while loop stack bashing detected

Hi all, i have written a program to demonstrate while loops and arrays however when i exit the program by pressing n i get the following error message.
*** stack smashing detected ***: <unknown> terminated Aborted (core dumped)
process returned 134 (0x86) etc

this only happens when i try to quit the program after the first time round.

here is the code

Code:

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#include <stdio.h>
#include <stdlib.h>

int main()
{
    int a=0;
    int i;
    char answer;
    char words[4];

    while (a != 2)
    {
        //check to see if first time run and ask the appropriate question
        if (a == 0)
        {
            printf("Would you like to enter a word y/n\n");
            scanf(" %c",&answer);
        }
        else if (a == 1)
        {
            printf("Would you like to enter another word y/n\n");
            scanf(" %c",&answer);
        }
        else
        {
            printf("Sorry I didnt understand that Please enter "y" to enter a word or "n" to quit\n");
            scanf(" %c",&answer);
        }

        //check what the answer was and run for loop if need be
        if (answer == 'y' || answer == 'Y')
        {
            printf("Please enter a five letter word\n");
            scanf(" %s",words);
            for (i = 0; i < 5; i++)
            {
                printf("The letter in posistion %d is %c\n",i ,words[i]);
            }
            printf("the word entered was %s\n",words);
            a=1;
        }
        else if (answer == 'n' || answer == 'N')
        {
            printf("goodbye!!\n");
            a=2;
        }
        else // somthing else typed in // todo error trapping for more than one character entered
        {
            a=3;
        }
    }

    return 0;
}

---

many thanks
Coop

Code:

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char words[4];

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Is not large enough to hold a 5 letter word - You will need 6 chars (5 + an extra for the nul char

Quote:

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Originally Posted by Click_here

Code:

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char words[4];

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Is not large enough to hold a 5 letter word - You will need 6 chars (5 + an extra for the nul char

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thanks for the reply is this the same for an array of integers ie to hold 5 integers i need myarray[5]

many thanks
coop

An array of int to hold 5 integers is indeed something like myarray[5]. Your problem is that you don't seem to understand the difference between an array of char and a C-string. A C-string requires a termination nul character so in order to hold a C-string with 5 characters you need an array of size 6 so that there is room for that termination character that the "%s" specifier will automatically adds to the characters entered. Also you should never use a function that will try to retrieve more characters than the size string, the scanf() function can accomplish this limiting by using the proper width specifier. This specifier must be at least one smaller than the size of the array. For example:

Code:

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char mystring[6];  // A C-string large enough to hold 5 charcters.
scanf("%5s", mystring); // Limit scanf() from overflowing the bounds of the array and save room for the end of string character.

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Quote:

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Originally Posted by click_here

You will need 6 chars (5 + an extra for the nul char

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For more info, see C Strings - Cprogramming.com