How do I typecast for void *

Question

1.00/5 (1 vote)

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i want to typecast personal structure.
but i dont know how do i typecast.
is anyone to solve this problem?

C++

typedef struct _data_type
{

	INT	   n;
	BOOL   b;
	double db;
	DWORD  dw;
	DOUBLE DB;
	
}DATATYPE;

#define CAST(data,str) (##str##*##)##data

void * data;
str ="DATATYPE"

CAST(data,str);

What I have tried:

is it impossible? plz, answer my question.

Posted 14-May-16 8:06am

Member 10395626

Updated 14-May-16 9:41am

Patrice T

v3

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Comments

Homero Rivera 14-May-16 14:10pm

Isn't void supposed to mean, "no result to return" for methods? Maybe you need null?
nullptr ?

Sergey Alexandrovich Kryukov 14-May-16 14:43pm

To start with, why?
—SA

PJ Arends 14-May-16 15:18pm

Do not use void* if you can at all avoid it. You loose all type safety if you use it.

The cast you are trying to do is better done with reinterpret_cast.

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CPallini · Answer 1 · 2016-05-14T09:41:00

I assume you are using C programming language (since typedef before struct isn't necessary in C++).

In C programming language, the cast could be written

C

void * data;
DATATYPE * d0 = (DATATYPE *) data;

Hence your macro would become

C

#define CAST(data, DATATYPE)  (DATATYPE*)(data)

Usage:

C

DATATYPE * d1 = CAST(data, DATATYPE);

However, I question the usefulness of such a macro.