C++ pointer-to-char question

Question

char *ps;
ps = &anotherChar;
cout << ps;

Why this displays the value of anotherChar not just the address?.

Answer 1

The operator << assumes you want to print the string not the address of the string.
Since ps points to the string, if you want to print the address of ps use "<< &ps"

Answer 2

There is an operator<< overload for char * which interprets it as a string to be output. If you want an address, you could to cast to void*.

Answer 3

Because char*'s are c-style strings. It does what it thinks you want, which is print out the c-string pointed to at ps.

You could print the address with:

cout << static_cast<void*>(ps) << endl;

Answer 4

The reason why is that a char* in C/C++ is a dual purposed type. It can be both a pointer to a single character or a C style string. The cout function chooses to treat this as a C style string and hence will print the values instead of the address.

This is also a potential problem for your application. C style strings end with a null terminator value. Hence this has the potential to read illegal memory since it's not properly initialized.

Answer 5

The << operator has numerous overloads for working with streams.

One of these prints a zero terminated string for you. The type of the variable used to point to a zero terminated string is a char*. This is what you've used. If you wrote:

char* ps = "hello";
cout << ps;

This overload would print "hello". It's probably just luck that there is a zero in memory after anotherChar so that the overload of the << stops printing characters after printing anotherChar. There is a good chance that if this isn't a zero you'd print out random characters from memory until a zero was encountered.

Answer 6

If you want the pointer value:

cout << (void *) ps;

Answer 7

C++ does not have a built in type for strings. It uses a null terminated (array that ends in '\0') char * (plus or minus const qualifiers) or char [] as a convention to represent strings. All the Standard C libraries use this convention.

So, the STL ostreams also use this convention.