How to access values returned by object's reference to void pointer?

Question

I am trying to access the object's value returned in the void pointer, how can i access those values x and y through void pointer variable.

How can I access the value returned in that void pointer ???

#include <iostream.h>
void* getDetector();

class DETECTOR
{
public:

    int x;
    int y;

    DETECTOR();
    ~DETECTOR();    
};

void* getDetector()
{
    return &DetectorObj;
}

DETECTOR::DETECTOR()
{
    x = 10;
    y = 20;
}

DETECTOR::~DETECTOR(){}

DETECTOR DetectorObj;

int main()
{
    void * getPtr = getDetector();

    cout<<getPtr->x;

    return 0;
}

I know it's common to use Turbo C++ in Indian schools, but you should really take some time to learn modern C++, or even C++ after it was standardized in 1998 (Turbo C++ is *not* using standard C++, Turbo C++ is to old). [Here's a list of good books](https://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list/388282#388282), and there are also many free compilers and environments available to download. — Some programmer dude, Jul 15 '19 at 13:50
Try not to use `void*` at all in C++. There's *very rarely* any good reason to. — Jesper Juhl, Jul 15 '19 at 13:51
And don't forget that in C++ (even before it was standardized) you needed to declare things before you use them. — Some programmer dude, Jul 15 '19 at 13:52

Hatted Rooster · Accepted Answer · 2019-07-15T14:02:49.837

4

You can read the members of the object by returning the objects type and storing a pointer to that instead of a void*.

#include <iostream>
//DETECTOR definition..


DETECTOR DetectorObj;

DETECTOR* getDetector()
{
    return &DetectorObj;
}

int main()
{
    DETECTOR* getPtr = getDetector();

    cout<<getPtr->x;

    return 0;
}

There's no reason to return a void* here.

If you truly want to access them through void* then you can cast the result:

cout << static_cast<DETECTOR*>(getDetector())->x;

But this is considered bad bad practice. Please don't do this unless you're being held at gunpoint by your lecturer.

Also, it's <iostream>, not <iostream.h>, this is 2019.

edited Jul 15 '19 at 14:02

answered Jul 15 '19 at 13:53

Hatted Rooster

35,759
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If the code used ``, there's no danger of it `using namespace std;`. And the code used plain `cout` because there was no namespace support in the compiler. – Jonathan Leffler Jul 15 '19 at 13:56
@JonathanLeffler Ah right, good point. What even are namespaces. – Hatted Rooster Jul 15 '19 at 13:57
@SombreroChicken Is it possible to access those values with void* variable. – Nikunj Chaklasiya Jul 15 '19 at 14:01
1

@NikunjChaklasiya Yes, by casting it. – Hatted Rooster Jul 15 '19 at 14:01
@SombreroChicken How can I cast that? – Nikunj Chaklasiya Jul 15 '19 at 14:02
1

Your mum is considered bad bad practice. – Lightness Races in Orbit Jul 15 '19 at 14:22

score 1 · Answer 2 · answered Jul 15 '19 at 13:50

1

Assuming DetectorObj is of type DETECTOR and has an instance:

int main()
{
    DETECTOR* getPtr = (DETECTOR*)getDetector();

    cout<<getPtr->x;

    return 0;
}

answered Jul 15 '19 at 13:50

EylM

5,967
2
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3

yeah, but WHY on earth would the object be casted to void* to be returned ? Change the interface instead and return the actual type, or a base type, or something sane. – Jeffrey Jul 15 '19 at 13:51

How to access values returned by object's reference to void pointer?

2 Answers2