Pointers to Structs

[John Durant]

Trying to figure out why the addresses of &aStruct and ptr are different from the value of &strr in the function ListStruct:


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

using namespace std;


struct jTest {
string first;
string second;
string third;
};

void ListStruct(jTest *strr);

int _tmain(int argc, _TCHAR* argv[])
{
jTest aStruct = { "omr", "two", "three"};
jTest * ptr;
ptr = &aStruct;

//Checks on value of &aStruct and ptr;
cout << "The value of &aStruct is " << &aStruct << " and the value of ptr is " << ptr << endl;

ListStruct(ptr);

cin.get();
}

void ListStruct(jTest *strr) {
cout << "The address of &strr is " << &strr << endl;
}

Returns:
The value of &aStruct is 0026F6CC and the value of ptr is 0026F6CC
The address for &strr is 0026F5E8

Why the difference in addresses?

BTW the difference is always 228 bytes.

Thanks,
John

[sohguanh]

void ListStruct(jTest *strr) {
cout << "The address of &strr is " << &strr << endl;
}

I believe sttr is storing the contents of ptr in the main program. If you do a cout << *strr << "\n" you should see back 0026F6CC.

[John Durant]

Changed it and this came back:


Error 1 error C2679: binary '<<' : no operator found which takes a right-hand operand of type 'jTest' (or there is no acceptable conversion) c:\users\bridget\documents\visual studio 2010\projects\structsand ptrs\structsand ptrs\structsand ptrs.cpp 34


Thanks,
John

[D80DSL3A]

Try cout << "The address pointed to by strr is " << strr << endl;. This should agree with the value of ptr.

&strr would be the address where the pointer strr is stored - not the value (strr) contained there.
Note: &strr won't = &ptr since strr is a local copy of ptr. They are 2 distinct pointers but the same value is stored in them.

[John Durant]

Thank you fun2code. Your explanation makes perfect sense. The value stored in ptr being the address of where the struct lives. Now I have to play with it some more to get it right in my mind. Thanks again for your help.