Related
I thought this would be really simple, but it's presenting some difficulties. If I have
std::string name = "John";
int age = 21;
How do I combine them to get a single string "John21"?
In alphabetical order:
std::string name = "John";
int age = 21;
std::string result;
// 1. with Boost
result = name + boost::lexical_cast<std::string>(age);
// 2. with C++11
result = name + std::to_string(age);
// 3. with FastFormat.Format
fastformat::fmt(result, "{0}{1}", name, age);
// 4. with FastFormat.Write
fastformat::write(result, name, age);
// 5. with the {fmt} library
result = fmt::format("{}{}", name, age);
// 6. with IOStreams
std::stringstream sstm;
sstm << name << age;
result = sstm.str();
// 7. with itoa
char numstr[21]; // enough to hold all numbers up to 64-bits
result = name + itoa(age, numstr, 10);
// 8. with sprintf
char numstr[21]; // enough to hold all numbers up to 64-bits
sprintf(numstr, "%d", age);
result = name + numstr;
// 9. with STLSoft's integer_to_string
char numstr[21]; // enough to hold all numbers up to 64-bits
result = name + stlsoft::integer_to_string(numstr, 21, age);
// 10. with STLSoft's winstl::int_to_string()
result = name + winstl::int_to_string(age);
// 11. With Poco NumberFormatter
result = name + Poco::NumberFormatter().format(age);
is safe, but slow; requires Boost (header-only); most/all platforms
is safe, requires C++11 (to_string() is already included in #include <string>)
is safe, and fast; requires FastFormat, which must be compiled; most/all platforms
(ditto)
is safe, and fast; requires the {fmt} library, which can either be compiled or used in a header-only mode; most/all platforms
safe, slow, and verbose; requires #include <sstream> (from standard C++)
is brittle (you must supply a large enough buffer), fast, and verbose; itoa() is a non-standard extension, and not guaranteed to be available for all platforms
is brittle (you must supply a large enough buffer), fast, and verbose; requires nothing (is standard C++); all platforms
is brittle (you must supply a large enough buffer), probably the fastest-possible conversion, verbose; requires STLSoft (header-only); most/all platforms
safe-ish (you don't use more than one int_to_string() call in a single statement), fast; requires STLSoft (header-only); Windows-only
is safe, but slow; requires Poco C++ ; most/all platforms
In C++11, you can use std::to_string, e.g.:
auto result = name + std::to_string( age );
If you have Boost, you can convert the integer to a string using boost::lexical_cast<std::string>(age).
Another way is to use stringstreams:
std::stringstream ss;
ss << age;
std::cout << name << ss.str() << std::endl;
A third approach would be to use sprintf or snprintf from the C library.
char buffer[128];
snprintf(buffer, sizeof(buffer), "%s%d", name.c_str(), age);
std::cout << buffer << std::endl;
Other posters suggested using itoa. This is NOT a standard function, so your code will not be portable if you use it. There are compilers that don't support it.
#include <iostream>
#include <sstream>
std::ostringstream o;
o << name << age;
std::cout << o.str();
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
string itos(int i) // convert int to string
{
stringstream s;
s << i;
return s.str();
}
Shamelessly stolen from http://www.research.att.com/~bs/bs_faq2.html.
This is the easiest way:
string s = name + std::to_string(age);
If you have C++11, you can use std::to_string.
Example:
std::string name = "John";
int age = 21;
name += std::to_string(age);
std::cout << name;
Output:
John21
It seems to me that the simplest answer is to use the sprintf function:
sprintf(outString,"%s%d",name,age);
#include <string>
#include <sstream>
using namespace std;
string concatenate(std::string const& name, int i)
{
stringstream s;
s << name << i;
return s.str();
}
#include <sstream>
template <class T>
inline std::string to_string (const T& t)
{
std::stringstream ss;
ss << t;
return ss.str();
}
Then your usage would look something like this
std::string szName = "John";
int numAge = 23;
szName += to_string<int>(numAge);
cout << szName << endl;
Googled [and tested :p ]
This problem can be done in many ways. I will show it in two ways:
Convert the number to string using to_string(i).
Using string streams.
