Thread: A question about pointers in C

Hi everybody, when reading and testing the source code of a program I had a question about pointers and addresses.

My OS is Ubuntu 8.04. The program I am talking about is SQLite ODBC driver (and its source code can be downloaded from here). However, to understand my problem, it is not important to know what this program does. I just want to ask about the way it used pointers in its source code.

Basically the source code of this program consists of 2 files: sqliteodbc.c and sqliteodbc.h. Inside sqliteodbc.h it defines a struct like this:

Code:

typedef struct {			
    int ov3;			
} ENV;

and a type like this:

Code:

typedef void * 	SQLHENV;

And then inside sqliteodbc.c it has 2 important functions, and the code of the 2 functions is as follow:

The first one is:

Code:

SQLAllocHandle(SQLHENV *env)
{
ENV *e;
e = (ENV *) malloc(sizeof (ENV));
*env = (SQLHENV) e;
}

And the second one is:

Code:

SQLSetEnvAttr(SQLHENV env)
{
ENV *e;
e = (ENV *) env;
e->ov3 = 1;
}

Then other functions in sqliteodbc.c can obtain the value of ov3 just by being passed to the parameter env. I means they only know the address the void pointer points to, and then they typecast it to ENV*, and then they can access its member, something like this:

Code:

foo (SQLHENV env)
{
ENV *e;
e = (ENV *) env;
if (e->ov3 == 1)
/*do something here*/
}

That is the way functions in sqliteodbc.c 'communicate'. And I think it is quite easy to understand.

However, when I tried to reproduce this communication mechanism, I failed.

After being built, this program will be a library, libsqliteodbc.so, and by linking to this library you can use its functions in your code. So I wrote a simple test code like this:

Code:

#include "sqliteodbc.h"

main() {

   SQLHENV env;
   ENV *ptest;
   
   SQLAllocHandle(&env);

   SQLSetEnvAttr(env);
   
   ptest = (ENV *) env;
   
   printf("%d\n", ptest->ov3);
 	
}

I expected the output from the above printf command would be '1', but it would not. Could you please explain to me why I did not get the value '1'?

Thanks in advance!

Thank you very much for your answer!

Originally Posted by bithub

What that library is doing is a common occurance in C. A void pointer is typedef'd to some name, and then used as a generic handle to hide the actual data structure implementation. I personally do not like this technique since it prevents type safety checking. A much better way is to forward declare a structure in the header file, and leave its implementation in the C file. This provides the same effect of hiding the implementation, but it also provides type checking.

Yes, before posting this question I also guessed that the author used void* pointers to hide the actual data structure implementation.

Originally Posted by bithub

Anyways, on to your question. The reason your code doesn't work as expected is probably due to it not knowing what a ENV is. ENV is a type that is hidden (not defined in sqlite3odbc.h), and therefore it cannot be used just by including the header file. You need to also provide the definition of ENV in that header file (which would defeat the purpose of hiding its implementation in the first place).

I am sorry, it was a typo, the file I included is sqliteodbc.h. And I managed to compile and run the test code. The program did not encounter any error, but it did not produce the result I expected either.