How does std::cout work with char pointers?

Question

Recently I've been learning about C++ pointers and I'm confused when it comes to char.

char *c = "sampletext";

cout<< *c <<endl;

cout<< c <<endl;

cout<< &c <<endl;

To be more precise: why does cout<< c <<endl; print whole C-string instead of address (as it takes place in int pointers for example), and why does cout<< *c <<endl; print first letter instead of whole C-string. Thank you in advance

Answer 1

This happens because the compiler checks the type of the argument and calls the overloaded version of the operator that receives that type.

In the C language, strings are implemented as char arrays terminated with the '\0' (nul) character, and arrays are referred to using pointers. So, when the argument received is c , the overloaded operator that receives a char* prints the content of the array up to the nul terminator.

When the argument received is *c , this dereferences the pointer to access its data, and since this is a pointer to char , the overloaded operator that receives a char prints it as a single char .

To print the address of the string array, you can cast the char* pointer to void* , like follows:

std::cout << (void*) c << std::endl;