Code:
#include <string>
#include <sstream>
#include <bits/stdc++.h>
#include <iostream>
using namespace std;
int main() {
string name = "John";
int age = 21;
string answer1 = "";
// Method 1). string s1 = to_string(age).
string s1=to_string(age); // Know the integer get converted into string
// where as we know that concatenation can easily be done using '+' in C++
answer1 = name + s1;
cout << answer1 << endl;
// Method 2). Using string streams
ostringstream s2;
s2 << age;
string s3 = s2.str(); // The str() function will convert a number into a string
string answer2 = ""; // For concatenation of strings.
answer2 = name + s3;
cout << answer2 << endl;
return 0;
}
In C++20 you'll be able to do:
auto result = std::format("{}{}", name, age);
In the meantime you can use the {fmt} library, std::format is based on:
auto result = fmt::format("{}{}", name, age);
Disclaimer: I'm the author of the {fmt} library and C++20 std::format.
If you'd like to use + for concatenation of anything which has an output operator, you can provide a template version of operator+:
template <typename L, typename R> std::string operator+(L left, R right) {
std::ostringstream os;
os << left << right;
return os.str();
}
Then you can write your concatenations in a straightforward way:
std::string foo("the answer is ");
int i = 42;
std::string bar(foo + i);
std::cout << bar << std::endl;
Output:
the answer is 42
This isn't the most efficient way, but you don't need the most efficient way unless you're doing a lot of concatenation inside a loop.
If you are using MFC, you can use a CString
CString nameAge = "";
nameAge.Format("%s%d", "John", 21);
Managed C++ also has a
string formatter.
As a one liner: name += std::to_string(age);
The std::ostringstream is a good method, but sometimes this additional trick might get handy transforming the formatting to a one-liner:
#include <sstream>
#define MAKE_STRING(tokens) /****************/ \
static_cast<std::ostringstream&>( \
std::ostringstream().flush() << tokens \
).str() \
/**/
Now you can format strings like this:
int main() {
int i = 123;
std::string message = MAKE_STRING("i = " << i);
std::cout << message << std::endl; // prints: "i = 123"
}
As a Qt-related question was closed in favour of this one, here's how to do it using Qt:
QString string = QString("Some string %1 with an int somewhere").arg(someIntVariable);
string.append(someOtherIntVariable);
The string variable now has someIntVariable's value in place of %1 and someOtherIntVariable's value at the end.
There are more options possible to use to concatenate integer (or other numerric object) with string. It is Boost.Format
#include <boost/format.hpp>
#include <string>
int main()
{
using boost::format;
int age = 22;
std::string str_age = str(format("age is %1%") % age);
}
and Karma from Boost.Spirit (v2)
#include <boost/spirit/include/karma.hpp>
#include <iterator>
#include <string>
int main()
{
using namespace boost::spirit;
int age = 22;
std::string str_age("age is ");
std::back_insert_iterator<std::string> sink(str_age);
karma::generate(sink, int_, age);
return 0;
}
Boost.Spirit Karma claims to be one of the fastest option for integer to string conversion.
std::ostringstream
#include <sstream>
std::ostringstream s;
s << "John " << age;
std::string query(s.str());
std::to_string (C++11)
std::string query("John " + std::to_string(age));
boost::lexical_cast
#include <boost/lexical_cast.hpp>
std::string query("John " + boost::lexical_cast<std::string>(age));
Here is an implementation of how to append an int to a string using the parsing and formatting facets from the IOStreams library.
#include <iostream>
#include <locale>
#include <string>
template <class Facet>
struct erasable_facet : Facet
{
erasable_facet() : Facet(1) { }
~erasable_facet() { }
};
void append_int(std::string& s, int n)
{
erasable_facet<std::num_put<char,
std::back_insert_iterator<std::string>>> facet;
std::ios str(nullptr);
facet.put(std::back_inserter(s), str,
str.fill(), static_cast<unsigned long>(n));
}
int main()
{
std::string str = "ID: ";
int id = 123;
append_int(str, id);
std::cout << str; // ID: 123
}
Common Answer: itoa()
This is bad. itoa is non-standard, as pointed out here.
