How to convert int to string in C++?

Asked 2023-09-20 20:57:41 View 247,718

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();
  • I think both methods you gave are good solutions. it depends on the context where you need to do it. If you're already working with streams, for example reading or writing a file, then your second method is the best. If you need to pass an int as a string to a function argument, then itoa could be an easy way. But most of the time, int to string conversion occurs when dealing with files, so streams are appropriate. - anyone
  • How does option 1 even work for you at all? It's my understanding that itoa() takes three parameters. - anyone
  • itoa will be faster than the stream equivalent. There are also ways of re-using the string buffer with the itoa method (avoiding heap allocations if you are frequently generating strings. e.g. for some rapidly updating numerical output). Alternatively you can generate a custom streambuf to reduce some of the allocation overhead etc. Constructing the stream in the first place is also not a low cost venture. - anyone
  • @Pete: Once you start worrying about which is faster, you'll want to look at stackoverflow.com/questions/4351371/… - anyone
  • Note that itoa() is not part of the standard and therefore using it renders your code not portable since not all compilers support it. For Linux you are most certainly out unless you are using something else than GCC, which does not support this function. If you have C++0x, go with what @Matthieu has suggested in his answer. If that's not the case, go with stringstream since it is a well supported feature and your code should be compatible with every C++ compiler out there. As an alternative you can always go with sprintf(). - anyone

Answers

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   2023-09-20 20:57:41

  • to_string not a member of std fix: stackoverflow.com/questions/12975341/… - anyone
  • Or depending on your compiler, just set the right language standard: g++ -std=c++11 someFile.cc - anyone
  • @Steve: it's supposed to be. It's a member of std in every compiler I know of except for one. - anyone
  • @Matthiew M. I am using the same which you suggest but i am getting this error : Error : No instance of overloaded function "std::to_string" matches the argument list i am using VS2010 c++ - anyone
  • @Flying: under VS2010 you have to explicitly cast the converting integer to one of the following types [_Longlong, _ULonglong, long double]; i.e: string s = to_string((_ULonglong)i); - anyone

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.

Answered   2023-09-20 20:57:41

  • I am not very familiar with dynamic_cast but I am using clang to compile so it complains about it. If I just omit the dynamic_cast then it compiles fine; what purpose does the dynamic_cast serve in this case? We are already creating an ostringstream, so why cast it? - anyone
  • @Mathew: The link in my answer leads to a detailed description of each part of the construct. While we created a ostringstream, we called operator<<() on it, which returns ostream & -- for which .str() is not defined. I really wonder how clang would make this work without the cast (or why it generates an error with it). This construct is published in many places, and I've used it for over a decade on many different compilers, including MSVC, GCC, and XLC, so I am rather surprised clang balks at it. - anyone
  • Just came to the party for curiosity, and downvoted. Reason : too much votes for a solution that's un-elegant, and likely slow. 1. macro usage. I don't systematically frown on any macro, but this one is too short, and end-clients always fear repetition of the argument, on top of fear for unprotected multilines macros. (not protected by do{}while(0)) 2. dynamic_cast. it seems you only need a static_cast here, unless you want to assert that the library indeed is implemented as you hope. in which case you should use boost::polymorphic_downcast instead. - anyone
  • @v.oddou: You're free to critizise, of course. But 1. is invalid -- the macro is a single statement, do { } while( 0 ) would not add anything. With 2. and 3. you probably got a point -- this could be done with a static cast, and perhaps one of you template wizards out there could come up with a "nicer" interface. But as I said, this is by no means an invention of myself. Look around, this macro (macro!) is quite ubiquitous. That's a case of POLA in itself. I might toy with this a bit to make it more "streamlined". - anyone
  • @v.oddou: Look at what I found among the things C++17 brought us. :-) I hope you like the updated answer. - anyone

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).

Answered   2023-09-20 20:57:41

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.

