Why does printing a char* give a string rather than an address?

Question

I'm going through the basics of learning C++, but keep hitting a wall when trying to decipher the following about chars and pointers. Included are line comments giving my current understanding of what's going on. Given that I have code like below:

    using namespace std;
    int main()
    {
          //String literal is an array of chars
          //Array address gets assigned to a ptr of char
          char myletters[] = {'h','i'};
          char* lp = myletters;
          cout << *lp << endl;

          //Logically equivalent to above statements
          char* letters2 = "hi";
          cout << *letters2 << endl;

          //String literal turns into array of chars
          //Array of chars gets assigned to a ptr of chars 
          //Each ptr of chars gets stored into letters array
          char* letters[] = {"hi","hello"};
          cout << *letters << endl;   
    }

My output will be:

h
h
hi

My question is: when I use the final cout to print the contents of *letters, why do I get the string "hi" rather than the address of "hi" or the address of the first character in "hi"? I get that the first uses of cout are printing a char, and that the last cout is printing a char*, but I'm still wondering why it prints the complete string rather than the address as I would generally expect from a pointer.

Thanks kindly.

There's a special overload for the `operator<<()` with `char*`. — πάντα ῥεῖ, Mar 19 '16 at 18:11
@πάνταῥεῖ You should make that an answer because that *is* the answer. — BrainStone, Mar 19 '16 at 18:13
All string functions work on string pointers. If you need a memory location cast the pointer to a long __(long)stringPtr__ — Arif Burhan, Mar 19 '16 at 18:13
@ArifBurhan - cast it to `void*`. There's no guarantee that `long` is big enough to hold an address. — Pete Becker, Mar 19 '16 at 18:14

score 0 · Accepted Answer · answered Mar 19 '16 at 18:15

0

the << operator has a special definition for char* that prints the C-string it refers.

In your case, *letters has char* type (being letters a char*[], same as char**) and not char as *lp have.

answered Mar 19 '16 at 18:15

Emilio Garavaglia

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I would add that to print the adress, you just have to cast to a `void*` so that the overload for `const char *` is not used anymore: `cout << static_cast(*letters) << endl;` – Colin Pitrat Mar 19 '16 at 18:37

Why does printing a char* give a string rather than an address?

1 Answers1