Why does scanf("%s", val); return null?

Question

I'm trying to read a single character into a char* called val and when I use scanf() it returns null. Here is my code:

#include <stdio.h>

int main(int argc, char *argv[])
{
    char *val;
    if (argc == 2)
    {
        val = argv[1];
    }
    else
    {
        scanf("%s", val);
    }
    printf("Val = %s", val);
}

If I try to use malloc(), it will continuously read from standard input if the input is one character long, which is how long I want my input to be.

Answer 1

Why does scanf("%s", val); return null?

scanf() needs to read input into a valid location.

---

With scanf("%s", val);, code passes to scanf() an uninitialized pointer val to scanf(). This leads to undefined behavior (UB).

Instead, pass a pointer to memory ready to accept the input.

• Pass a pointer to a location ready to accept characters.

• Use a width to avoid overrun.

• check the return value.

    char *val;
    char s[100+1]; 
    if (argc == 2) {
      val = argv[1];
    } else {
      if (scanf("%100s", s) != 1) {
        fprintf(stderr, "Invalid input\n");
        return -1;
      }
      val = s;
    }

To read a single character use:

char ch; 
if (scanf("%c", &ch) != 1) {
  fprintf(stderr, "Invalid input\n");
  return -1;
}