C++ char array pointer confusion

Question

I need some help understanding pointers:

Basic Pointers:

int i = 3;           
cout << i << endl;   //prints 3
cout << &i << endl;  //prints address of i

int * p = &i;
cout << p << endl;   //prints the address p points to
cout << &p << endl;  //prints the address of p
cout << *p << endl;  //prints the value stored at the address p points to

Now the confusion:

char *buf = "12345"; 
cout << &buf[2] << endl;  //prints 345
cout << buf+2 << endl;    //prints 345
cout << *buf << endl;     //prints 1
cout << *(buf+2) << endl; //prints 3
cout << buf << endl;      //prints 12345
cout << &buf << endl;     //prints 001CFA6C

How do I print the address of buf[3]?

Actually, you shouldn't use this initialization: The type of "12345" is char const[6]. To avoid backward compatibility issues it is allowed to decay this type to char*` but you loose the const. The initialization should be char const* buf = "12345";. — Dietmar Kühl
– Dietmar Kühl, Commented Sep 2, 2012 at 23:58

Charles Salvia · Accepted Answer · 2012-09-03 02:36:01Z

3

char pointers are somewhat special in the sense that historically they have been used as strings in C. C++ is mostly backwards compatible with C, so it has support for C strings. So if you print a char pointer, rather than printing the address, it prints each character in the string until it reaches a NULL char, just like in C.

To print the actual address, cast the pointer to void*.

cout << static_cast<void*>(&buf[3]) << endl;

edited Sep 3, 2012 at 2:36

answered Sep 2, 2012 at 23:50

Charles Salvia

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Kerrek SB · Accepted Answer · 2012-09-02 23:49:44Z

1

The iostreams have a special overload for char pointers (treating them as pointers to a null-terminated array and printing the entire array). Bypass the overload by converting the pointer to a void pointer, which is printed as a numeric value:

std::cout << static_cast<void*>(buf + 3) << std::endl;

answered Sep 2, 2012 at 23:49

Kerrek SB

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Dietmar Kühl · Accepted Answer · 2012-09-02 23:50:12Z

1

Your problem is that you try to understand pointers using char. However, char is special. In particular char const* isn't treated like any other pointer but it is treated like a C-string. That is &buf[2] is indeed the address of the third character but this address is considered to be a the start of a null-terminated sequence of characters and printing this pointer doesn't cause the pointer to be printed but rather the string starting at this address. Try the same with ints to avoid this interaction.

answered Sep 2, 2012 at 23:50

Dietmar Kühl

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