Thread: Noob code question

Just a thought...

Code:

#include <stdio.h>
#include <ctype.h>

int main(void)
{
   char line[80] = "y"; /* Buffer size excessive? I'm being lazy. */
   while ( tolower(line[0]) != 'n' )
   {
      /* ... */
      printf("Would you like to play again? (y/n) ");
      fflush(stdout);
      fgets(line, sizeof line, stdin);
   }
   return 0;
}

/*
Would you like to play again? (y/n) sure
Would you like to play again? (y/n) yes
Would you like to play again? (y/n) why not
Would you like to play again? (y/n) no
*/

You could use this too:

Code:

int main() {

char ans;
printf("Play again? (y/n");
ans = fgetc;

if (ans == 'y') {
// play again
} else {
//exit or whatever
}

return(0);
}

Originally Posted by octoc

You could use this too:

For your own practice, try making compilable, working code that repeats the prompt in a loop.

what i pick up from your code is that u need

&playagain when using scanf or will crash
playagain=='y' because its evaluationg playagain to see if its (y) when u put playagain = 'y'. this mean y is giving to playagain

Code:

#include <stdio.h>
#include <conio.h>
int playagain;

int main(void)
{
    char playagain;
    printf("Would you like to play again? (y/n)");
    scanf("\n&#37;c", &playagain); 
 
  
    if ( playagain == 'y') 
    {
        printf("Would you like to play again? (y/n)");
         scanf("\n%c", &playagain); 
     if ( playagain == 'y') 
    {
        printf("Would you like to play again? (y/n)");
         scanf("\n%c", &playagain);
         
     }
    
    }
    else if ( playagain == 'n')
     {
        printf("Maybe another time");
     }
     
 
    getch();
    
    return 0;
}

you are missing fflush(stdout) between printf and scanf
also conio.h and getch could be fully avoided here - use standard equiivalent instead