You can concatenate int to string by using the given below simple trick, but note that this only works when integer is of single digit. Otherwise, add integer digit by digit to that string.
string name = "John";
int age = 5;
char temp = 5 + '0';
name = name + temp;
cout << name << endl;
Output: John5
There is a function I wrote, which takes the int number as the parameter, and convert it to a string literal. This function is dependent on another function that converts a single digit to its char equivalent:
char intToChar(int num)
{
if (num < 10 && num >= 0)
{
return num + 48;
//48 is the number that we add to an integer number to have its character equivalent (see the unsigned ASCII table)
}
else
{
return '*';
}
}
string intToString(int num)
{
int digits = 0, process, single;
string numString;
process = num;
// The following process the number of digits in num
while (process != 0)
{
single = process % 10; // 'single' now holds the rightmost portion of the int
process = (process - single)/10;
// Take out the rightmost number of the int (it's a zero in this portion of the int), then divide it by 10
// The above combination eliminates the rightmost portion of the int
digits ++;
}
process = num;
// Fill the numString with '*' times digits
for (int i = 0; i < digits; i++)
{
numString += '*';
}
for (int i = digits-1; i >= 0; i--)
{
single = process % 10;
numString[i] = intToChar ( single);
process = (process - single) / 10;
}
return numString;
}
In C++ 20 you can have a variadic lambda that does concatenate arbitrary streamable types to a string in a few lines:
auto make_string=[os=std::ostringstream{}](auto&& ...p) mutable
{
(os << ... << std::forward<decltype(p)>(p) );
return std::move(os).str();
};
int main() {
std::cout << make_string("Hello world: ",4,2, " is ", 42.0);
}
see https://godbolt.org/z/dEe9h75eb
using move(os).str() guarantees that the ostringstream object's stringbuffer is empty next time the lambda is called.
You can use the C function itoa() like this:
char buf[3];
itoa(age, buf, 10);
name += buf;
I was looking for solutions to appending strings with other primitives and found that stringstream was the easiest solution. However, I wanted to streamline the process so I wanted to make a function to ease its use. In case you are proposing alternate methods for concatenation, i need the final result to be char*. I used a loop (i) with:
std::stringstream ss;
ss << "test" << i;
char* name = new char[ss.str().size() + 1];//allocate
strcpy(name, ss.str().c_str());//copy and put (char*)c_str in name
So the output is something link test1test2test3... This was the most reasonable solution I could muster. I was trying to put it into a function for ease of use, but am running into problems. I wanted to do something like:
char* string_to_pointer( char* dest, std::stringstream* _ss ) {
char* result = new char[_ss->str().size() + 1];
strcpy(result, _ss->str().c_str());
return result;
}
I could then do something like:
std::stringstream ss;
ss << "test" << i;
char* name = string_to_pointer( name, &ss );
I'm pretty new to c++ and this seems like the correct use syntactically, but I am running into runtime issues and would welcome solutions on how to get this in an easy to use function without resulting to Boost.
What about something like this:
#include <string>
#include <sstream>
class ToString {
std::ostringstream stream;
public:
template<typename T>
inline ToString &operator<<(const T&val) {
stream << val;
return *this;
}
inline operator std::string() const {
return stream.str();
}
};
You can use it like this:
std::string str = ToString() << "Test " << 5 << " and " << 4.2;
Use the std::stringstream::str() function to retrieve the contents of the string.
Example:
int foo = 42;
double bar = 12.67;
std::stringstream ss;
ss << "foo bar - " << foo << ' ' << bar;
std::string result = ss.str();
If you dont want to modify the string further, you can now simple call result.c_str() to acquire a const char*. However, if you really need a modifyable char* you have to copy the contents of the string to a cstring:
std::unique_ptr<char[]> cstring = new char[result.size() + 1];
strcpy(cstring.get(), result.c_str());
char* string_to_cstring ( const std::string &_ss )
Would be cleaner! Use with string_to_cstring(ss.str())
Need the returned C-String to be changeble? Because if not, just use ss.str().c_str() wherever you need it!
Or use:
char* result = new char[ss.str().size() + 1] (); // value initialized
ss.str().copy(result,std::string::npos);
For example only and not the actual code:
stringstream ss;
ss << " world!";
string hello("Hello");
// insert hello to beginning of ss ??
