I am learning about C++ pointers and need to clear up a few things. My understanding is that a simple pointer holds the address location of an object or an array and when you want to know the content of the address its pointing to, you put the * in front of it. But what I dont understand is, when you try to print the address location of what it holds, it instead prints the contents of that location instead. This is of course valid for char type pointers. If I make the pointer a type of int, then it prints the address location instead. So, my question is why isnt this consistent among pointer types? I think you might be already confused and want an example, so here it is:

#include <iostream>

using namespace std;


int main()
{
	char cArr[] = "abcdefgh";
	int  Arr[] = {1,2,3,4,5};

	cout << cArr << endl;
	cout << Arr << endl;


	return 0;
}

Ok, here I set up an array of chars and an array of ints. When I print the content of cArr I get the full string instead of the address. But when I print the content of Arr, I get the actual address location instead of 1. Obviously, if I have *cArr and *Arr, i get the contents of that location. So, my question is, depending on the type of Array, why is the output inconsistent? thanks EDIT: how can I print out the address of cArr, like I did for Arr? by using &?

actually i dont think i need to put a null character at the end of the string because i think the compiler does that for you.

i tried &cArr[0] and got the string output instead. but when I do &cArr, that seems to print out the address location. but that still really doesnt answer my question as to why char* is treated differently than int*

"char* is not handled the same way other pointers are by the cout object"

I wonder why?

&cArr[0]...

All that does is dereference the pointer and then make it back into a pointer.

The reason they are treated differently is that the "<<" is an overloaded operator. Depending on the type sent on the right, the function evalulates differently.

MSDN Documentation

oh ok. so in other words "cout" re-formats or re-interprets the output?

No. It doesn't reformat and reinterpret anything. It's just that char * , signed char * and unsigned char * are treated in a special way. In C there is no String type like in Java and C++, so all functions like printf treat char * as null-terminated strings. C++ inherited this behaviour, so cout.operator<< treates char * as a null-terminated strings. If you want to print the actual address cast it to void *.

And use std::string in C++ instead of char *.

thanks guys for your help. I was further experimenting about pointers when something didnt make sense to me. I thought this was odd. Consider the following code:

#include <iostream>using namespace std;void main(){	char *cp, Arr[] = "abcdefgh";	int *ip, Irr[] = {1,2,3,4,5};	cp = Arr;	ip = Irr;	cout << *(cp) << " " << *(cp+1) << endl;	}

Well the output is obvious, its "a" and "b".

But if i change the cout line to this instead:

cout << *(cp) << " " << *(cp++) << endl;

I would have expected "a" and "a" but instead I get "b" and "a".

Is this correct?? How does "b" get output first? I havent incremented the pointer until the second output and then again thats post-fix increment not prefix! Am i screwing up somewhere thinking it should be "a" and "a"?