Related
I want to read an input string and connect their values to variables in my class.
Some example Inputs might be:
78 C 15.48
3 B
87 P 15
0
..
The first argument is an int from 0-100, second a char and third int or float. A String can consist of one, two or three arguments which are separated by space. After reading a line, this program does some calculations and then expects another input until a 0 is entered or a break occurred.
To read an input String, i'm currently using
std::string str;
std::getline(std::cin, str);
My program already has the variables
int firstArgument;
char secondArgument;
float thirdFloatArgument;
int thirdIntArgument;
now, lets say str is: 46 C 87.3
after reading the line my variables should be:
firstArgument = 46;
secondArgument = 'C';
thirdFloatArgument = 87.3;
How can I extract the Information from the input String?
I was thinking about counting the spaces to see how much values are given and then separating the string via this delimiter,as the amount of arguments might vary. So:
int count = 0;
int length = str.length();
for(int i = 0; i < length; i++){
int c = str[i];
if(isspace(c)){
count++;
}
}
with space count being 2 I now know that 3 arguments were passed, but I don't know how to go on from there. Using std:istringstream might be an option but from what I've seen online it is mostly used in a while loop to print each word of a string in a new line or like that. But my input can vary in the amount of arguments so a loop would not work.
I think I need something like: "String before first ' ' is firstArgument, String between first and second ' ' is secondArgument, string after second ' ' is either thirdFloatArgument or thirdIntArgument (respectively if only one or two arguments are given, which can be determined with the amount of spaces). But how would I do this? Or are there some easier approaches?
Big thanks in advance!
As Some programmer dude mentioned it is a good idea to use std::istringstream to convert values from string to other data types. It allows you to treat input string the same way as you treat std::cin. Here is a code snippet that you can use in your case:
#include <string>
#include <iostream>
#include <algorithm>
#include <sstream>
struct Arguments {
int first{};
char second{};
double third{};
};
Arguments parseArgs(const std::string& inputLine) {
Arguments args;
const int argc = std::ranges::count(inputLine, ' ');
std::istringstream stream(inputLine);
if (argc >= 0) {
stream >> args.first;
}
if (argc >= 1) {
stream >> args.second;
}
if (argc >= 2) {
stream >> args.third;
}
return args;
}
int main() {
std::string inputLine{};
std::getline(std::cin, inputLine);
const auto args = parseArgs(inputLine);
std::cout << args.first << ", " << args.second << ", " << args.third << "\n";
}
Note that you have to compile this example using C++20 because I used std::ranges. If you do not have compiler that supports this standard you can use std::count(inputLine.cbegin(), inputLine.cend(), ' ');
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.
How can I parse a string that looks like "xxxx-xxxx" and get those xxxx parts as a number? For an example, the user will type in "9349-2341" and I will get those numbers as two different integers.
I need to do that for a random number generator, which chooses the number between these xxxx variables.
Thanks.
You can use std::stringstream to extract numbers from string. It looks like that:
std::stringstream str_stream;
std::string str_to_parse = "1234-5678";
int num[2];
str_stream << str_to_parse;
str_stream >> num[0];
str_stream.ignore(1); // otherwise it will extract negative number (-5678)
str_stream >> num[1];
Also, there is C functions, like sscanf(). For example, your pattern can be extracted with this format: "%d-%d".
std::string str = "1234-5678";
std::string str1 = str.substr (0,4);
std::string str2 = str.substr(5, 4);
int n1 = std::stoi(str1);
int n2 = std::stoi(str2);
// do your random number generation between n1 and n2
Using regular expression
If your input is assured to resemble "xxxx-xxxx" where 'x' represents a digit, you can simply ultilize the following function:
#include <iostream>
#include <string>
#include <regex>
using namespace std;
int main()
{
string input = "9349-2341";
// This pattern matches any string begining with 4 digits and ending with 4 digits, both parts seperated by a slash
string pattern = "([0-9]{4})-[0-9]{4}";
smatch matcher;
regex prog (pattern);
if (regex_search(input, matcher, prog))
{
auto x = matcher[1];
cout << x << " " << endl;
input = matcher.suffix().str();
}
else
{
cout << "Invalid input!" << endl;
}
return 0;
}
As for how to convert string to number, check out this article, from which the following segment is quoted:
string Text = "456";//string containing the number
int Result;//number which will contain the result
stringstream convert(Text); // stringstream used for the conversion initialized with the contents of Text
if ( !(convert >> Result) )//give the value to Result using the characters in the string
Result = 0;//if that fails set Result to 0
//Result now equal to 456
Or, simply as followed:
Using sscanf
#include <cstdio>
using namespace std;
int main(int argc, char ** argv)
{
char input[] = "1234-5678";
int result, suffix;
sscanf(input, "%i-%i", &result, &suffix);
printf("Output: '%i-%i'.\n", result, suffix);
return 0;
}
You should check out C++ reference websites, such as CPlusPlus.
I am trying to convert an integer to char array and I came across this piece of code
int i = 5;
std::string s;
std::stringstream out;
out << i;
s = out.str();
But when I try to print the value of s it still prints 5. I don't know if its supposed to do that or am I doing something wrong? Besides I would prefer if I could convert the same int to char array. But I would appreciate any help in the matter.
Thanks!
Code taken from: Alternative to itoa() for converting integer to string C++?
Yes, it's supposed to do that. You'd (primarily) notice the difference from just printing a number out directly if you do some other string-type manipulation on the result (e.g., concatenating it with other strings, searching for characters in the string).
Just for example:
std::cout << i+i; // should print "10"
std::cout << s+s; // should print "55"
Besides I would prefer if I could convert the same int to char array.
char *charPtr = new char[ s.length() + 1 ] ; // s is the string in the snippet posted
strcpy( charPtr, s.c_str() ) ;
// .......
delete[] charPtr ; // Should do this, else memory leak.
If you would like to stop worrying about issues like that you might be interested in boost/lexical_cast.hpp.
#include <boost/lexical_cast.hpp>
#include <string>
#include <iostream>
int main() {
const int i=5;
const char* s = boost::lexical_cast<std::string>(i).c_str();
std::cout << s << std::endl;
}
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.