How to concatenate a std::string and an int?
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);
#include <string> ) #include <sstream> (from standard C++) 在C ++ 11中，你可以使用std::to_string ，例如：
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.