Printing character pointer versus printing character by dereferencing

Question

char c = 'a';

char* p = &c;

cout << p << endl;

cout << *p << endl;

This is a code written in C++ language.

Why in the first cout statement the program tries to print until it finds a null character and in the second statement it just prints a single character?

Answer 1

In C++, pointers to char are often considered a "C string" by convention -- which is a null-terminated string of char s. This is an expected convention in much of the C++ standard library -- and this holds true in std::ostream as well.

In particular, in your two calls:

1. The pointer p will call std::ostream 's operator<<(const char*) overload -- which prints it as a null-terminated character string, and
2. The char *p will call std::ostream 's operator<<(char) overload for characters, which prints the singular character

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Note: If you want to print the address of the pointer itself, you will want to call the operator<<(const void*) overload -- which requires an explicit cast to void* :

std::cout << static_cast<void*>(p) << std::endl;

This operator is not discovered automatically since the language sees that operator<<(const char*) does not require any type conversions to take place for p .

Answer 2

It's specific thing of C and C++. Pointer of char means string with zero symbol at end.