Why does %c output me and %s throws an error?

Question

C++ / C#

I am writing code, and for some reason, if %s is replaced with %c, it outputs the rules, but if %s gives this error:
An exception was thrown: read access violation.
it was 0x4F.
Please help me decide, xs what it is, most likely something elementary, but I xs

#include <stdio.h>
#include <conio.h> //підключення бібліотеки текстового інтерфейсу
#include <locale>
void main()
{
  setlocale(LC_ALL, "Russian");

  char b = '|';
  char n = '#';
  char x = 'FIO';
  char y = 'kurs';
  char z = 'grup';
  char f;
  int k;
  char g;
  printf_s("\n%c%c%c%10s%10c", b,n,b,x,b);
}

Answer 1

char x = 'FIO';
Mistake.
Multi-character constants are implementation defined, used for heavy optimization, and are extremely unreliable in Visual C++. In GCC it's kind of ba-me, but in any case, converting to char 'FIO', which should take three bytes, will give 'F' or 'O'.
You need not a character constant, but a string one: And, accordingly, %s. And how does printf and any other function with a variable number of arguments work? These arguments are dumped in bulk into an area of ​​memory called the “call stack”, and printf starts parsing this bulk. To shift the pointer by one byte and say: this is char - and the %c format is used. Of course, an incorrect format will result in data type misinterpretation and failure on one or all platforms.
char x[] = "FIO";
%c - there is one byte in memory, interpret it as a single character
%s - there is a pointer to a zero-terminated string in memory, 4/8 bytes. (All arrays in native C are passed as pointers anywhere.)