C ++中的简单字符验证

时间:2018-08-09 18:21:38

标签: c++

我是一名初学者,是C ++的新手,并且希望学习。这是我到目前为止的内容:

这是教师提供给我的伪代码版本:

if str.find("@")==string::npos OR <-- @ not found

str.find(".")==string::npos OR <-- . not found

str.find(" ")!=string::npos) <-- space found

我对功能要求的理解:

 check emails for @, ., or space  function,

 if @ exists, no output
 if @ doesn't exist, output "This (email) is invalid."
 if . exists, no output
 if ' ' exists, output "This (email) is invalid."

这是对不起的尝试,不起作用:

void Roster::printIncorrectEmails()
{
char at = '@';
char pd = '.';
char sp = ' ';

for (int i = 0; i < sizeof(RepoArray); i++) {

    string email = Student::RepoArray.getEmailAddress();

    size_t found = email.find(at, pd);
    if (found != string::npos)
    {
        cout << "Invalid Email: " << email << '\n';
    }

    found = email.find(sp);
    if (found == string::npos)
    {

        cout << "Invalid Email: " << email << '\n';
    }
}
}

如果有人愿意看到我,我也有很多其他尝试。

0 个答案:

没有答案