可能有点过分了，但是 boost::lexical_cast<int>(theString)应该是作业 很好。

我的代码:

#include <iostream>
using namespace std;

int main()
{
    string s="32";  //String
    int n=stoi(s);  //Convert to int
    cout << n + 1 << endl;

    return 0;
}

Boost.Lexical_cast呢?

下面是他们的例子:

下面的例子将命令行参数视为数值数据序列:

int main(int argc, char * argv[])
{
    using boost::lexical_cast;
    using boost::bad_lexical_cast;

    std::vector<short> args;

    while(*++argv)
    {
        try
        {
            args.push_back(lexical_cast<short>(*argv));
        }
        catch(bad_lexical_cast &)
        {
            args.push_back(0);
        }
    }
    ...
}

可能有点过分了，但是 boost::lexical_cast<int>(theString)应该是作业 很好。

诚然，我的解决方案不适用于负整数，但它将从包含整数的输入文本中提取所有正整数。它使用numeric_only locale:

int main() {
        int num;
        std::cin.imbue(std::locale(std::locale(), new numeric_only()));
        while ( std::cin >> num)
             std::cout << num << std::endl;
        return 0;
}

输入文本:

 the format (-5) or (25) etc... some text.. and then.. 7987...78hjh.hhjg9878

输出整数:

 5
25
7987
78
9878

类numeric_only定义为:

struct numeric_only: std::ctype<char> 
{
    numeric_only(): std::ctype<char>(get_table()) {}

    static std::ctype_base::mask const* get_table()
    {
        static std::vector<std::ctype_base::mask> 
            rc(std::ctype<char>::table_size,std::ctype_base::space);

        std::fill(&rc['0'], &rc[':'], std::ctype_base::digit);
        return &rc[0];
    }
};

完整在线演示:http://ideone.com/dRWSj

如果你不懂硬代码:)

bool strCanBeInt(std::string string){
    for (char n : string) {
        if (n != '0' && n != '1' && n != '2' && n != '3' && n != '4' && n != '5'
            && n != '6' && n != '7' && n != '8' && n != '9') {
            return false;
        }
    }
    return true;
}

int strToInt(std::string string) {
    int integer = 0;
    int numInt;
    for (char n : string) {
        if(n == '0') numInt = 0;
        if(n == '1') numInt = 1;
        if(n == '2') numInt = 2;
        if(n == '3') numInt = 3;
        if(n == '4') numInt = 4;
        if(n == '5') numInt = 5;
        if(n == '6') numInt = 6;
        if(n == '7') numInt = 7;
        if(n == '8') numInt = 8;
        if(n == '9') numInt = 9;
        if (integer){
            integer *= 10;
        }
        integer += numInt;
    }
    return integer;
}