undefined behaviour while using fgetc(stdin)

Question

Before asking this question, I have read: fgetc(stdin) in a loop is producing strange behaviour. I am writing a program to ask the user if they want to exit. Here is my code:

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

int ask_to_exit()
{
    int choice;
    while(choice!='y'||choice!='n')
    {   printf("Do you want to continue?(y/n):");
        while((fgetc(stdin)) != '\n');
        choice = fgetc(stdin);
        if(choice=='y') return 0;
        else if(choice=='n') return 1;
        else printf("Invalid choice!\n");
    }
}

int main()
{
    int exit = 0;
    while(!exit)
    {
        exit = ask_to_exit();
    }
    return 0;
}

Since fflush(stdin) is undefined behaviour, I did not use it. After following the solution in the other question, I still get the error. Here is a test run of the above program:

$./a.out
Do you want to continue?(y/n):y<pressed enter key>
<pressed enter key>
Invalid choice!
Do you want to continue?(y/n):y<pressed enter key>
<pressed enter key>
Invalid choice!
Do you want to continue?(y/n):n<pressed enter key>
<pressed enter key>
Invalid choice!
Do you want to continue?(y/n):n<pressed enter key>
n<pressed enter key>
<program exits>

does that matter?? I am checking with while(choice!='y'||choice!='n') , so even if it is not initialized ,it will not be 'y' or 'n',so it should work,right?? — pro neon, Mar 13 '18 at 05:31
`do... while()` loop is more appropriate for such cases, instead of `while` loop. — H.S., Mar 13 '18 at 05:31
Yes it matters. Using a local variable without initializing is undefined behavior — H.S., Mar 13 '18 at 05:33
`choice!='y'||choice!='n'` is *always* true. If it is one, it can't possibly be the other. If it is neither, still true. If it is *both*, call your university astrophysics department, because you've discovered a worm hole. — WhozCraig, Mar 13 '18 at 05:35
@WhozCraig: the first time through the loop, choice has not been initialised so its value is indeterminate. And an indeterminate value might compare equal to `'y`'` and then immediately compare equal to `'n'`. No astrophysics required; see http://www.open-std.org/jtc1/sc22/wg14/www/docs/dr_260.htm — rici, Mar 13 '18 at 06:53
Note that the choice of `exit` as the name for a variable isn't a particularly good idea. It means you can no longer call the function of that name in the function where you define the variable. It isn't technically wrong; it is ill-advised. — Jonathan Leffler, Mar 13 '18 at 08:02

Stephen Docy · Accepted Answer · 2018-03-13T05:43:52.007

1

You need to get some input before checking the value of choice or it will be uninitialized. And also, you are draining the left-over input before grabbing the first character, you should do it after:

int ask_to_exit()
{
    int choice;

    do
    {
        printf("Do you want to continue?(y/n):");
        choice = fgetc(stdin);

        while (fgetc(stdin) != '\n');

        if (choice == 'y') {
            return 0;
        } else if (choice == 'n') {
            return 1;
        } else {
            printf("Invalid choice!\n");
        }
    } while (1);
}

ouput:

Do you want to continue?(y/n):g
Invalid choice!
Do you want to continue?(y/n):h
Invalid choice!
Do you want to continue?(y/n):j
Invalid choice!
Do you want to continue?(y/n):k
Invalid choice!
Do you want to continue?(y/n):m
Invalid choice!
Do you want to continue?(y/n):y
Do you want to continue?(y/n):y
Do you want to continue?(y/n):y
Do you want to continue?(y/n):n
Press any key to continue . . .

edited Mar 13 '18 at 05:43

answered Mar 13 '18 at 05:37

Stephen Docy

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That `||` should be `&&`. – WhozCraig Mar 13 '18 at 05:39
after changing || to && it has still the same problem. – pro neon Mar 13 '18 at 05:43
@WhozCraig...true that is the correct logic, though since both `y` and `n` cause the function to return, a simple `true` for the while check would actually be best – Stephen Docy Mar 13 '18 at 05:43
Check this:https://ideone.com/LFCbwn. It's the same code you suggested.Still it doesn't work. – pro neon Mar 13 '18 at 05:49
Did you try my exact code or did you just modify yours to match mine? Maybe you forgot to move the `while (fgetc(stdin)...` code to after `choice = fgetc(stdin);` – Stephen Docy Mar 13 '18 at 05:49
Sorry I just had forgot to put n in the input. Thanks! – pro neon Mar 13 '18 at 05:50
sorry but it still has the same problem and I just pasted your code. – pro neon Mar 13 '18 at 05:52
Hmmm, it works fine for me, specifically how is your code failing? – Stephen Docy Mar 13 '18 at 05:53
it works in my computer but fails in the online ide. You can check this yourself:https://ideone.com/LFCbwn.Anyways sorry for any trouble. – pro neon Mar 13 '18 at 05:55
Quite possibly because it looks like the language is set to `C++` in the ide? – Stephen Docy Mar 13 '18 at 05:58
might be but since there was no option to change it, I didn't knew it was set to c++. – pro neon Mar 13 '18 at 05:59
I'm not sure if that is the issue or not. I threw the code in a cpp file in Visual Studio and it still worked fine. Might be some weird behavior with respect to how the ide is feeding the input to the program. – Stephen Docy Mar 13 '18 at 06:00

R Sahu · Answer 2 · 2018-03-13T05:40:26.053

I see the following problems in your code.

You are using choice before initializing it.
You are skipping a line of input before reading anything into choice.

Error noticed by @WhozCraig:

choice!='y'||choice!='n' is always true.

I suggest:

int ask_to_exit()
{
   int choice;
   while ( 1 )
   {
      printf("Do you want to continue?(y/n):");
      choice = fgetc(stdin);
      if ( choice == 'y' )
      {
         return 0;
      }
      if ( choice == 'n' )
      {
         return 1;
      }

      printf("Invalid choice!\n");

      // Skip rest of the line.
      int c;`
      while( (c = fgetc(stdin)) != EOF && c != '\n');

      // If EOF is reached, return 0 also.
      if ( c == EOF )
      {
         return 0;
      }
   }
}

The return values are ok, `0` is returned for `y`, which makes `while(!exit)` true, and so the program continues. Returning `1` for `n` causes `while(!exit)` to be false, and the program exits. — Stephen Docy, Mar 13 '18 at 05:39

undefined behaviour while using fgetc(stdin)

2 Answers2