Answered   2023-09-20 20:57:41

  • @lifebalance: You do not need to clear() a newly created ostringstream object. clear() resets the error/eof flags, and there has not been any error/eof condition generated yet. - anyone
  • @Rasoul NumberToString(23213.123) produces 23213.1 while std::to_string(23213.123) produces 23213.123000 What happens there? - anyone
  • I wouldn't normally write a function for three lines, but it does make it self documenting via the name. Clarity vs self documenting -- those shouldn't conflict, right? - anyone
  • @KillzoneKid This is because std' ostream are stateful (this means that any previous state change is kept, like the number of decimal digit) while this method starts with a default state. - anyone
  • Use .flags(...) to read & clear formatting flags, and .str("") to clear an existing string. - anyone

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.

Answered   2023-09-20 20:57:41

  • I was curious about the claim of not having any dynamic memory allocation while remain threadsafe (re-entrant), so I read your code -- the c_str() returns a pointer to a buffer declared inside the fmt::FormatInt class -- so the pointer returned will be invalid at the semicolon -- see also stackoverflow.com/questions/4214153/lifetime-of-temporaries - anyone
  • Yes, the same behavior as with std::string::c_str() (thus the naming). If you want to use it outside of the full expression construct an object FormatInt f(42); Then you can use f.c_str() without a danger of it being destroyed. - anyone
  • I get something weird when I try to convert from int to string using std::to_string(num). If I store the result in a variable and try to Access it like stringNum[1] or stringNum[n] as n increases, I get garbage. - anyone

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   2023-09-20 20:57:41

  • Agreed on boost installation. I think that more than often one would format the string. For this purpose I prefer boost::format e.g format("%02d", number ).str() - anyone

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();
}

Answered   2023-09-20 20:57:41

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   2023-09-20 20:57:41

  • itoa() is not a standard function! - anyone
  • @cartoonist: Then what is it? - anyone
  • This function is not defined in ANSI-C and C++. So it's not supported by some compiler such as g++. - anyone
  • cplusplus.com/reference/cstdlib/itoa: "Portability This function is not defined in ANSI-C and is not part of C++, but is supported by some compilers." gcc.godbolt.org/z/MaEEf4evW - anyone
  • @cartoonist I think it has to do with glibc, not g++. I wish they just added all these functions to glibc. Would make things easier for people - anyone

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   2023-09-20 20:57:41

  • Heh, yes. However, I usually rely on snprintf() and friends for anything of consequence when handling C strings. - anyone
  • @MatthieuM. Your comment proves further, that you are not. If the output was truncated due to this limit then the return value is the number of characters (excluding the terminating null byte) which would have been written to the final string if enough space had been available. Thus, a return value of size or more means that the output was truncated. So you call it with a NULL and zero size to get the necessary buffer size. - anyone
  • @user1095108: I think you are mistaking snprintf (note the SNP prefix) and sprintf (note the SP prefix). You pass the size to the former, and it takes care not to overflow, however the latter knows not the size of the buffer and thus may overflow. - anyone
  • The idea is to call a snprintf variant first and a sprintf variant after that. As the buffer size is known by then, calling sprintf becomes entirely safe. - anyone
  • @user1095108 Ah, yes. If the first buffer to snprintf is insufficient, the return value tells you what would be sufficient. I would still use snprintf for the second call, because relying on the implementations of sprintf and snprintf to match is unnecessarily dangerous. - anyone

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.

Answered   2023-09-20 20:57:41

  • I agree that this is a serious problem if you need to exchange data. Unfortunately, also std::to_string uses the current locale (see en.cppreference.com/w/cpp/string/basic_string/to_string , the 'Notes' section). Almost all standard tools (from stringstreams to sprintf, but also sscanf etc) are using the current locale. I wasn't aware of this until recently when it hit me hard. Currently using home-grown stuff, not hard to make. - anyone
  • In the link above it is also statet that C++17 provides std::to_chars as a higher-performance locale-independent alternative. - anyone
  • Unfortunately, I am stuck with C++11 for the coming year(s) (quite an improvement already, luckily). - anyone
  • from_chars and to_chars would be perfect but unfortunately they didn't offer a wchar_t variant. - anyone

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!."

Answered   2023-09-20 20:57:41

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

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

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

    return 0;
}

Answered   2023-09-20 20:57:41

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

Answered   2023-09-20 20:57:41

  • For performance intensive applications you need this one and not the stringstream route. - anyone

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.

Answered   2023-09-20 20:57:41

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;
}

Answered   2023-09-20 20:57:41

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.

