Scanning character caused problem in devC

Question

So my code does the following:

1. Ask what's the option
2. If option is 1: Scan some numbers
3. If option is 2: Print those numbers
4. After each option, ask if user wanted to continue choosing (Y/N)

This is my main code

while(yesnocheck==1)
{
    printf("What's your option?: ");
    scanf("%d",&b);

    switch(b){
        case 1:
            printf("How many numbers?: ");
            scanf(" %d",&n);
            a=(struct sv*)malloc(n*sizeof(struct sv));

            for(int i=0;i<n;i++)
                scanf("%d",&((a+i)->num));
            break;
        case 2:
            for(int i=0;i<n;i++)
                printf("%d\n",(a+i)->num);
            break;
    }
    yesnocheck==yesnochecker();
}

And this is the yesnochecker function:

int yesnochecker()
{
    char yesorno;
    printf("Do you want to continue? (Y/N)");

        while(scanf("%s",&yesorno))
    {
        if(yesorno=='Y')
            return 1;
        if(yesorno='N')
            return 0;   
        printf("*Wrong input. Please reenter (Y/N): ");
    }
}

So on dev C++, my code won't run correctly. After it's done option 1, when I enter "Y" then choose option 2, case 2 will display some weird numbers. However it works well on online C compilers.

And then, when I change the char yesorno in yesnochecker() function to char yesorno[2] and treat it as a string, the code does work.

Can someone shed some light?

Answer 1

It is a bad idea to read a char c with scanf("%s", &c);. "%s" requires a buffer to store a string. The only string which fits into a char is an empty string (consisting only of a terminator '\0' – not very useful). Every string with 1 character requires 2 chars of storage – 1 for the character, 1 for the terminator ('\0'). Providing a char for storage is Undefined Behavior.

So, the first hint was to use the proper formatter instead – "%c".

This is better as it removes the Undefined Behavior. However, it doesn't solve another problem as the following sample shows:

#include <stdio.h>

int cont()
{
  char c; do {
    printf("Continue (y/n): ");
    scanf("%c", &c);
    printf("Input %c\n", c);
  } while (c != 'y' && c != 'n');
  return c == 'y';
}

int main()
{
  int i = 0;
  do {
    printf("Loop iteration %d.\n", ++i);
  } while (cont());
  /* done */
  return 0;
}

Output:

Loop iteration 1.
Continue (y/n): y↵
Input 'y'
Loop iteration 2.
Continue (y/n): 
Input '
'
Continue (y/n): n↵
Input 'n'

Live Demo on ideone

WTH?

The scanf("%c") consumes one character from input. The other character (inserted for the ENTER key) stays in input buffer until next call of any input function.

Too bad, without ENTER it is hard to confirm input on console.

A possible solution is to read characters until the ENTER key is received (or input fails for any reasons). (And, btw., getc() or fgetc() can be used as well to read a single character.):

#include <stdio.h>

int cont()
{
  int c;
  do {
    int d;
    printf("Continue (y/n): ");
    if ((c = fgetc(stdin)) < 0) {
      fprintf(stderr, "Input failed!\n"); return 0;
    }
    printf("Input '%c'\n", c);
    for (d = c; d != '\n';) {
      if ((d = fgetc(stdin)) < 0) {
        fprintf(stderr, "Input failed!\n"); return 0;
      }
    }
  } while (c != 'y' && c != 'n');
  return c == 'y';
}

int main()
{
  int i = 0;
  do {
    printf("Loop iteration %d.\n", ++i);
  } while (cont());
  /* done */
  return 0;
}

Output:

Loop iteration 1.
Continue (y/n): y↵
Input 'y'
Loop iteration 2.
Continue (y/n): Hello↵
Input 'H'
Continue (y/n): n↵
Input 'n'

Live Demo on ideone

Please, note, that I changed the type for the read character to int. This is because getc()/fgetc() return an int which is capable to store any of the 256 possible char values as well as -1 which is returned in case of failing.

However, it isn't any problem to compare an int with a character constant (e.g. 'y'). In C, the type of character constants is just int (SO: Type of character constant).