Prove Union has Single Address

Peter_APIIT · #1 28-Apr-2008, 21:44

Hello to dear expert programmer, i have wrote a program that test the address of union member.

I cannot print out address of union member.

Below is what i have wrote :

CPP / C++ / C Code:

#include<iostream>
#include<set>
#include<windows.h>
 
using namespace std;
 
// ------------------------------------------------
 
 
union Package
{
    char i;
    short j;
    int k;
    float f;
    long double mathematicalEquation;
};
 
/*
    Sizeof Union determines by largest sotrage 
    class in an union
*/
 
// ------------------------------------------------
int main(int argc, char *argv[])
{
    cout << "Sizeof Package " << sizeof(Package);
 
    Package aPackage;
 
    cout << "\n" << "Address of Package : " << &aPackage;
    cout << "\n" << "Address of I : " << &aPackage.i;
 
 
    return 0;
}
// ------------------------------------------------

Thanks for your help.

davekw7x · #2 28-Apr-2008, 21:54

Quote:

Originally Posted by Peter_APIIT

I cannot print out address of union member.

To print the value of a pointer, cast it to (void *):

CPP / C++ / C Code:

#include<iostream>

using namespace std;

union Package
{
    char i;
    short j;
    int k;
    float f;
    long double mathematicalEquation;
};


int main()
{
    cout << "Size of Package = " << sizeof(Package) << endl << endl;
    
    Package aPackage;

    cout << "Addr of aPackage   : " << reinterpret_cast<void *>(&aPackage)
         << endl;
    cout << "Addr of aPackage.i : " << reinterpret_cast<void *>(&aPackage.i)
         << endl;
    cout << "Addr of aPackage.j : " << reinterpret_cast<void *>(&aPackage.j)
         << endl;
    cout << "Addr of aPackage.k : " << reinterpret_cast<void *>(&aPackage.k)
         << endl;
    cout << "Addr of aPackage.f : " << reinterpret_cast<void *>(&aPackage.f)
         << endl;
    cout << "Addr of aPackage.mathematicalEquation : " 
         << reinterpret_cast<void *>(&aPackage.mathematicalEquation)
         << endl;
    return 0;
}

Output (from GNU g++)

Code:

Size of Package = 12

Addr of aPackage   : 0xbf9d6848
Addr of aPackage.i : 0xbf9d6848
Addr of aPackage.j : 0xbf9d6848
Addr of aPackage.k : 0xbf9d6848
Addr of aPackage.f : 0xbf9d6848
Addr of aPackage.mathematicalEquation : 0xbf9d6848

Regards,

Dave

Peter_APIIT · #3 29-Apr-2008, 04:36

Why u say aPackage is a pointer ?

As far as i know, pointer contains address of another variable.

Thanks for your help.

davekw7x · #4 29-Apr-2008, 07:21

Quote:

Originally Posted by Peter_APIIT

Why u say aPackage is a pointer

Look at the code. My code is the same as yours except for the cast.

The variable aPackage is a struct, so &aPackage is a pointer whose value is the address of aPackage.

In fact, it is not necessary to cast anything except the pointer to char. (Try my program again without the casts.) I just put the casts there for emphasis.

The "<<" operator for pointers to just about everything except pointer to char is overloaded to print the value of the pointer.

The "<<" operator for pointer to char is overloaded to treat a pointer to char as the address of a C-Style "string", so the cast is necessary in that case to make sure it prints the value of the pointer, and not try to print print a zero-byte terminated sequence of chars pointed to by that pointer.

Regards,

Dave

Peter_APIIT · #5 06-May-2008, 02:41

I understand half of your explanation.

Why my union is contain a single char and not pointer to char but still need to cast it to pointer ?

Thanks.

davekw7x · #6 06-May-2008, 07:42

Quote:

Originally Posted by Peter_APIIT

Why my union is contain a single char and not pointer to char but still need to cast it to pointer

I didn't cast the union to a pointer.

Regards,

Dave

Peter_APIIT · #7 08-May-2008, 04:00

Quote:

Why my union is contain a single char and not pointer to char but still need to cast it to pointer ?

What i mean is char and not char * ?
Why i need to cast since since i only have char inside the union?

Thanks.

Peter_APIIT · #8 08-May-2008, 04:01

Quote:

Why my union is contain a single char and not pointer to char but still need to cast it to pointer ?

What i mean is char and not char * ?
Why i need to cast since since i only have char inside the union?

Thanks.

davekw7x · #9 08-May-2008, 07:39

Quote:

Originally Posted by Peter_APIIT

What i mean is char and not char * ?
Why i need to cast since since i only have char inside the union?

Thanks.

You want to print the address of a char variable. The address of a char variable has type that is "pointer to char"

I tried to explain that the '<<' operator treats pointer to char like a C-style 'string'.

Here's an example:

CPP / C++ / C Code:

#include <iostream>

using namespace std;

int main()
{
    char c1 = '1';
    char *c2;
    c2 = "Hi";
    cout << "The '<<' operator is overloaded in such a way that"<< endl
         << "if you feed it a pointer to char, it thinks that" << endl
         << "you have a C-style 'string'" << endl << endl;
    
    cout << "c2 is a pointer to char." << endl
         << "The value of c2 is the address of the string " << endl
         << "literal, so the string is printed." << endl << endl;

    cout << "     c2: "
         << c2 << endl << endl; // The C-style "string is printed

    cout << "If we want to know the value of the pointer, c2," << endl
         << "then cast it to a pointer to void:" << endl;

    cout << "     c2: "
         << reinterpret_cast<void *>(c2) << endl << endl;

    // If you want use '<<' to print the address of a char variable
    // you have to tell the '<<' operator that you are not printing
    // a c-style string. To print the value of a pointer to char
    // you cast it to a pointer to void

    c2 = &c1;
    cout << "New the value of c2 is the address of a char. If" << endl
         << "we tried 'cout<<c2', the operator would try to print" << endl
         << "out a 'string', but there is no 'string' to print." << endl << endl;

    cout << "So, to print the value of the pointer, cast the variable" << endl
         << "to (void *):" << endl;

    cout << "     c2: "
         << reinterpret_cast<void *>(c2) << endl;

    return 0;
}

Output:

Code:

The '<<' operator is overloaded in such a way that
if you feed it a pointer to char, it thinks that
you have a C-style 'string'

c2 is a pointer to char.
The value of c2 is the address of the string 
literal, so the string is printed.

     c2: Hi

If we want to know the value of the pointer, c2,
then cast it to a pointer to void:
     c2: 0x8048aa0

New the value of c2 is the address of a char. If
we tried 'cout<<c2', the operator would try to print
out a 'string', but there is no 'string' to print.

So, to print the value of the pointer, cast the variable
to (void *):
     c2: 0xbfd692df

Regards,

Dave

Peter_APIIT · #10 10-May-2008, 01:46

Not really understand what u talking here.

But my understand is like this :

CPP / C++ / C Code:

cout << "\n" << "Address of I : " << &aPackage.i;

When i put execute this statement, my intention is to print the address of the char but << operator treat this as pointer to char, in the meanwhile << operator treat this to C style string.

Therefore, i need dereference to get the address of char.

For instance, u declare int a;

but you pass it as (&a); you also can treat this as pointer in function definition.

I think i understand(May be).

By the way, thanks for your help.

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Prove Union has Single Address

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