Answered   2023-09-20 20:57:41

  • While this code may resolve the OP's issue, it is best to include an explanation as to how your code addresses the OP's issue. In this way, future visitors can learn from your post, and apply it to their own code. SO is not a coding service, but a resource for knowledge. Also, high quality, complete answers are more likely to be upvoted. These features, along with the requirement that all posts are self-contained, are some of the strengths of SO as a platform, that differentiates it from forums. You can edit to add additional info &/or to supplement your explanations with source documentation. - anyone

Use:

#define convertToString(x) #x

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

Answered   2023-09-20 20:57:41

  • Works only with literal numbers, doesn't evaluate variable content, though useful sometimes. - anyone
  • Right. But definitely handy at time - anyone
  • Only works at compile time with literal constants numbers, i think the OP asks for a dynamic conversion, using variable integers - anyone

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

Answered   2023-09-20 20:57:41

  • Hi! Could you add an explanation on why and how this provides an answer to the question? - anyone

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
std::string ✔️ universal method (for formattable types)
✔️ supports common bases, and locale
❌ slow to compile
❌ slow (forward to std::vformat)
std::to_chars
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
std::string ✔️ concise and self-explanatory
✔️ zero overhead (if you need a std::string)
❌ only works for fundamental types
❌ base 10 only
std::ostringstream
std::string ✔️ universal method (for types with << operator)
✔️ considers locale (e.g. can change base)
❌ slow, and high overhead of streams
std::sprintf
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"
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;

Answered   2023-09-20 20:57:41

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   2023-09-20 20:57:41

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.

Answered   2023-09-20 20:57:41

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

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

Answered   2023-09-20 20:57:41

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;
}

Answered   2023-09-20 20:57:41

  • Thank you for this code snippet, which might provide some limited short-term help. A proper explanation would greatly improve its long-term value by showing why this is a good solution to the problem, and would make it more useful to future readers with other, similar questions. Please edit your answer to add some explanation, including the assumptions you've made. - anyone
  • Completely fails for n≤0 - anyone
  • Add comments, explain your answer, read how to answer. - anyone

Use:

#include<iostream>
#include<string>

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;
    bool isNegative = num < 0;
    if(isNegative) num*=-1;

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

       num /= 10;
    }while (num);
  
    return isNegative? numAsStr.insert(0, 1, '-') : numAsStr;
}

Answered   2023-09-20 20:57:41

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

Answered   2023-09-20 20:57:41

  • Can you provide a reference to toInt()? Where is "String" (uppercase "s") defined? Part of some vendor-specfic library? Are you sure it is C++? It is in the Arduino environment, but that may not count. It doesn't appear to be part of the standard string class. - anyone
  • What did you test it on? Compiler, incl. version. Environment (operating system, IDE, etc.)? C++ version? - anyone

This worked for me -

My code:

#include <iostream>
using namespace std;

int main()
{
    int n = 32;
    string s = to_string(n);
    cout << "string: " + s  << endl;
    return 0;
}

Answered   2023-09-20 20:57:41

char * bufSecs = new char[32];
char * bufMs = new char[32];
sprintf(bufSecs, "%d", timeStart.elapsed()/1000);
sprintf(bufMs, "%d", timeStart.elapsed()%1000);

Answered   2023-09-20 20:57:41

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   2023-09-20 20:57:41

  • While this solution works, it is highly discouraged! Names starting with underscore and a capital letter are reserved for the compiler, you shouldn't ever use them. Injecting functions into the std namespace is not something you should ever do, either. Also, it doesn't seem like _MAX_INT_DIG is a standard macro, so if it is defined wrongly, this code has the great potential of inducing undefined behaviour. -1 - anyone
  • What is _MAX_INT_DIG and why is it doubled? - anyone

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.

Answered   2023-09-20 20:57:41

  • It's not very helpful to describe an algorithm, rather than show an example in code. - anyone

I think using stringstream is pretty easy:

 string toString(int n)
 {
     stringstream ss(n);
     ss << n;
     return ss.str();
 }

 int main()
 {
    int n;
    cin >> n;
    cout << toString(n) << endl;
    return 0;
 }

Answered   2023-09-20 20:57:41

  • This was mentioned in the question, and unfortunately is quite slow. - anyone