我原以为这很简单,但它出现了一些困难。如果我有
std::string name = "John";
int age = 21;
我如何结合他们得到一个单一的字符串“John21”?
我原以为这很简单,但它出现了一些困难。如果我有
std::string name = "John";
int age = 21;
我如何结合他们得到一个单一的字符串“John21”?
当前回答
在c++ 20中,你可以有一个可变变量lambda,它可以在几行内将任意可流类型连接到字符串:
auto make_string=[os=std::ostringstream{}](auto&& ...p) mutable
{
(os << ... << std::forward<decltype(p)>(p) );
return std::move(os).str();
};
int main() {
std::cout << make_string("Hello world: ",4,2, " is ", 42.0);
}
参见https://godbolt.org/z/dEe9h75eb
使用move(os).str()可以保证下次调用lambda时ostringstream对象的stringbuffer为空。
其他回答
如果你使用MFC,你可以使用CString
CString nameAge = "";
nameAge.Format("%s%d", "John", 21);
托管c++也有一个 字符串格式化程序。
在c++ 20中,你可以有一个可变变量lambda,它可以在几行内将任意可流类型连接到字符串:
auto make_string=[os=std::ostringstream{}](auto&& ...p) mutable
{
(os << ... << std::forward<decltype(p)>(p) );
return std::move(os).str();
};
int main() {
std::cout << make_string("Hello world: ",4,2, " is ", 42.0);
}
参见https://godbolt.org/z/dEe9h75eb
使用move(os).str()可以保证下次调用lambda时ostringstream对象的stringbuffer为空。
#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]
作为一个与Qt相关的问题,下面是如何使用Qt:
QString string = QString("Some string %1 with an int somewhere").arg(someIntVariable);
string.append(someOtherIntVariable);
字符串变量现在有someIntVariable的值代替%1,someOtherIntVariable的值在结尾。
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