Easiest way to convert int to string in C++

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Top 5 Answer for Easiest way to convert int to string in C++

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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)

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

C++17 update:

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.

// variadic template template < typename... Args > std::string sstr( Args &&... args ) {     std::ostringstream sstr;     // fold expression     ( 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 (C++98) 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.

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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.

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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).

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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); 

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