checking if a string contains numbers

Question

I want to check a passcode if it contain at least 2 numbers. I tried it with a double for loop like this:

for(int i = 0; i <= count; i++)
{
    for(int k = 0; k < 10; k++)
    {
        if(passcode[i] == k)
        {
            printf("yee\n");
            break;
        }
        
    }
}

but it doesn't work

I thought for each 'round', k would be a number from 0 to 9, and if it would be equal to a number the user gave, it would (in this case) print something

(passcode is an input from the user and count is the amount of signs given in)

Answer 1

int sum = 0;

for(int i = 0; i <= count; i++)
    {
        if (passcode[i] >= '0' && passcode[i] <= '9')
            {
                sum++
            }
    }

if (sum >=2)
  printf("At least two digits");

As mentioned in some comments, you can also use isdigit(), if you have access to this function.
One other important point (also in the comments): characters and their number meaning might be different: '2' is different from 2: '2' is a character, there are 256 ASCII characters. If you want to verify this with a number, you are generally dealing with the ASCII code of that character (which is 50 for the character '2').

Answer 2

the digits are just ascii values you need single comparision like.

int len = LENGHT_OF_STRING;
char * s = "the string";

for(int i = 0; i < len; i++){
    if(s[i] >= '0' && s[i] <= '9'){
        prinf("Found digit.");
    }
}

your comparision of k belongs to [0, 10) is not a valid implementaion. This is doing comparision of ascii value of i'th char to the above range.

Answer 3

You have another option -- let strpbrk() do the work for you. strpbrk() (in string.h) locates the first occurrence within a string of any of the characters specified in the second accept string. So in the case of digits, your accept string is simply "0123456789". The function returns a pointer to the first occurrence found, so you simply loop twice adding 1 to the returned pointer address before the second iteration, considering any NULL return a failure to find two digits in the string.

A short working example is:

#include <stdio.h>
#include <string.h>

#define DIGITS "0123456789"
#define NDIGITS 2

int main (int argc, char **argv) {
  
  int ndigits = NDIGITS;
  char *p = argv[1];
  
  if (argc < 2 ) {    /* validate 1 argument given for password */
    fprintf (stderr, "error: insufficient input,\n"
                     "usage: %s password\n", argv[0]);
    return 1;
  }
  
  while (ndigits--) {                     /* loop NDIGITS times */
    if (!(p = strpbrk (p, DIGITS))) {     /* get ptr to digit (or NULL) */
      fputs ("error: password lacks 2 digits.\n", stderr);
      return 1;
    }
    p += 1;     /* increment pointer by 1 for next search */
  }
  
  puts ("password contains at least 2 digits");
}

Example Use/Output

$ ./bin/passwd2digits some2digitpass1
password contains at least 2 digits

or

$ ./bin/passwd2digits some2digitpass
error: password lacks 2 digits.

Let strpbrk() worry about how to find the digits -- all you need to do it use it. man 3 strpbrk