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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 an 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();
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)
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.
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).
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.
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.
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);
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();
}
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.
sprintf() is pretty good for format conversion. You can then assign the resulting C string to the C++ string as you did in 1.
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.
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!."
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;
}
C++17 provides std::to_chars as a higher-performance locale-independent alternative.
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.
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;
}
Use:
#define convertToString(x) #x
int main()
{
convertToString(42); // Returns const char* equivalent of 42
}
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.
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.
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 the required format.
You can convert a char * array to a string as shown in the third line.
If you're using MFC, you can use CString:
int a = 10;
CString strA;
strA.Format("%d", a);
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
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;
}
All you have to do is use String when defining your variable (String intStr). Whenever you need that variable, call whateverFunction(intStr.toInt())
Using the plain standard stdio header, you can cast the integer over sprintf into a buffer, like so:
#include <stdio.h>
int main()
{
int x = 23;
char y[2]; // The output buffer
sprintf(y, "%d", x);
printf("%s", y)
}
Remember to take care of your buffer size according to your needs (the string output size).
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;
}
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;
}
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;
}
char * bufSecs = new char[32];
char * bufMs = new char[32];
sprintf(bufSecs, "%d", timeStart.elapsed()/1000);
sprintf(bufMs, "%d", timeStart.elapsed()%1000);
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).
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.
I'm working on Eclipse, on Fedora. I want to make a number more understandable using a dot to separate the thousands. This number is an integer Value that can be 0<Value<99.999
So... if Value = 1000 it shows 1000 and what I want is to show 1.000 (note the dot). The code I need to change is the next one:
char str[12];
sprintf(str, "%d", (int)(Value));
m_Text.SetText(str);
And what I thought is to do something like:
if (Value > 999)
{
int RightPart = (int)Value % 1000;
int LeftPart = Val/1000 ;
}
And then append in a string LeftPart + "." + RightPart so if Value = 1563 it will be 1+.+563 which is 1.563
My problem is that it's very ugly to do this and I was wondering if there were a better way of doing it. I've searched on google and found imbueand localebut they are only for cout. I've see too some posts like this and this but this doesn't help me with my problem.
Thank you.
NOTE: I want to remark that I DON'T WANT to change the output format. I want to change the int I receive so I can have the dots in the str var with which I will work later.
NOTE 2: Basically the code has to: receive an integer (Value), and send it like a string to setText(). setText() will basically print it on the screen where and when it has to, and I want it to print 1.563 and not 1563 which is more difficult to read.
Use stringstream and same imbue technique for it.
For example:
struct dotted : std::numpunct<char> {
char do_thousands_sep() const { return '.'; } // separate with dots
std::string do_grouping() const { return "\3"; } // groups of 3 digits
static void imbue(std::ostream &os) {
os.imbue(std::locale(os.getloc(), new dotted));
}
};
and then:
std::stringstream ss;
dotted::imbue(ss);
ss << Value;
std::cout << ss.str();
Demo
#include <iomanip>
#include <locale>
#include <iostream>
#include <sstream>
using namespace std;
template<class T>
string format(T value)
{
stringstream ss;
ss.imbue(locale(""));
ss << fixed << value;
return ss.str();
}
int main(int argc, char argv[])
{
cout.imbue(locale(""));
cout << 1000000 << endl;
return 0;
}
Prints:
1.000.000
You should probably look at which locale is used when "" is passed and change it with a more appropriate one.
Or vice versa:
int str_to_nr(string str)
{
int val;
stringstream ss(str);
ss.imbue(locale(""));
ss >> val;
return val;
}
int main(int argc, char argv[])
{
cout << str_to_nr("1.000") << endl;
return 0;
}
Prints:
1000
#gandgandi's answer basically gives you a technique to print out the value with dots on demand using C++ streams.
sprintf (which is generally not safe to use, BTW) is a C library function. C does not provide a mechanism to alter the locale behavior in the same way C++ provides. In C, you will have to define the locale behavior with mechanisms/tools provided by your operating system and assign it a name. Then, you can use the C function setlocale to alter the locale in the program to the one you have created to get it to print how you want. However, this will only work on machines that have the locale you have created installed. It won't work on any other machine.
With only a little bit of work and almost no creative effort, you can just use the C++ solution already provided combined with a call to sprintf.
char str[12];
stringstream ss;
dotted::imbue(ss);
ss << Value;
sprintf(str, "%s", ss.str());
m_Text.SetText(str);
But, there is really no reason to use sprintf at all:
stringstream ss;
dotted::imbue(ss);
ss << Value;
m_Text.SetText(ss.str());
I need to create files with generated names. I use boost::lexical_cast to transform integers to std::string. Is it a possibility to get string with padding zeros;
I have no c++11 tools, just everything that MSVS 2008 supports.
Example :
int i = 10;
std::string str = boost::lexical_cast<std::string>(i);
// str = "10"
// expect str = "000010"
p.s. don't suggest to use sprintf please.
Why boost::lexical_cast? Use std::stringstream
std::ostringstream ss;
ss << std::setw(6) << std::setfill('0') << i;
const std::string str = ss.str();
You could use std::ostringstream with the normal stream manipulators for formatting.
I have tried to convert a color code by reading a file, retrieve the color code and store it as a string. This works, but when I tried to just simply convert it to an int, it doesn't work - always getting 0 when I do a cout.
string value = "0xFFFFFF";
unsigned int colorValue = atoi(value.c_str());
cout << colorValue << endl;
as you can see, the color I've got is 0xFFFFFF, but converting it to an int will only give me 0. Can someone please tell me what I'm missing or what i'm doing wrong?
Thanks
I suggest using stringstreams:
std::string value = "0xFFFFFF";
unsigned int colorValue;
std::stringstream sstream;
sstream << std::hex << value;
sstream >> colorValue;
cout << colorValue << endl;
As #BartekBanachewicz says, atoi() is NOT the C++ way of doing this. Leverage the power of C++ streams and use std::istringstream to do it for you. See this.
An excerpt:
template <typename DataType>
DataType convertFromString(std::string MyString)
{
DataType retValue;
std::stringstream stream;
stream << std::hex << MyString; // Credit to #elusive :)
stream >> retValue;
return retValue;
}
What is the easiest way to convert from int to equivalent string in C++? I am aware of two methods. Is there an 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();
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)
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.
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).
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.
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.
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);
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();
}
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.
sprintf() is pretty good for format conversion. You can then assign the resulting C string to the C++ string as you did in 1.
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.
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!."
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;
}
C++17 provides std::to_chars as a higher-performance locale-independent alternative.
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.
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;
}
Use:
#define convertToString(x) #x
int main()
{
convertToString(42); // Returns const char* equivalent of 42
}
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.
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.
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 the required format.
You can convert a char * array to a string as shown in the third line.
If you're using MFC, you can use CString:
int a = 10;
CString strA;
strA.Format("%d", a);
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
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;
}
All you have to do is use String when defining your variable (String intStr). Whenever you need that variable, call whateverFunction(intStr.toInt())
Using the plain standard stdio header, you can cast the integer over sprintf into a buffer, like so:
#include <stdio.h>
int main()
{
int x = 23;
char y[2]; // The output buffer
sprintf(y, "%d", x);
printf("%s", y)
}
Remember to take care of your buffer size according to your needs (the string output size).
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;
}
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;
}
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;
}
char * bufSecs = new char[32];
char * bufMs = new char[32];
sprintf(bufSecs, "%d", timeStart.elapsed()/1000);
sprintf(bufMs, "%d", timeStart.elapsed()%1000);
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).
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.