How to convert int to string in C++?

Question

How can I convert from int to the equivalent string in C++? I am aware of two methods. Is there another way?

(1)

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

(2)

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

Answer 1

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

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

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

std::string s = fmt::format_int(42).str();

Or a C string:

fmt::format_int f(42);
const char* s = f.c_str();

The latter doesn't do any dynamic memory allocations and is more than 70% faster than libstdc++ implementation of std::to_string on Boost Karma benchmarks. See Converting a hundred million integers to strings per second for more details.

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

Answer 2

In C++11 we can use the "to_string()" function to convert an int into a string:

#include <iostream>
#include <string>

int main() {
    int x = 1612;
    std::string s = std::to_string(x);
    std::cout << s << std::endl;

    return 0;
}

Answer 3

C++ has evolved over time, and with it the methods to convert an int to a string. I will provide a summary in this answer. Note that some methods don't give you a std::string directly, but a char*. You can easily convert char* to the former, and in some cases it's beneficial to avoid std::string.

Comparison

The following table compares all options (only C++ standard options, no third-party libraries) from most recent to least recent.

Method	Result	Pros & Cons
std::format c++20	std::string	✔️ universal method (for formattable types) ✔️ supports common bases, and locale ❌ slow to compile ❌ slow (forward to std::vformat)
std::to_chars c++17	written to char[]	✔️ fast and zero overhead (no dynamic allocations) ✔️supports ANY base as run-time argument ❌ only works for fundamental types ❌ interface is not ergonomic
std::to_string c++11	std::string	✔️ concise and self-explanatory ✔️ zero overhead (if you need a std::string) ❌ only works for fundamental types ❌ base 10 only
std::ostringstream c++98	std::string	✔️ universal method (for types with << operator) ✔️ considers locale (e.g. can change base) ❌ slow, and high overhead of streams
std::sprintf c++98	written to char[]	✔️ smallest assembly output ✔️ supports some bases ✔️ compatible with C, unlike all other methods ❌ only works for fundamental types ❌ interface is not ergonomic ❌ no type safety

Recommended Practice

Use std::to_string if you just need to turn an int into a decimal string. It's simple, elegant, and correct.

If you can't use std::to_string, choose another option based on the features you need. Prefer more modern solutions like std::to_chars over older solutions like std::sprintf.

Examples

std::format

std::string d = std::format("{}", 100);   // d = "100"
std::string h = std::format("{:#x}", 15); // h = "0xf"

std::to_chars

std::array<char, 5> a;
auto [ptr, ec] = std::to_chars(a.data(), a.data() + a.size(), 1234);
// a = {'1', '2', '3', '4', indeterminate} (no null terminator!)
// ptr points to 4, and ec == std::errc{}
// notice that there is no null terminator
std::string_view view(a.data(), ptr); // string_view doesn't require null terminators
std::string s(a.data(), ptr); // wrapping in a std:string kinda defeats the point

std::to_string

std::string d = std::to_string(100); // d = "100"

std::ostringstream

std::string d = (std::ostringstream() << 100).str();            // d = "100"
std::string h = (std::ostringstream() << std::hex << 15).str(); // h = "0xf"

Note: these one-liners rely on LWG 1203 (C++20), but recent compilers allow it in C++11 mode too. Update your compiler, or create a separate std::ostringstream stream variable if it doesn't work.

sprintf / snprintf

char a[20];
sprintf(a, "%d", 15);                // a = {'1', '5', '\0', ?, ?, ?, ...}
snprintf(a, sizeof(a), "%#x", 15);   // a = {'0', 'x', 'f', '\0', ?, ?, ...}
std::string s = a;

Answer 4

If you're using the Microsoft Foundation Class library, you can use CString:

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

Answer 5

Here's another easy way to do it:

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

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

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

Answer 6

int i = 255;
std::string s = std::to_string(i);

In C++, to_string() will create a string object of the integer value by representing the value as a sequence of characters.

Answer 7

Using stringstream for number conversion is dangerous!

See std::ostream::operator<< where it tells that operator<< inserts formatted output.

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

E.g., int = 1000 could be converted 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.

From std::to_string:

C++17 provides std::to_chars as a higher-performance locale-independent alternative.

Answer 8

If you need fast conversion of an integer with a fixed number of digits to char* left-padded with '0', this is the example for little-endian architectures (all x86, x86_64 and others):

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. In this example n is a given integer. After conversion it's string representation can be accessed as (char*)&s:

std::cout << (char*)&s << std::endl;

Note: If you need it on big-endian byte order, though I did not tested it, but here is an example: for three-digit number it is int32_t s = 0x00303030 | (n/100)<< 24 | (n/10%10)<<16 | (n%10)<<8; for four-digit numbers (64 bit arch): int64_t s = 0x0000000030303030 | (n/1000)<<56 | (n/100%10)<<48 | (n/10%10)<<40 | (n%10)<<32; I think it should work.

Answer 9

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);

Answer 10

C++20: std::format would be the idiomatic way now.

---

C++17:

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

// variadic template
template < typename... Args >
std::string sstr( Args &&... args )
{
    std::ostringstream sstr;
    // fold expression
    ( sstr << std::dec << ... << args );
    return sstr.str();
}

Usage:

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 ) );

---

C++98:

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.

Answer 11

All you have to do is use String when defining your variable (String intStr). Whenever you need that variable, call whateverFunction(intStr.toInt())

Answer 12

Current C++

Starting with 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 = std::to_string(a);
// or: auto s = std::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.

Old C++

For older (pre-C++11) compilers, probably the most common easy way wraps essentially your second choice into a template that's usually 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).

Answer 13

C++17 provides std::to_chars as a higher-performance locale-independent alternative.

Answer 14

C++11 introduced std::to_string() for numeric types:

int n = 123; // Input, signed/unsigned short/int/long/long long/float/double
std::string str = std::to_string(n); // Output, std::string

Answer 15

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 the pattern

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

Answer 16

string number_to_string(int x) {

    if (!x)
        return "0";

    string s, s2;
    while(x) {
        s.push_back(x%10 + '0');
        x /= 10;
    }
    reverse(s.begin(), s.end());
    return s;
}

Answer 17

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.

Answer 18

It would be easier using stringstreams:

#include <sstream>

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

Or make a function:

#include <sstream>

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

Answer 19

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).

Answer 20

I usually use the following method:

#include <sstream>

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

It is described in details here.

Answer 21

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)

Answer 22

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.

Example:

#include <iostream>

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

Answer 23

Use:

#define convertToString(x) #x

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

Answer 24

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:

NumberToString(69);

or

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

Answer 25

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