pointers as function arguments

I am using a pointer to a double as an argument for a function that prints out the memory address, contents and name of the variable that the pointer points to, but I can't figure out how to output the name of the variable that the pointer points to.

Here is the code:

Code:

---


#include <iostream>
 #include <string.h>

 void iaddress(double* num);

 int main()
 {
        double num1 = 1.23;
        double num2 = 100.75;
        char *str1 = new char[26];
        strcpy(str1, "Pointers are interesting!");
        double num3 = -35.5;
        double *ptr4 = new double;
        int i = 0;

        // str1 output
        cout << "str1 - main: address: "  << &str1 << " contents: " << str1 << endl;

        //        num1 output and function call
        cout << "num1 - main: address: " << &num1 << " contents: " << num1 << endl;
        ptr4 = &num1;
        iaddress(ptr4);

        //        num2 output and function call
        cout << "num2 - main: address: " << &num2 << " contents: " << num2 << endl;
        ptr4 = &num2;
        iaddress(ptr4);

        //        num3 output and function call
        cout << "num3 - main: address: " << &num3 << " contents: " << num3 << endl;
        ptr4 = &num3;
        iaddress(ptr4);

        //        ptr4 output and function call
        cout << "ptr4 - main: address: " << ptr4 << " contents: " << *ptr4 << endl;
        *ptr4 = 123.4;
        iaddress(ptr4);


        delete ptr4;
        delete [] str1;



        return 0;

 }

 //        FUNCTION HEADER AND CODE

 void iaddress(double* num)
 {
        cout << "        " << num << "- main: address: " << num << " contents: " << num << endl;
        delete num;
 }

---

Thanks for any help.

Brian

>>but I can't figure out how to output the name of the variable that the pointer points to.

Thats because you cant. And thats because there is no name to output. The pointer just points to a section of memory. The variable names in your code don't exist in the compiled product (at least, not in a form you would recognise). Its all just memory addresses. So your pointer can obtain the value of what it points to, and you can display the address of what it points to. You can even display the address of the pointer, but thats it. There's nothing else to see.

Interesting question. Answer: You can't - with some qualification.

The compiler doesn't recognize your variable name, as such. It equates the "human" name you've given it with an address. Part of the compilation process.

'int1' may be '0063FDE8' (stole that from your output :) ). As "low" as C/C++ may be, or seem to be, it's still high-level in machine-language terms.

An incredibly coarse explanation? At compile time, a #define directive, of sorts, is initiated. Something along the lines of:

#define int1 *0063FDE8

(Smart people, go flame someone else! :D )

From there on, whenever 'int1' is encountered, the contents of memory location '0063FDE8' are accessed. In effect, 'int1' becomes '0063FDE8'.

-Skipper

P.S. Thanks, lightatdawn, for typing faster than me! :D