C++ 如何连接 std::string 和 int？

How to concatenate a std::string and an int?

提问by Obediah Stane

I thought this would be really simple but it's presenting some difficulties. If I have

我认为这会很简单，但它带来了一些困难。如果我有

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

How do I combine them to get a single string "John21"?

我如何将它们组合成一个字符串"John21"？

回答by DannyT

In alphabetical order:

按字母顺序：

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);

1. is safe, but slow; requires Boost(header-only); most/all platforms
2. is safe, requires C++11 (to_string()is already included in #include <string>)
3. is safe, and fast; requires FastFormat, which must be compiled; most/all platforms
4. (ditto)
5. is safe, and fast; requires the {fmt} library, which can either be compiled or used in a header-only mode; most/all platforms
6. safe, slow, and verbose; requires #include <sstream>(from standard C++)
7. 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
8. is brittle (you must supply a large enough buffer), fast, and verbose; requires nothing (is standard C++); all platforms
9. is brittle (you must supply a large enough buffer), probably the fastest-possible conversion, verbose; requires STLSoft(header-only); most/all platforms
10. safe-ish (you don't use more than one int_to_string()call in a single statement), fast; requires STLSoft(header-only); Windows-only
11. is safe, but slow; requires Poco C++; most/all platforms

1. 安全，但速度慢；需要Boost（仅标题）；大多数/所有平台
2. 是安全的，需要 C++11（to_string()已经包含在 中#include <string>）
3. 安全、快捷；需要FastFormat，它必须被编译；大多数/所有平台
4. （同上）
5. 安全、快捷；需要{fmt} 库，它可以被编译或在仅标头模式下使用；大多数/所有平台
6. 安全、缓慢且冗长；需要#include <sstream>（来自标准 C++）
7. 脆弱（您必须提供足够大的缓冲区）、快速且冗长；itoa() 是一个非标准扩展，不保证适用于所有平台
8. 脆弱（您必须提供足够大的缓冲区）、快速且冗长；什么都不需要（是标准的 C++）；所有平台
9. 很脆弱（您必须提供足够大的缓冲区），可能是最快的转换，冗长；需要STLSoft（仅标头）；大多数/所有平台
10. 安全（在单个语句中您不会使用多个int_to_string()调用），快速；需要STLSoft（仅标头）；仅限 Windows
11. 安全，但速度慢；需要Poco C++；大多数/所有平台

回答by Jeremy

In C++11, you can use std::to_string, e.g.:

在 C++11 中，您可以使用std::to_string，例如：

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

回答by Jay Conrod

If you have Boost, you can convert the integer to a string using boost::lexical_cast<std::string>(age).

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

Another way is to use stringstreams:

另一种方法是使用字符串流：

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

A third approach would be to use sprintfor snprintffrom the C library.

第三种方法是使用sprintf或snprintf来自 C 库。

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.

其他海报建议使用itoa. 这不是标准功能，因此如果您使用它，您的代码将不可移植。有些编译器不支持它。

回答by Ben Hoffstein

#include <iostream>
#include <sstream>

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

回答by tloach

#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.

从http://www.research.att.com/~bs/bs_faq2.html无耻地被盗。

回答by Kevin

This is the easiest way:

这是最简单的方法：

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

回答by 0x499602D2

If you have C++11, you can use std::to_string.

如果你有 C++11，你可以使用std::to_string.

Example:

例子：

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

name += std::to_string(age);

std::cout << name;

Output:

输出：

John21

回答by user12576

It seems to me that the simplest answer is to use the sprintffunction:

在我看来，最简单的答案是使用sprintf函数：

sprintf(outString,"%s%d",name,age);

回答by Seb Rose

#include <string>
#include <sstream>
using namespace std;
string concatenate(std::string const& name, int i)
{
    stringstream s;
    s << name << i;
    return s.str();
}

回答by Zing-

#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

然后你的用法看起来像这样

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

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