Thanks for all the responses, I also found this code, which works:
ostringstream& insert( ostringstream& oss, const string& s )
{
streamsize pos = oss.tellp();
oss.str( s + oss.str() );
oss.seekp( pos + s.length() );
return oss;
}
You cannot do it without making at least one copy. One way:
std::stringstream ss;
ss << " world!";
const std::string &temp = ss.str();
ss.seekp(0);
ss << "Hello";
ss << temp;
This relies on the "most important const" to extend the lifetime of the temporary and avoid making an extra copy.
Or, simpler and possibly faster:
std::stringstream ss;
ss << " world!";
std::stringstream temp;
temp << "Hello";
temp << ss.rdbuf();
ss = std::move(temp); // or ss.swap(temp);
This borrows the rdbuf approach from this answer, since the interesting problem here is how to minimize copies.
the only way i can see is to create the string from stream and prefix your other string
string result = hello + ss.str();
its called a stream for a reason.
Assuming ss1 contains "hello"
ss1 << ss.rdbuf();
or
ss1 << "hello" << ss;
Refer this URL for more info:-
stringstream
i have a string and i need to add a number to it i.e a int. like:
string number1 = ("dfg");
int number2 = 123;
number1 += number2;
this is my code:
name = root_enter; // pull name from another string.
size_t sz;
sz = name.size(); //find the size of the string.
name.resize (sz + 5, account); // add the account number.
cout << name; //test the string.
this works... somewhat but i only get the "*name*88888" and... i don't know why.
i just need a way to add the value of a int to the end of a string
There are no in-built operators that do this. You can write your own function, overload an operator+ for a string and an int. If you use a custom function, try using a stringstream:
string addi2str(string const& instr, int v) {
stringstream s(instr);
s << v;
return s.str();
}
Use a stringstream.
#include <iostream>
#include <sstream>
using namespace std;
int main () {
int a = 30;
stringstream ss(stringstream::in | stringstream::out);
ss << "hello world";
ss << '\n';
ss << a;
cout << ss.str() << '\n';
return 0;
}
You can use string streams:
template<class T>
std::string to_string(const T& t) {
std::ostringstream ss;
ss << t;
return ss.str();
}
// usage:
std::string s("foo");
s.append(to_string(12345));
Alternatively you can use utilities like Boosts lexical_cast():
s.append(boost::lexical_cast<std::string>(12345));
Use a stringstream.
int x = 29;
std::stringstream ss;
ss << "My age is: " << x << std::endl;
std::string str = ss.str();
you can use lexecal_cast from boost, then C itoa and of course stringstream from STL
I thought this would be really simple, but it's presenting some difficulties. If I have
std::string name = "John";
int age = 21;
How do I combine them to get a single string "John21"?
In alphabetical order:
std::string name = "John";
int age = 21;
std::string result;
// 1. with Boost
result = name + boost::lexical_cast<std::string>(age);
// 2. with C++11
result = name + std::to_string(age);
// 3. with FastFormat.Format
fastformat::fmt(result, "{0}{1}", name, age);
// 4. with FastFormat.Write
fastformat::write(result, name, age);
// 5. with the {fmt} library
result = fmt::format("{}{}", name, age);
// 6. with IOStreams
std::stringstream sstm;
sstm << name << age;
result = sstm.str();
// 7. with itoa
char numstr[21]; // enough to hold all numbers up to 64-bits
result = name + itoa(age, numstr, 10);
// 8. with sprintf
char numstr[21]; // enough to hold all numbers up to 64-bits
sprintf(numstr, "%d", age);
result = name + numstr;
// 9. with STLSoft's integer_to_string
char numstr[21]; // enough to hold all numbers up to 64-bits
result = name + stlsoft::integer_to_string(numstr, 21, age);
// 10. with STLSoft's winstl::int_to_string()
result = name + winstl::int_to_string(age);
// 11. With Poco NumberFormatter
result = name + Poco::NumberFormatter().format(age);
is safe, but slow; requires Boost (header-only); most/all platforms
is safe, requires C++11 (to_string() is already included in #include <string>)
is safe, and fast; requires FastFormat, which must be compiled; most/all platforms
(ditto)
is safe, and fast; requires the {fmt} library, which can either be compiled or used in a header-only mode; most/all platforms
safe, slow, and verbose; requires #include <sstream> (from standard C++)
is brittle (you must supply a large enough buffer), fast, and verbose; itoa() is a non-standard extension, and not guaranteed to be available for all platforms
is brittle (you must supply a large enough buffer), fast, and verbose; requires nothing (is standard C++); all platforms
is brittle (you must supply a large enough buffer), probably the fastest-possible conversion, verbose; requires STLSoft (header-only); most/all platforms
safe-ish (you don't use more than one int_to_string() call in a single statement), fast; requires STLSoft (header-only); Windows-only
is safe, but slow; requires Poco C++ ; most/all platforms
In C++11, you can use std::to_string, e.g.:
auto result = name + std::to_string( age );
If you have Boost, you can convert the integer to a string using boost::lexical_cast<std::string>(age).
