Problem with LPVOID C++

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Hi Experts,

C++

#include "stdafx.h"
#include <Windows.h>
void Test(LPVOID lpvBuffer);
int _tmain(int argc, _TCHAR* argv[])
{
	DWORD dwCode1 = 0;
	Test(&dwCode1);
	return 0;
}


void Test(LPVOID lpvBuffer)
{
	DWORD code = 200;
	lpvBuffer = &code;
}

In the method Test(), i am assigning value 200 to the LPVOID parameter, But i always getting a value 0 in dwCode1. What is the problem with this code ?

-Satheesh

Posted 28-Dec-13 20:03pm

Satheesh1546

Updated 28-Dec-13 21:00pm

v2

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Albert Holguin 29-Dec-13 17:28pm

I noticed you tried to set the pointer to the address of "code" in your Test() function. Don't even try to do that... you're setting the address of a local variable to pass back, that's not a good idea, because it will cease to exist once the function returns.

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nv3 · Accepted Answer · 2013-12-28T22:33:00

Solution 1

You are not assigning a value to the location where lpvBuffer points, but to the pointer itself.

And why did you use a pointer to void if you want to pass a DWORD as in/out parameter? Use a pointer to DWORD instead. So you code should look like this:

C++

void Test (DWORD* pArg)
{
	DWORD code = 200;
	*pArg = code;
}

Posted 28-Dec-13 22:33pm

nv3

Comments

Satheesh1546 29-Dec-13 4:43am

Thank you ! How to assign a value to the location where lpvBuffer points with out changing the parameter to DWORD ?

nv3 29-Dec-13 5:05am

In order for the comiler to know how large (and of which type) the location is that you want to write a value to, it must have a type. Hence you cannot assign a value via a void pointer. If that is all you have and you know that the pointer in fact points to a DWORD you can cast it, e.g.:

*((DWORD*) lpvBuffer) = code;

Using this technique without exactly knowing what you are doing will however get you in trouble and can produce errors that are extremely difficult to debug.

Problem with LPVOID C++

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Solution 1

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Problem with LPVOID C++

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