validating input using scanf

The following piece of code checks the users password, if it's not equal to the users password then it just asks them to enter their password again. The only problem is when i enter a character i.e. # or A or , the while loop i just runs continuously. I'm guessing it's because i've asked scanf to read in digits not chars, I'm not sure as to how i'm suppose to check both chars and ints using scanf at once?

Code:

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  int password;
  printf("Please enter your password:");
  scanf("%d", &password);


  while (password != 233)
  {
      printf("%s", "Wrong password, please try again:");
      scanf("%d", &password);
  }

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See http://www.eskimo.com/~scs/C-faq/q12.19.html

Your best bet is to use fgets to retrieve input and then sscanf to parse the input. scanf is a problematic function.

Quote:

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Originally Posted by anonytmouse

Your best bet is to use fgets to retrieve input and then sscanf to parse the input. scanf is a problematic function.

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Hmm it doesn't seem to work:

Code:

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  char  password[2];
  char password2;

  fgets(password, 2, stdin);
  scanf(password, &password2);
  printf("%s", "Password is:");
  printf("%c", password2);

---

it just prints out nothing.

anonytmouse didn't make a typo. He really did mean sscanf() as opposed to scanf(). You can see the differences here. scanf() reads from a stream but sscanf() reads from a string.

Ok i tried using sscanf if i put sscanf it doesn't seem to ask for any input which is strange?

Code:

---


  fgets(password, 2, stdin);
  sscanf(password, &password2);
  printf("%s", "Password is:");
  printf("%c", password2);

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I think you want something like this:

Code:

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#include <stdio.h>
int main()
{
    char p_str[256];
    int p_num;
    do {
        printf("Enter your password: ");
        fflush(stdout);
        fgets(p_str, sizeof p_str, stdin);
    } while (!sscanf(p_str, "%d", &p_num));
    printf("Password is: %d\n", p_num);
    return 0;
}

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The while loop is necessary because sscanf will return 0 if it doesn't find a number, so you need to ask again.