Another way is to use stringstreams:
std::stringstream ss;
ss << age;
std::cout << name << ss.str() << std::endl;
A third approach would be to use sprintf or snprintf from the C library.
char buffer[128];
snprintf(buffer, sizeof(buffer), "%s%d", name.c_str(), age);
std::cout << buffer << std::endl;
Other posters suggested using itoa. This is NOT a standard function, so your code will not be portable if you use it. There are compilers that don't support it.
#include <iostream>
#include <sstream>
std::ostringstream o;
o << name << age;
std::cout << o.str();
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
string itos(int i) // convert int to string
{
stringstream s;
s << i;
return s.str();
}
Shamelessly stolen from http://www.research.att.com/~bs/bs_faq2.html.
This is the easiest way:
string s = name + std::to_string(age);
If you have C++11, you can use std::to_string.
Example:
std::string name = "John";
int age = 21;
name += std::to_string(age);
std::cout << name;
Output:
John21
It seems to me that the simplest answer is to use the sprintf function:
sprintf(outString,"%s%d",name,age);
#include <string>
#include <sstream>
using namespace std;
string concatenate(std::string const& name, int i)
{
stringstream s;
s << name << i;
return s.str();
}
#include <sstream>
template <class T>
inline std::string to_string (const T& t)
{
std::stringstream ss;
ss << t;
return ss.str();
}
Then your usage would look something like this
std::string szName = "John";
int numAge = 23;
szName += to_string<int>(numAge);
cout << szName << endl;
Googled [and tested :p ]
This problem can be done in many ways. I will show it in two ways:
Convert the number to string using to_string(i).
Using string streams.
Code:
#include <string>
#include <sstream>
#include <bits/stdc++.h>
#include <iostream>
using namespace std;
int main() {
string name = "John";
int age = 21;
string answer1 = "";
// Method 1). string s1 = to_string(age).
string s1=to_string(age); // Know the integer get converted into string
// where as we know that concatenation can easily be done using '+' in C++
answer1 = name + s1;
cout << answer1 << endl;
// Method 2). Using string streams
ostringstream s2;
s2 << age;
string s3 = s2.str(); // The str() function will convert a number into a string
string answer2 = ""; // For concatenation of strings.
answer2 = name + s3;
cout << answer2 << endl;
return 0;
}
In C++20 you'll be able to do:
auto result = std::format("{}{}", name, age);
In the meantime you can use the {fmt} library, std::format is based on:
auto result = fmt::format("{}{}", name, age);
Disclaimer: I'm the author of the {fmt} library and C++20 std::format.
If you'd like to use + for concatenation of anything which has an output operator, you can provide a template version of operator+:
template <typename L, typename R> std::string operator+(L left, R right) {
std::ostringstream os;
os << left << right;
return os.str();
}
Then you can write your concatenations in a straightforward way:
std::string foo("the answer is ");
int i = 42;
std::string bar(foo + i);
std::cout << bar << std::endl;
Output:
the answer is 42
This isn't the most efficient way, but you don't need the most efficient way unless you're doing a lot of concatenation inside a loop.
If you are using MFC, you can use a CString
CString nameAge = "";
nameAge.Format("%s%d", "John", 21);
Managed C++ also has a
string formatter.
