Question

int main() {


  char **k;
  char *s ="abc"; 
  char *b ="def";    

  k = &s;
  k++;
  k = &b;
  cout<<*(k - 1)<<endl; // nothing but newline. Shouldn't I get "abc"? 
  //EDIT: corrected a typo should be *(k - 1)  
}

I got nothing but a newline from cout. When I looked at the behavior of char* I get the impression that since I have the address of of the first character I could use char* like an array, which is true. However, for char** this behavior seems to be totally different, when I tried k++ it doesn’t seem to behave like an array. Why is that?

Also when I tried (K + 1) = &b; I got an error, why couldn’t I do that?

Answer 1

k is a pointer to a pointer to a char, so when you advance it, it does not walk down the string as you’re expecting it to. Instead, it now points to the char * that is located adjacent to the one it was pointing to. In your case there is none, so you’re de-referencing a random memory location when you print the value of *(k - 1).

(K + 1) = &b;

The above is an error because you must have an l-value on the left side of an assignment, it cannot be a temporary expression (r-value).

EDIT:
Here’s an example that’ll hopefully be easier to follow than reading about the mistake you’ve made.

int main()
{
  char **k;
  char *s[2] = {"abc", "qwerty"};

  k = &s[0];
  std::cout << *(k + 1) << std::endl; // prints "qwerty"

  k++;
  std::cout << *(k - 1) << std::endl; // prints "abc"

  std::cin.get();

  return 0;
}