Having a problem with while function with 3 conditions

Question

So I have a void function called phone where I want to ask the user to input a 11 digit number starting with 0 where it's possible to have a single gap (eg. 01234567890 or 01234567 890).

void phone()
{
    char number[20];
    printf("Please enter your phone number: ");
    fgets(number, 20, stdin);
    while(number[0] != 0 && (strlen(number) != 11 || strlen(number) != 11 + ' '))
    {
        printf("please try again");
        fgets(number, 20, stdin);
    }
}

This is my current function that I have where I can make it work to accept numbers starting with 0 I can't get the second either or part of the condition to work. I'm relatively new to c so I'm not really sure if there is an easier way to do this. Also if you look at my output it jumps from the first print statement to the second before asking for input, not sure where the fault is there as I have fgets before the while function. output

Answer 1

There are several problems in your code.

For instance:

number[0] != 0

doesn't check if the first character is the digit/character 0. It checks whether the numeric value is zero. Two very different things...

To check for the digit/character 0 you must do:

number[0] != '0'

Further, this strlen(number) != 11 + ' ' doesn't do anything near what you expect.

In general the check you want to do is far to complex to be a condition in a while statement. I'll suggest that you use a function for the validation. Like:

int number_is_invalid(const char* number)
{
    // add code to validate number
    // return 1 if invalid
    // return 0 if valid
}

void phone()
{
  char number[20];
  printf("Please enter your phone number: ");
  fgets(number, 20, stdin);
  number[strcspn(number, "\n")] = 0;   // remove the trailing \n

  while (number_is_invalid(number))
  {
    printf("please try again");
    fgets(number, 20, stdin);
    number[strcspn(number, "\n")] = 0;      // remove the trailing \n
  }
}

So how could the function look?

Well, perhaps like:

// Function to validate a number
// - A valid number has 11 digit and optionally 1 space
// - A valid number must start with the digit 0
int number_is_invalid(const char* number)
{
    // Must start with the digit 0
    if (*number != '0') return 1; // invalid - doesn't start with digit 0

    int number_of_spaces = 0;
    int number_of_digits = 0;
    while(*number)
    {
        if (*number == ' ')
        {
            ++number_of_spaces;
        }
        else if (*number >= '0' && *number <= '9')
        {
            ++number_of_digits;
        }
        else
        {
            return 1;  // invalid character in input
        }
        ++number;
    }
    if (number_of_digits != 11) return 1; // invalid - wrong number of digits
    if (number_of_spaces > 1) return 1;   // invalid - too many spaces
    return 0;
}

Answer 2

You have the explanation in the code:

int countSpaces(char numbers[])
{
/*
This function counts the spaces from a given char and returns it
*/
int counter=0;
for(int i=0;i<strlen(numbers);i++)
    if(numbers[i]==' ')
        counter++;
return counter;
}

void phone()
{
char number[20];
printf("Please enter your phone number: ");
fgets(number, 20, stdin);
while(number[0] != '0' || strlen(number)-1-countSpaces(number)!= 11 /*|| countSpaces(number)>1*/)
{
    //[1]You compared number[0] != 0 which is always true number[0] is char and is different from 0 because it s int;
    //[2]When you input a char array there will be added '\0' at the end of the array
    //'\0'-represents the end of char array and will be counted so we add -1;
    //[3]We decrease the number of spaces found in the array;
    //[4]If you really want to have exact one space delete the comments from condition
    printf("%d\n",countSpaces(number));
    printf("please try again");
    fgets(number, 20, stdin);
}
}

Answer 3

You basically want this:

#include <stdio.h>
#include <string.h>           // needed for strcspn

// counts the number of spaces
int spacecount(const char* str)
{
  int count = 0;
  for (int i = 0; str[i] != 0; i++)
  {
    if (str[i] == ' ')
      count++;
  }

  return count;
}

void phone()
{
  char number[20];
  printf("Please enter your phone number: ");
  fgets(number, 20, stdin);
  number[strcspn(number, "\n")] = 0;   // remove the trailing \n

  while (number[0] != '0' || strlen(number) != 11 || spacecount(number) > 1)
  {
    printf("please try again");
    fgets(number, 20, stdin);
    number[strcspn(number, "\n")] = 0;      // remove the trailing \n
  }
}

• fgets leaves the string with a trailing \n which needs to be removed, that's what number[strcspn(number, "\n")] = 0 does.
• you need to count the number of spaces your phone number contains and make sure there is no more than one whitespace and that its length is exactly 11 and that it starts with a '0' (ASCII value of 0 digit) and not a 0 which is the NUL character. That's exactly what the number[0] != '0' || strlen(number) != 11 || spacecount(number) > 1 condition does.

You still need to ensure that the user hasn't entered anything other than digits such as letters or other characters. I leave this to you as an exercise.

This addresses your immediate problems, but there is still room for improvement though such as:

• writing your own fgets that removes the trailing \n
• doing the validation in a separate function