Are arrays of character pointers treated specially?

mirizar · #1 02-Mar-2005, 18:34

Hi!

I have created an array of character pointers
char *myArray[2] = {"test", "test2"};

If I print myArray[0] it prints "test" instead of the address of the test string.

Is there any way to print the address where "test" is stored instead of the string "test"?

Any help would be greatly appreciated.

Thanks, Michelle

LuciWiz · #2 03-Mar-2005, 00:48

Pointers to char can be really tricky in C/C++.
I wrote a simple test that you can run and figure it all out, hopefully. Please note that I wouldn't normally write a program in this manner; what I mean is if I saw a piece of code written like this by anybody else I would probably flip

So, this is just a test:

CPP / C++ / C Code:

	int a = 5, b = 10;
	char * myPointerTOCharArray[2] = {"test1", "test2"};
	int * myPointerTOIntArray[2] = {&a, &b};;
	int myIntArray[2] = {2, 4};
	
	//This is tricky :)
	char * test3 = "test3";
	char * test4 = "test4";
	char ** myPointerTOPointerToCharArray[2] = {&test3, &test4};

	cout << "The first element's value" << endl;

	cout << myPointerTOCharArray[0] << endl;
	cout << myIntArray[0] << endl;
	cout << * myPointerTOIntArray[0] << endl;
	cout << * myPointerTOPointerToCharArray[0]	<< endl;
	
	cout << "\nThe first element's address" << endl;

	cout << & myPointerTOCharArray[0] << endl;
	cout << & myIntArray[0] << endl;
	cout << myPointerTOIntArray[0] << endl;
	cout << myPointerTOPointerToCharArray[0] << endl;

If you use a C compiler, you'll have to change the printing.

Anyway, the quick answer to your question would be to print & myArray[0] to get the address. I hope this is fine.

Best regards,
Lucian

davekw7x · #3 03-Mar-2005, 07:52

Quote:

Originally Posted by mirizar

Hi!

I have created an array of character pointers
char *myArray[2] = {"test", "test2"};

If I print myArray[0] it prints "test" instead of the address of the test string.

Is there any way to print the address where "test" is stored instead of the string "test"?

Any help would be greatly appreciated.

Thanks, Michelle

cout is an example of an ostream. The extraction operator, <<, treats pointer to char as a c-style string (a sequence of chars terminated by a zero byte).

If you want to see the address of a pointer, cast it to "pointer to void".

You can play around with this:

CPP / C++ / C Code:

#include <iostream>
using std::cout;
using std::endl;

int main()
{
  char *messages[2] = {"First Message", "Second Message"};
  char *pChar;

  pChar = messages[1];

  cout << "Here's cout << pChar:         <" << pChar << ">" << endl << endl;
  cout << "Value of the pointer pChar   = " << (void *)pChar << endl;
  cout << "Address of the pointer pChar = " << &pChar << endl << endl;

  for (int i = 0; i < 2; i++) {
    cout << "messages[" << i << "]: <" << messages[i] << ">" << endl;

    cout << "Value of pointer messages[" << i 
         << "] = " << (void *)messages[i] << endl;

    cout << "Address of messages[" << i << "]       = " 
         << &messages[i] << endl << endl;

  }


  return 0;
}

Regards,

Dave

davekw7x · #4 03-Mar-2005, 08:23

Quote:

Originally Posted by davekw7x

cout is an example of an ostream. The extraction operator, <<, treats pointer to char as a c-style string (a sequence of chars terminated by a zero byte).

If you want to see the address of a pointer, cast it to "pointer to void".

You can play around with this:

I am responding to my own response (again).

I gave the "C++" approach (that is, I used cout, as the Original Poster did). Actually, real C++ programmers would not use the static cast (void *), but it's perfectly valid in C++, and is, in my opinion, appropriate for this simple case.

As a further note, I would like to note that people who have learned C before getting into C++ sometimes yearn for good old printf(), and really hate to learn all the ins and outs of cin>> and cout << (a little C++ pun here, folks).

Well, the good news is that printf() and all of its relatives are part of the standard C++ library also, and its really (really) OK to use them in C++. Why printf()? Well printf supports a format specifier that prints values of pointers without requiring casts. You can try this for comparison:

CPP / C++ / C Code:

#include <iostream>
#include <cstdio>

using std::cout;
using std::endl;
using std::printf;

int main()
{
  char *messages[2] = {"First Message", "Second Message"};
  char *pChar;

  pChar = messages[1];

  printf("Using printf pChar with %%s:   <%s>\n", pChar);
  printf("Using printf pChar with %%p:    %p\n", pChar);
  printf("Address of   pChar        :    %p\n\n", &pChar);
  cout << "Here's cout << pChar:         <" << pChar << ">" << endl;
  cout << "Value of the pointer pChar   = " << (void *)pChar << endl;
  cout << "Address of the pointer pChar = " << &pChar << endl << endl;

  return 0;
}

Regards,

Dave

LuciWiz · #5 03-Mar-2005, 08:51

Ooops!

Now that I read Dave's posts I understand the error in mine. Disregard it please, his solution is the correct one.

Regards,
Lucian

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