C-strings，如何确定字符的长度是否在 6 到 10 之间？

Question

I am beginner programmer. I am writing a program to check if the password, as a c-string, is between 6 and 10 characters long. If not, the user has to re-enter the password until it meets the requirement. For the input validation part of the program, it works when the password is less than 6 characters -- telling the user to re-enter the password. But, when it is more than 10 characters long, it doesn't display an error that it is more than 10 characters long. How would I fix this? Thanks for your input.

#include <iostream>
#include <cstring>

using namespace std;

int main()
{
const int SIZE = 12; // Maximum size for the c-string
char pass[SIZE];   // to hold password c-string.
int length;


// get line of input
cout << "Enter a password between 6 and " << (SIZE - 2) << "characters long:\n";
cin.getline(pass, SIZE);


length = strlen(pass);

while (length < 6 || length > 10)
{
cout << "Error: password is not between 6 and " << (SIZE - 2) << " characters long.\n"
     << "Enter the password again: ";
cin.getline(pass, SIZE);
    length = strlen(pass);
}

return 0;
}

Answer 1

I'm not sure why you're using cstrings, since normal strings work just as well. Here is how I would rewrite it:

string pass;
cout << "Enter password: ";
cin  >> pass;
while (pass.length() > 10 || pass.length() < 6)
{
  cout << "Not the right length, try again: ";
  cin  >> pass;
}

If you need to have the password in an array, use this:

//To use the c_string version of it, type 
pass.c_str();
return 0;

This is untested however, but it should work.

Answer 2

Your not allowing a password over 10 characters. C strings are null terminated so 10 chars plus null byte is 11. You will only ever get a 10 length. Ps I would suggest

    char password[SIZE+2]

Makes it clearer that the constant is the maximum length of the password

EDIT

I agree with the other posts that std::string is a better option but it's also important especially for a biginer to understand problems not just accept solutions because that's what should be done

Answer 3

const int SIZE = 11;
char password[SIZE];

There is no way the password can store more than 11 chars (where the last char is NULL terminator '\\0' ).

cin.getline(password, SIZE);

There is no way you will get more than you asked, which is SIZE .

A quick fix would be changing it to char password[SIZE + 1]; . Why? Because it will be capable to store more chars than your wanted, thus allowing you to check if input is too long.

Of course, you have to change your cin.getline(... to cin.getline(password, SIZE + 1); as well.

You forgot to update your length in that if too. I would suggest do something like this instead, making it cleaner:

const int SIZE = 11;
char password[SIZE + 1];

while (true) // Create a loop
{
    cout << "Enter a password between 6 and "
         << (SIZE - 1) << " characters long:\n";

    // If bad input, discard the remaining input
    if (!cin.getline(password, SIZE + 1))
    {
        cin.clear();
        cin.ignore(INT_MAX, '\n');
    }

    int length = strlen(password);

    // Check input validation if password is bet. 6 and 10 characters
    if (length < 6 || length > 10)
    {
        cout << "Password is not between 6 " << (SIZE - 1)
             << " characters.\n Please enter your password again:\n";
    }
    else { break; } // Break loop if input vaild
}