40 microseconds (on my machine) on the first conversion, and then speeds up considerably afterwards if done redundantly.


This is the C++ library's main way of converting integers to strings, and vice versa. There are similar sister functions to stringstream that further limit the intended use of the stream, such as ostringstream. Using ostringstream specifically tells the reader of your code that you only intend to use the << operator, essentially. This function is all that's particularly necessary to convert an integer to a string. See this question for a more elaborate discussion.

#include <sstream>
#include <string>

int main() {
    int x = 5;
    std::ostringstream stream;
    stream << x;
    std::string x_str = stream.str();
    return 0;

As for runtime, the ostringstream operation takes about 71 microseconds (on my machine), and then speeds up considerably afterwards if done redundantly, but not by as much as the previous functions.

Of course there are other options, and you can even wrap one of these into your own function, but this offers an analytical look at some of the popular ones.

answered 10 months ago YesThatIsMyName #20

Using stringstream for number conversion is dangerous!

See http://www.cplusplus.com/reference/ostream/ostream/operator%3C%3C/ where it tells that operator<< inserts formatted output.

Depending on your current locale an integer greater than 3 digits, could convert to a string of 4 digits, adding an extra thousands separator.

E.g., int = 1000 could be convertet to a string 1.001. This could make comparison operations not work at all.

So I would strongly recommend using the std::to_string way. It is easier and does what you expect.

answered 5 months ago AndreyS Scherbakov #21

It's rather easy to add some syntactical sugar that allows one to compose strings on the fly in a stream-like way

#include <string>
#include <sstream>

struct strmake {
    std::stringstream s;
    template <typename T> strmake& operator << (const T& x) {
        s << x; return *this;
    operator std::string() {return s.str();}

Now you may append whatever you want (provided that an operator << (std::ostream& ..) is defined for it) to strmake() and use it in place of an std::string.


#include <iostream>

int main() {
    std::string x =
      strmake() << "Current time is " << 5+5 << ":" << 5*5 << " GST";
    std::cout << x << std::endl;

answered 4 months ago Natesh bhat #22

Here's another easy way to do

char str[100] ; 
sprintf(str , "%d" , 101 ) ;  
string s = str; 

sprintf is a well known one to insert any data into a string of required format .

You can convert char * array to string as shown in the third line.

answered 3 weeks ago Tayyab Mazhar #23

#include "stdafx.h"

std::string intToString(int num);

int main()

    int integer = 4782151;

    std::string integerAsStr = intToString(integer);

    std::cout << "integer = " << integer << std::endl;
    std::cout << "integerAsStr = " << integerAsStr << std::endl;

    return 0;

std::string intToString(int num)
    std::string numAsStr;

    while (num)
        char toInsert = (num % 10) + 48;
        numAsStr.insert(0, 1, toInsert);

        num /= 10;

    return numAsStr;

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