按字母顺序排列:

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

#include <iostream>
#include <sstream>

std::ostringstream o;
o << name << age;
std::cout << o.str();

如果有Boost，可以使用Boost::lexical_cast<std::string>(age)将整数转换为字符串。

另一种方法是使用stringstreams:

std::stringstream ss;
ss << age;
std::cout << name << ss.str() << std::endl;

第三种方法是使用C库中的sprintf或snprintf。

char buffer[128];
snprintf(buffer, sizeof(buffer), "%s%d", name.c_str(), age);
std::cout << buffer << std::endl;

其他海报建议使用itoa。这不是一个标准函数，所以如果你使用它，你的代码将不能移植。有些编译器不支持它。

在c++ 11中，你可以使用std::to_string，例如:

auto result = name + std::to_string( age );

#include <sstream>

template <class T>
inline std::string to_string (const T& t)
{
   std::stringstream ss;
   ss << t;
   return ss.str();
}

那么你的用法应该是这样的

   std::string szName = "John";
   int numAge = 23;
   szName += to_string<int>(numAge);
   cout << szName << endl;

谷歌[并测试:p]

您可以使用下面给出的简单技巧将int连接到string，但请注意，这仅适用于integer为个位数时。否则，向该字符串逐位添加整数。

string name = "John";
int age = 5;
char temp = 5 + '0';
name = name + temp;
cout << name << endl;

Output:  John5