Why doesn't the program throw segfault error on editing fixed string?

Question

I am newbie to Unions. In the following code, according to my understanding, we are editing a fixed string literal "GeeksQuiz" by trying to replace with 'S' inplace of G. Shouldn't it throw some segfault error. If not then why isn't the answer "SeeksQuiz" and is "GeeksQuiz" ?

# include <iostream> 
# include <string.h> 
using namespace std; 
  
struct Test 
{ 
  char str[20]; 
}; 
  
int main() 
{ 
  struct Test st1, st2; 
  strcpy(st1.str, "GeeksQuiz"); 
  st2 = st1; 
  st1.str[0] = 'S'; 
  cout << st2.str; 
  return 0; 
} 

Answer 1

Let me breakdown what is happening here line by line:

1. Struct Test st1, st2; creates two local struct Tests. These are not pointers / references, but the structs themselves. if you try to print out the structs, they will have random characters since str is not initialized.

2. strcpy(st1.str, "GeeksQuiz"); Copies the string literal GeeksQuiz to st1.str, so st1.str should have GeeksQuiz with null terminating character.

3. st2 = st1; We set st1 to be equal to st2. Both are not pointers or references, (flat objects) so its like

   int a = 5;
   int b = 6;
   a = b; // a = 6
   a = 7; // a = 7, b is still 6
   

So st2.str = GeeksQuiz, st1.str = GeeksQuiz, but they (st1, st2) both are separate entities.

4. st1.str[0] = 'S'; st1.str = SeeksQuiz, st2.str = GeeksQuiz.

5. cout << st2.str; st2.str = GeeksQuiz, which is what gets outputted.

We are nowhere trying to change the actual literal itself, anytime we use literals to initialize something, it is copied to that variable. I hope all this clarifies your question.

If st1 and st2 were pointers to structs, then setting st1 = st2 both will point to the same place in memory, so changing something through *st1 will be reflected in *st2 as well, but thats not the case here.