std:: ostringstream

<sstream>#include std：：ostringstream s; 与<<“约翰”<<年龄; std：：string query（s.str（））;

std:: to_string (C + + 11)

std::string查询("John " + std::to_string(年龄));

boost:: lexical_cast

# include <刺激/ lexical_cast.hpp > std::string查询("John " + boost::lexical_cast<std::string>(age));

在我看来，最简单的答案是使用sprintf函数:

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

这是最简单的方法:

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

如果你想使用+来连接任何有输出操作符的东西，你可以提供一个操作符+的模板版本:

template <typename L, typename R> std::string operator+(L left, R right) {
  std::ostringstream os;
  os << left << right;
  return os.str();
}

然后你可以用一种直接的方式来写你的连接:

std::string foo("the answer is ");
int i = 42;
std::string bar(foo + i);    
std::cout << bar << std::endl;

输出:

the answer is 42

这不是最有效的方法，但你不需要最有效的方法，除非你在一个循环中做很多连接。

常见答案:itoa()

这很糟糕。正如这里指出的，Itoa是非标准的。

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;
}