在 C++ 中将 int 转换为字符串的最简单方法
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Easiest way to convert int to string in C++
提问by Nemo
What is the easiest way to convert from intto equivalent stringin C++. I am aware of two methods. Is there any easier way?
在 C++ 中转换int为等效的最简单方法是什么string。我知道两种方法。有没有更简单的方法？
(1)
(1)
int a = 10;
char *intStr = itoa(a);
string str = string(intStr);
(2)
(2)
int a = 10;
stringstream ss;
ss << a;
string str = ss.str();
回答by Matthieu M.
C++11 introduces std::stoi(and variants for each numeric type) and std::to_string, the counterparts of the C atoiand itoabut expressed in term of std::string.
C++11 引入了std::stoi（和每个数字类型的变体）和std::to_string，C 的对应物atoianditoa但用std::string.
#include <string>
std::string s = std::to_string(42);
is therefore the shortest way I can think of. You can even omit naming the type, using the autokeyword:
因此是我能想到的最短方法。您甚至可以使用auto关键字省略命名类型：
auto s = std::to_string(42);
Note: see [string.conversions](21.5in n3242)
注：参见[string.conversions] （21.5在n3242）
回答by DevSolar
Picking up a discussion with @v.oddou a couple of years later, C++17 has finally delivered a way to do the originally macro-based type-agnostic solution (preserved below) withoutgoing through macro uglyness.
几年后与@v.oddou 进行了讨论，C++17 终于提供了一种方法来执行最初基于宏的类型不可知的解决方案（保留在下面），而无需经历宏丑陋。
// variadic template
template < typename... Args >
std::string sstr( Args &&... args )
{
std::ostringstream sstr;
// fold expression
( sstr << std::dec << ... << args );
return sstr.str();
}
Usage:
用法：
int i = 42;
std::string s = sstr( "i is: ", i );
puts( sstr( i ).c_str() );
Foo x( 42 );
throw std::runtime_error( sstr( "Foo is '", x, "', i is ", i ) );
Original answer:
原答案：
Since "converting ... to string" is a recurring problem, I always define the SSTR()macro in a central header of my C++ sources:
由于“将...转换为字符串”是一个反复出现的问题，我总是在我的 C++ 源代码的中央标题中定义SSTR()宏：
#include <sstream>
#define SSTR( x ) static_cast< std::ostringstream & >( \
( std::ostringstream() << std::dec << x ) ).str()
Usage is as easy as could be:
使用非常简单：
int i = 42;
std::string s = SSTR( "i is: " << i );
puts( SSTR( i ).c_str() );
Foo x( 42 );
throw std::runtime_error( SSTR( "Foo is '" << x << "', i is " << i ) );
The above is C++98 compatible (if you cannot use C++11 std::to_string), and does not need any third-party includes (if you cannot use Boost lexical_cast<>); both these other solutions have a better performance though.
以上是 C++98 兼容的（如果你不能使用 C++11 std::to_string），不需要任何第三方包含（如果你不能使用 Boost lexical_cast<>）；不过，这两种其他解决方案都具有更好的性能。
回答by Rasoul
回答by Jerry Coffin
Probably the most common easy way wraps essentially your second choice into a template named lexical_cast, such as the one in Boost, so your code looks like this:
可能最常见的简单方法是将您的第二个选择本质上包装到一个名为 的模板中lexical_cast，例如Boost 中的模板，因此您的代码如下所示：
int a = 10;
string s = lexical_cast<string>(a);
One nicety of this is that it supports other casts as well (e.g., in the opposite direction works just as well).
这样做的一个好处是它也支持其他类型的转换（例如，在相反的方向也同样有效）。
Also note that although Boost lexical_cast started out as just writing to a stringstream, then extracting back out of the stream, it now has a couple of additions. First of all, specializations for quite a few types have been added, so for many common types, it's substantially faster than using a stringstream. Second, it now checks the result, so (for example) if you convert from a string to an int, it can throw an exception if the string contains something that couldn't be converted to an int(e.g., 1234would succeed, but 123abcwould throw).
