Why pointer returns value, not adress?

Question

Why in my code I see value, which stored in the address, but not adress?

char *fortunes[] =
    {
        "1",
        "2",
        "3",
        "4"
    };

    cout << *fortunes[2];   // result 3
    cout << fortunes[2];    // result 3, but I expected to see adress

@KarolyHorvath: It is a pointer, so an address is not an unreasonable expectation. — Benjamin Lindley, Mar 11 '15 at 10:37
@KarolyHorvath Yes, and OP does not know the overload for that. Thus the question. — Baum mit Augen, Mar 11 '15 at 10:38
I'm sure he *knows* it, he just hasn't realised that that's being used. — Karoly Horvath, Mar 11 '15 at 10:39

score 4 · Accepted Answer · answered Mar 11 '15 at 10:40

4

There is an overload for std::ostream& operator<< which takes a const char* and interprets it as a null-terminated string, printing out characters until it finds the null terminator. If you want to print the value of the pointer (the address it holds), you can cast it to void*.

For example

cout << reinterpret_cast<const void*>(fortunes[2]) << endl;

answered Mar 11 '15 at 10:40

juanchopanza

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or to make it more obvious, as we use this overload all the time: `cout << "3";` - you wouldn't expect an address to be printed, would you? – Karoly Horvath Mar 11 '15 at 10:45

Why pointer returns value, not adress?

1 Answers1