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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 trying to split a string into an array of individual characters. However, I would like the string to be input by the user, for which I need to define the string using a variable.
My question is, why does this work:
#include <iostream>
using namespace std;
int main() {
char arr [] = {"Giraffe"};
cout << arr[0];
return 0;
}
But this doesn't?
#include <iostream>
using namespace std;
int main() {
string word;
word = "Giraffe";
char arr [] = {word};
cout << arr[0];
return 0;
}
Thanks
Your example doesn't work because you're trying to put a std::string into an array of char. The compiler will complain here because std::string has no type conversion to char.
Since you're just trying to print the first character of the string, just use the array accessor overload of std::string, std::string::operator[] instead:
std::string word;
word = "Giraffe";
std::cout << word[0] << std::endl;
In your second example, the type of word is a std::string and there are no default type conversions from std::string to the type char.
On the other hand, the first example works because it can be interpreted as an array of char (but actually its just c-style const char *).
If, for some reason, you would want to convert std::string into the c-style char array, you might want to try something like this...
#include <iostream>
#include <string>
#include <cstring>
int main(void)
{
std::string word;
word = "Giraffe";
char* arr = new char[word.length() + 1]; // accounting for the null-terminating character
strcpy(arr, word.data());
std::cout << arr[0] << std::endl;
delete[] arr; // deallocating our heap memory
return 0;
}
I have a question about a char array:
I have a form '"1"+lapcounter+":"+seconds' that must come in a char array.
How can i fill this array in this form?
Thanks
If you mean you have some numeric variables which you want to format into a string, use a string-stream for that:
std::stringstream ss;
ss << "1" << lapcounter << ":" << seconds";
Now you can extract a string from that:
std::string s = ss.str();
and if you really want a character array for some reason (which I'm sure you don't)
char const * cs = s.c_str();
Use sprintf, or snprintf. This function works similar to printf but instead of standard output, the output will go to char array you specified. For example:
char buffer[32];
snprintf(buffer, sizeof(buffer), "1%d:%d", lapcounter, seconds);
to_string is used like this:
#include <iostream>
#include <string>
int main()
{
int lapcounter = 23;
std::string str("1");
str.append(std::to_string(lapcounter ));
str.append(":seconds");
std::cout << str << std::endl;
}
prints
123:seconds
if you really need a char array you get that from ss.c_str()
I have a string which actually contains a number and a string, separated by ,, for instance "12,fooBar".
I would like to put it into separated variables, i.e. the number into unsigned int myNum and the string into std::string myStr.
I have the following snipped of code:
size_t pos1=value.find(',');
std::cout << value.substr(0, pos1) << " and "
<< (value.substr(0, pos1)).c_str() << std::endl;
This yields 12 and 1. Anything I missed here? What happend to the 2 in the second part?
Note: I isolated the problem to this snipped of code. I need c_str() to pass it to atoi to get the unsigend int. Here I don't want to print the second part.
Update: I actually get the string from levelDB Get. If I put a test string like I put here, it works.
The posted code produces the same substring: value.substr(0, pos1). Note that std::string::substr() does not modify the object, but returns a new std::string.
Example:
#include <iostream>
#include <string>
int main ()
{
std::string value ="12,fooBar";
unsigned int myNum;
std::string myStr;
const size_t pos1 = value.find(',');
if (std::string::npos != pos1)
{
myNum = atoi(value.substr(0, pos1).c_str());
myStr = value.substr(pos1 + 1);
}
std::cout << myNum << " and "
<< myStr << std::endl;
return 0;
}
Output:
12 and fooBar
EDIT:
If the unsigned int is the only piece required then the following will work:
unsigned int myNum = atoi(value.c_str());
as atoi() will stop at the first non-digit character (excluding optional leading - or +), in this case the ,.
The cleanest C++ style solution to this problem is to use a stringstream.
#include <sstream>
// ...
std::string value = "12,fooBar";
unsigned int myNum;
std::string myStr;
std::stringstream myStream(value);
myStream >> myNum;
myStream.ignore();
myStream >> myStr;
Your second substr should be value.substr(pos1+1,value.length())
One more option is using std::from_chars function from the 17th standard (< charconv > header):
int x;
from_chars(&s[i], &s.back(), x); // starting from character at index i parse
// the nearest interger till the second char pointer
There are different overloads for different types of value x (double etc.).
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.