c++ - 遍历一个字符串并将其与另一个进行比较以检查第一个中的字符是否进入第二个

Question

我该怎么做？一整天我一直在到处寻找解决方案:(它在c ++中，所有问题都在标题中，谢谢

if(fp.is_open())
{
    while(getline(fp, buff))
    {
        if(buff.length() == lung)
        {
            // check if the characters are in the string

            while(found != string::npos)
            {
                cout << i << "\r";

                for(int x = 0; x < buff.length(); ++x)
                {
                    found = lettere_possibili.find(buff[x]);

                    if(found == string::npos)
                    {
                        continue;
                    }
                }

                ++i;        
            }

            j = 0;
        }

        ++k;
    }

    fp.close();
}

Answer 1

这个问题并不完全清楚，但在我看来，您似乎正在寻找find_first_not_of()：

bool containsOnlyCharsFromPossibili = (buff.find_first_not_of(lettere_possibili) == string::npos);

Answer 2

#include <iostream>

void find_chars(const std::string &first, const std::string &second)
{
    std::string::const_iterator s;

    for (s = second.begin(); s != second.end(); ++s)
    {
        if (first.find(*s) != std::string::npos)
        {
            std::cout << *s << " is in " << second << "\n";
        }
    }
}

int main()
{
    find_chars("abc", "abcdef");
}

这是输出：

a is in abcdef
b is in abcdef
c is in abcdef