Difference between the two array pointer initialization methods

There is no difference between these cases.

They both assign the address of a string literal to a char pointer, but the latter uses an unusual syntax.

Likewise, int a = 42;

and int a = {42};

equivalent.

In the comments you mentioned char *a = "Hello";

and char a[] = "Hello";

.

They are completely different. The second creates an array. It means that

char a[] = {'H','e','l','l','o','\0'};

      

        
        
        
      

    

There is []

no room inside . becase the compiler can guess the size of the array for you ( 6

in this case).

And another case is completely different.

When you use "string literal"

outside the intialization of the array char

like in this case

printf("Hello");

      

        
        
        
      

    

or

char *a = "Hello";

      

        
        
        
      

    

the compiler implicitly creates an array const char

to hold your string. As you know, in these contexts, the name of the array decays to a pointer to its first element. So,

char *a = "Hello";

      

        
        
        
      

    

and

const char internal_array[] = "Hello";

char *a = internal_array; // same as  char *a = &internal_array[0];

      

        
        
        
      

    

are equivalent.

If you try to do something like

char *a = "Hello";
a[0] = 'A';

      

        
        
        
      

    

you get a failure because, despite the fact that it is a pointer to const-not char

, a

in fact, points to a constant string. It's not a good idea to change it.

As for another case,

char a[] = "Hello";
a[0] = 'A';

      

        
        
        
      

    

fine. In this case, you will get a new array char

that contains the string. Of course, this is not a const, so you can change it.