Question

Can someone explain the output of the following code

char* a[] = {"ABC123", "DEF456", "GHI789"};
char **p = a;
cout<<++*p<<std::endl;
cout<<*p++<<std::endl;
cout<<++*p<<std::endl;

Output:

BC123
BC123
EF456

What is confusing to me is the different behavior of ++*p and *p++. I was expecting the output to be:

ABC123
DEF456
GHI789

Answer 1

char* a[] = {"ABC123", "DEF456", "GHI789"};
char **p = a;
cout<<++*p<<std::endl; // 1
cout<<*p++<<std::endl; // 2
cout<<++*p<<std::endl; // 3

On line 1 first *p will be pointing to the element in the array "ABC123" and the ++ moves one forward and so ‘BC123’ is printed.

On line 2 *p is still pointing to BC123 so this is printed and then once printed the ++ is carried out. This moves the pointer to the 2nd element in the array

On line 3 it is the same as line 1. You have taken the contents of p (Now the 2nd element in the array) and moved one character in that string, thus printing EF456

(Also have a look at here Pointer arithmetic on string type arrays, how does C++ handle this? as I think it might be useful to get an understanding of what is happening)

To print what you expected the following would work:

cout<<*p++<<std::endl;
cout<<*p++<<std::endl;
cout<<*p++<<std::endl;

Or

cout<<*p<<std::endl;
cout<<*(++p)<<std::endl;
cout<<*(++p)<<std::endl;

Or various other ways (taking into account precedence as others have said)