How to use isdigit() for a pin while loop [closed] - c

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This program accepts any four digit number as the pin , so i used a strlen() to find out whether the pin is has four characters, but I need to make sure that user enters four numbers, how do I use isdigit() before the loop and for the loop condition?
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
char pin [4];
printf("Please enter your pin : \n") ;
scanf("%4s" , &pin) ;
system("cls") ;
while (strlen(pin) != 4) {
count = count + 1 ;
if(count < 3){
printf("Invalid Pin \n") ;
printf("Please enter your pin again \n") ;
scanf("%4s", &pin) ;
system("cls");
}
else if (count == 3){
printf("Sorry you can't continue , Please contact your bank for assistance !") ;
return 0 ;
}
}

Answer 1

There are so many issues with this code I may not have covered every one of them in this answer:
Where is your main function? You can't have code outside a function; that doesn't work. C code other than global variable declarations, assignments to constant values, typedefs, struct and enum definitions, must go in a function. In your case you probably want the main function to house the code starting at line 6.
When calling scanf with a string argument, don't take the address of the string - the array is a reference in and of itself.
Calling strlen(pin) before any value has been copied into pin is 100% undefined behavior. Since the memory is uninitialized the strlen function will keep looking for a null character and possibly go out of the array bounds.
C strings are null-terminated. When you declare a string intended to hold n characters, you need to declare the array with a size of n+1 to have room for the null character.

Answer 2

First to answer your question
I would write a helper function 'validatePin' which would check the pin length and validate that the pin is a numeric pin, you can extend this function to do any other validation you require. It might look something like the following
const int PIN_OK = 0;
const int PIN_INVALID_LEN = -1;
const int PIN_INVALID_CHARS = -2;
int validatePin(const char *pin)
{
// Valdiate PIN length
if (strlen(pin) != 4) return PIN_INVALID_LEN;
// Validate PIN is numeric
for (int i = 0; i < 4; ++i)
{
if (!isdigit(pin[i])) return PIN_INVALID_CHARS;
}
return PIN_OK;
}
Then you can adjust your while loop to look something like the following
while (validatePin(pin) != PIN_OK)
{
....
}
Other points regarding your code.
Your char buffer used for scanf does not account for the null terminator. You need to increase the buffer size.
The char array is already giving you the address of the buffer, there is no need to use & to take the address of the char array in scanf.
I have not run your code, so there might be other issues that I missed at first glance.

Answer 3

There are a number of ways to go about doing what you need to do. You can either check your input length, or just validate what it is you have read, your choice. One approach would be:
#include <stdio.h>
#include <ctype.h>
enum { TRIES = 3, NPIN };
int main (void) {
char pin[NPIN+1] = "";
size_t tries = 0, pinok = 0;
for (; tries < TRIES; tries++) { /* 3 tries */
printf ("Please enter your pin : ");
if (scanf (" %4[^\n]%*c", pin) == 1) { /* read pin */
pinok = 1;
size_t i;
for (i = 0; i < NPIN; i++) /* validate all digits */
if (!isdigit (pin[i]))
pinok = 0;
if (pinok) /* if pin good, break */
break;
}
}
if (!pinok) { /* if here and not OK, call bank */
fprintf (stderr, "Sorry you can't continue, Please contact "
"your bank for assistance.\n") ;
return 1;
}
printf ("\n correct pin : %s\n\n", pin); /* your pin */
return 0;
}
(note: to protect against an excessively long string entered at one time, you should empty stdin at the end of the outer for loop each iteration)
Example Use/Output
Failed case:
$ ./bin/pin
Please enter your pin : a555
Please enter your pin : 555a
Please enter your pin : 55a5
Sorry you can't continue , Please contact your bank for assistance.
Successful case:
$ ./bin/pin
Please enter your pin : 2345
correct pin : 2345
Look it over an let me know if you have any questions.