understanding character pointers

[ndcn541]

I was expecting the cout to print addresses since b and d are pointers, but only b prints the value and not the address, why is that?

int main()
{
  
  char a[] = "abcd";
  char* b = a;
  
  char c = 'c';
  char* d = &c;
    

    cout << b << d << endl;
    return 0;
}

[Repeater]

<< is effectively shorthand for a function, operator<< that accepts a char pointer.

There are many other versions of the same function that accept other kinds of parameter.

It was decided, when creating C++, that the operator<< here that accepts a char pointer will NOT output the value of the char pointer, but will instead output the character it's pointing at, and all the subsequent characters in memory until it hits a zero.

This was a deliberate choice, because it's so common to want to output an array of characters that has a zero on the end of it, because in C that's what a string is, and you keep track of them by holding onto a pointer to the first character.

Note that when you do this: cout << d you've got a problem because d is a pointer to a char that does NOT have a zero right after it, so you're going to keep reading whatever memory is after that char and outputting it. Might be nonsense, might be some other variable, might just crash.

[seeplus]

Just cast b d to say long int (or long long int for x64). It will then display the address. You might want to also use std::hex to get the address display in hex rather than decimal.

#include <iostream>
#include <iomanip>

int main()
{
	char a[] = "abcd";
	char *b = a;

	char c = 'c';
	char *d = &c;

	std::cout << std::hex << reinterpret_cast<unsigned long>(b) << "  " << reinterpret_cast<unsigned long>(d) << '\n';
}

18ff34  18ff33