如何在 C 中連接 std::string 和 Integer ？

連接 std::string 和 int

組合 std::string 和 int 形成單一字串可以是簡單的任務。以下是實現此目的的多種方法：

1。使用Boost

#include <boost/lexical_cast.hpp>

std::string name = "John";
int age = 21;

std::string result = name + boost::lexical_cast<std::string>(age);

2.使用 C 11

std::string name = "John";
int age = 21;

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

3。使用FastFormat

#include <fastformat/format.hpp>

std::string name = "John";
int age = 21;

std::string result;
fastformat::fmt(result, "{0}{1}", name, age); // FastFormat.Format
fastformat::write(result, name, age); // FastFormat.Write

4.使用{fmt} 庫

#include <fmt/core.h>

std::string name = "John";
int age = 21;

std::string result = fmt::format("{}{}", name, age);

5.使用IOStreams

std::string name = "John";
int age = 21;

std::stringstream sstm;
sstm << name << age;
std::string result = sstm.str();

5.使用IOStreams

std::string name = "John";
int age = 21;

char numstr[21]; // enough to hold all numbers up to 64-bits
std::string result = name + itoa(age, numstr, 10);

6.使用itoa

std::string name = "John";
int age = 21;

char numstr[21]; // enough to hold all numbers up to 64-bits
sprintf(numstr, "%d", age);
std::string result = name + numstr;

7.使用sprintf

#include <stlsoft/integer_to_string.hpp>

std::string name = "John";
int age = 21;

char numstr[21]; // enough to hold all numbers up to 64-bits
std::string result = name + stlsoft::integer_to_string(numstr, 21, age);

8.使用STLSoft的integer_to_string

#include <stlsoft/winstl/int_to_string.hpp>

std::string name = "John";
int age = 21;

std::string result = name + winstl::int_to_string(age);

9.使用STLSoft的winstl::int_to_string()

#include <Poco/NumberFormatter.h>

std::string name = "John";
int age = 21;

std::string result = name + Poco::NumberFormatter().format(age);

10。使用 Poco NumberFormatter

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