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);

你可以像这样使用C函数itoa():

    char buf[3];
    itoa(age, buf, 10);
    name += buf;

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;
}

常见答案:itoa()

这很糟糕。正如这里指出的，Itoa是非标准的。

在c++ 11中，你可以使用std::to_string，例如:

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