As a one liner: name += std::to_string(age);
The std::ostringstream is a good method, but sometimes this additional trick might get handy transforming the formatting to a one-liner:
#include <sstream>
#define MAKE_STRING(tokens) /****************/ \
static_cast<std::ostringstream&>( \
std::ostringstream().flush() << tokens \
).str() \
/**/
Now you can format strings like this:
int main() {
int i = 123;
std::string message = MAKE_STRING("i = " << i);
std::cout << message << std::endl; // prints: "i = 123"
}
As a Qt-related question was closed in favour of this one, here's how to do it using Qt:
QString string = QString("Some string %1 with an int somewhere").arg(someIntVariable);
string.append(someOtherIntVariable);
The string variable now has someIntVariable's value in place of %1 and someOtherIntVariable's value at the end.
There are more options possible to use to concatenate integer (or other numerric object) with string. It is Boost.Format
#include <boost/format.hpp>
#include <string>
int main()
{
using boost::format;
int age = 22;
std::string str_age = str(format("age is %1%") % age);
}
and Karma from Boost.Spirit (v2)
#include <boost/spirit/include/karma.hpp>
#include <iterator>
#include <string>
int main()
{
using namespace boost::spirit;
int age = 22;
std::string str_age("age is ");
std::back_insert_iterator<std::string> sink(str_age);
karma::generate(sink, int_, age);
return 0;
}
Boost.Spirit Karma claims to be one of the fastest option for integer to string conversion.
std::ostringstream
#include <sstream>
std::ostringstream s;
s << "John " << age;
std::string query(s.str());
std::to_string (C++11)
std::string query("John " + std::to_string(age));
boost::lexical_cast
#include <boost/lexical_cast.hpp>
std::string query("John " + boost::lexical_cast<std::string>(age));
Here is an implementation of how to append an int to a string using the parsing and formatting facets from the IOStreams library.
#include <iostream>
#include <locale>
#include <string>
template <class Facet>
struct erasable_facet : Facet
{
erasable_facet() : Facet(1) { }
~erasable_facet() { }
};
void append_int(std::string& s, int n)
{
erasable_facet<std::num_put<char,
std::back_insert_iterator<std::string>>> facet;
std::ios str(nullptr);
facet.put(std::back_inserter(s), str,
str.fill(), static_cast<unsigned long>(n));
}
int main()
{
std::string str = "ID: ";
int id = 123;
append_int(str, id);
std::cout << str; // ID: 123
}
Common Answer: itoa()
This is bad. itoa is non-standard, as pointed out here.
You can concatenate int to string by using the given below simple trick, but note that this only works when integer is of single digit. Otherwise, add integer digit by digit to that string.
string name = "John";
int age = 5;
char temp = 5 + '0';
name = name + temp;
cout << name << endl;
Output: John5
There is a function I wrote, which takes the int number as the parameter, and convert it to a string literal. This function is dependent on another function that converts a single digit to its char equivalent:
char intToChar(int num)
{
if (num < 10 && num >= 0)
{
return num + 48;
//48 is the number that we add to an integer number to have its character equivalent (see the unsigned ASCII table)
}
else
{
return '*';
}
}
string intToString(int num)
{
int digits = 0, process, single;
string numString;
process = num;
// The following process the number of digits in num
while (process != 0)
{
single = process % 10; // 'single' now holds the rightmost portion of the int
process = (process - single)/10;
// Take out the rightmost number of the int (it's a zero in this portion of the int), then divide it by 10
// The above combination eliminates the rightmost portion of the int
digits ++;
}
process = num;
// Fill the numString with '*' times digits
for (int i = 0; i < digits; i++)
{
numString += '*';
}
for (int i = digits-1; i >= 0; i--)
{
single = process % 10;
numString[i] = intToChar ( single);
process = (process - single) / 10;
}
return numString;
}
In C++ 20 you can have a variadic lambda that does concatenate arbitrary streamable types to a string in a few lines:
auto make_string=[os=std::ostringstream{}](auto&& ...p) mutable
{
(os << ... << std::forward<decltype(p)>(p) );
return std::move(os).str();
};
int main() {
std::cout << make_string("Hello world: ",4,2, " is ", 42.0);
}
see https://godbolt.org/z/dEe9h75eb
using move(os).str() guarantees that the ostringstream object's stringbuffer is empty next time the lambda is called.
You can use the C function itoa() like this:
char buf[3];
itoa(age, buf, 10);
name += buf;