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).
Sample query:
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?
(1)
int a = 10;
char *intStr = itoa(a);
string str = string(intStr);
(2)
int a = 10;
stringstream ss;
ss << a;
string str = ss.str();How to convert int to string in C++?
It’s very easy to convert an integer to string in C++ now.
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 to name the type, using the auto keyword:
auto s = std::to_string(42);Answer #2:
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();
}
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 ) );
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.
Answer #3:
I usually use the following method:
#include <sstream>
template <typename T>
std::string NumberToString ( T Number )
{
std::ostringstream ss;
ss << Number;
return ss.str();
}Answer #4:
You can use std::to_string available in C++11 as suggested above:
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:
fmt::format_int(42).str();
Or a C string:
fmt::format_int f(42);
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.
Answer #5:
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 #6:
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.
Answer #7:
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 #8:
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 #9:
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.
Hope you learned something from this post.
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