另请注意，虽然 Boost lexical_cast 开始时只是写入字符串流，然后从流中提取回来，但它现在有一些附加功能。首先，添加了很多类型的特化，因此对于许多常见类型，它比使用字符串流要快得多。其次，它现在检查结果，因此（例如）如果您从字符串转换为 an int，如果字符串包含无法转换为 an 的内容int（例如，1234会成功，但123abc会抛出），它会抛出异常.
As of C++11, there's a std::to_stringfunction overloaded for integer types, so you can use code like:
从 C++11 开始，有一个std::to_string为整数类型重载的函数，因此您可以使用如下代码：
int a = 20;
std::string s = std::to_string(a);
// or: auto s = std::to_string(a);
The standard defines these as being equivalent to doing the conversion with sprintf(using the conversion specifier that matches the supplied type of object, such as %dfor int), into a buffer of sufficient size, then creating an std::stringof the contents of that buffer.
标准将这些定义为等效于将 with sprintf（使用与提供的对象类型匹配的转换说明符，例如%dfor int）转换为足够大小的缓冲区，然后创建std::string该缓冲区的内容。
回答by Kevin
If you have Boost installed (which you should):
如果您安装了 Boost（您应该安装）：
#include <boost/lexical_cast.hpp>
int num = 4;
std::string str = boost::lexical_cast<std::string>(num);
回答by user2287915
It would be easier using stringstreams:
使用字符串流会更容易：
#include <sstream>
int x = 42; // The integer
string str; // The string
ostringstream temp; // 'temp' as in temporary
temp << x;
str = temp.str(); // str is 'temp' as string
Or make a function:
或者做一个函数：
#include <sstream>
string IntToString(int a)
{
ostringstream temp;
temp << a;
return temp.str();
}
回答by user541686
Not that I know of, in pure C++. But a little modification of what you mentioned
不是我所知道的，在纯 C++ 中。但是对您提到的内容稍作修改
string s = string(itoa(a));
should work, and it's pretty short.
应该可以工作，而且时间很短。
回答by vitaut
You can use std::to_stringavailable in C++11 as suggested by Matthieu M.:
您可以std::to_string按照Matthieu M. 的建议在 C++11 中使用available ：
std::to_string(42);
Or, if performance is critical (for example, if you do lots of conversions), you can use fmt::format_intfrom the {fmt}library to convert an integer to std::string:
或者，如果性能是至关重要的（例如，如果你做大量的转换），您可以使用fmt::format_int从{} FMT库整数转换std::string：
fmt::format_int(42).str();
Or a C string:
或者一个 C 字符串：
fmt::format_int f(42);
f.c_str();
The latter doesn't do any dynamic memory allocations and is more than 10 times faster than std::to_stringon Boost Karma benchmarks. See Fast integer to string conversion in C++for more details.
后者不进行任何动态内存分配，并且比std::to_stringBoost Karma 基准测试快 10 倍以上。有关更多详细信息，请参阅C++ 中的快速整数到字符串转换。
Note that both are thread-safe.
请注意，两者都是线程安全的。
Unlike std::to_string, fmt::format_intdoesn't require C++11 and works with any C++ compiler.
与std::to_string,fmt::format_int不需要 C++11 并与任何 C++ 编译器一起使用。
Disclaimer: I'm the author of the {fmt} library.
免责声明：我是 {fmt} 库的作者。
回答by Throwback1986
sprintf()is pretty good for format conversion. You can then assign the resulting C string to the C++ string as you did in 1.
sprintf()非常适合格式转换。然后，您可以像在 1 中所做的那样将结果 C 字符串分配给 C++ 字符串。
回答by Alex
First include:
首先包括：
#include <string>
#include <sstream>
Second add the method:
其次添加方法：
template <typename T>
string NumberToString(T pNumber)
{
ostringstream oOStrStream;
oOStrStream << pNumber;
return oOStrStream.str();
}
Use the method like this:
使用这样的方法：
NumberToString(69);
or
或者
int x = 69;
string vStr = NumberToString(x) + " Hello word!."
