Question

 #include <stdio.h>

  int main(void){

  char *p = "Hello";  
  p = "Bye";              //Why is this valid C code? Why no derefencing operator?

  int *z;
  int x;
  *z = x
  z* = 2                 //Works
  z  = 2                 //Doesn't Work, Why does it work with characters?

 char *str[2] = {"Hello","Good Bye"};

 print("%s", str[1]);      //Prints Good-Bye.     WHY no derefrencing operator?
                          // Why is this valid C code? If I created an array with pointers
                         // shouldn't the element print the memory address and not the string?
  return 0;

  }

My Questions are outlined with the comments. In gerneal I’m having trouble understanding character arrays and pointers. Specifically why I can acess them without the derefrencing operator.

Answer 1

When you type “Bye”, you are actually creating what is called a String Literal. Its a special case, but essentially, when you do

p = "Bye";

What you are doing is assigning the address of this String literal to p(the string itself is stored by the compiler in a implementation dependant way (I think) ). Technically address to the first element of a char array, as Richard J. Ross III explains.

Since it is a special case, it does not work with other types.

By the way, you should likely get a compiler warning for lines like char *p = "Hello";. You should be required to define them as const char *p = "Hello"; since modifying them is undefined as the link explains.

---

As to the printing code.

print("%s", str[1]); 

This doesnt need a dereferencing operation, since internally %s requires a pointer(specifically char *) to be passed, thus the dereferencing is done by printf. You can test this by passing a value when printf is expecting a pointer. You should get a runtime crash when it tries to dereference it.