Thread: a question about fflushing

Hi. I'm doing this loop:

Code:

     char answer[3];

       do {
              printf("Enter (y/n): ");
              fgets(answer, sizeof(answer), stdin);
       } while (!(answer[0] == 'y' && answer[1] == '\0') &&
                     !(answer[0] == 'n' && answer[1] == '\0'));

It keeps printing:

Enter (y/n): Enter (y/n):

twice after each typo someone makes.

I addes fflush() after the printf() since I'm not using the newline, and I still
get the same double output. What am I doing wrong?


Thanks

> answer[0] == 'y' && answer[1] == '\0'
This will never be true.
Your string at the very least is going to contain "y\n\0"

If there is room in the buffer, fgets() will store the newline, it not, it will leave the newline on the input stream for the next call.
There is always a \0 in the buffer.

Sorry that was a typo:

I meant, and in my program had:

[code]

while(!(answer[0] == 'y' && answer[1] == '\n') &&
!(answer[0] == 'n' && answer[1] == '\n'));


My question is about the double printing on typos within
the loop

Enter (y/n): Enter(y/n):

after every mistake the user makes:
Instead of

Enter (y/n): s
Enter (y/n):

Thanks

If the input doesn't fit (you only allow for one character of actual visible type), or there are previous calls to getchar() or scanf() in the code, then you'll see the effect you describe.