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

我写了一个函数,它以int数作为参数,并将其转换为字符串字面量。此函数依赖于另一个函数,该函数将单个数字转换为其char等价:

char intToChar(int num)
{
    if (num < 10 && num >= 0)
    {
        return num + 48;
        //48 is the number that we add to an integer number to have its character equivalent (see the unsigned ASCII table)
    }
    else
    {
        return '*';
    }
}

string intToString(int num)
{
    int digits = 0, process, single;
    string numString;
    process = num;

    // The following process the number of digits in num
    while (process != 0)
    {
        single  = process % 10; // 'single' now holds the rightmost portion of the int
        process = (process - single)/10;
        // Take out the rightmost number of the int (it's a zero in this portion of the int), then divide it by 10
        // The above combination eliminates the rightmost portion of the int
        digits ++;
    }

    process = num;

    // Fill the numString with '*' times digits
    for (int i = 0; i < digits; i++)
    {
        numString += '*';
    }


    for (int i = digits-1; i >= 0; i--)
    {
        single = process % 10;
        numString[i] = intToChar ( single);
        process = (process - single) / 10;
    }

    return numString;
}

其他回答

如果你有c++ 11,你可以使用std::to_string。

例子:

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

name += std::to_string(age);

std::cout << name;

输出:

John21

在c++ 11中,你可以使用std::to_string,例如:

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

作为一行:name += std::to_string(年龄);

#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]

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。