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Parameter passing to function with void * as Argument

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# 1  
Old 05-22-2014
Parameter passing to function with void * as Argument
Earlier I had one structure C
Code:
typedef struct c
{
    int cc;
}CS;

I used to call a library function say int GetData(CS *x) which was returning me the above structure C with data.
Code:
GetData(CS *x)

Function call used to be like:
Code:
CS CSobj;
GetData(&CSObj);

Now there are two structures say C and D
Code:
typedef struct c
{
    int cc;
}CS;
CS CSobj;

typedef struct d
{
    int dc;
    int dd;
}DS;
DS DSobj;

Function GetData() has been modified to:
Code:
GetData(void* x)

I have to call the library function say int GetData(void* x) which will be returning me one of the above structures through that void* paramter. The return type of the function tells which structure is returned.

Problem is while calling the function GetData() how and what parameter to pass since I am unaware of which struct the function will return. Any way out of this problem?
Last edited by rupeshkp728; 05-22-2014 at 08:54 AM.. Reason: adding more details
# 2  
Old 05-22-2014
Without knowing more, a NULL pointer is a good guess.

Regards,
Alister
# 3  
Old 05-22-2014
Passing a pointer to memory location which is big enough to hold the larger of the two structures will solve this problem.
Union will solve this issue.
# 4  
Old 05-22-2014
Yes, a union is useful in this situation. Specifically:

Code:
enum { TYPE_CS, TYPE_DS };

typedef struct CS {
        int type;
        int cc;
} CS;

typedef struct DS {
        int type;
        int dc;
        int dd;
} DS;

typedef union S {
        int type;
        DS ds;
        CS cs;
} S;

CS mycs={TYPE_CS, 0};
DS myds={TYPE_DS, 1, 1};

void myfunc(S *data)
{
        switch(s->type) {
        case TYPE_DS:
                s->ds.dc=0;
                s->ds.dd=0;
                break;
        case TYPE_CS:
                s->cs.cc=0;
                break;
        }
}

myfunc(&mycs);
myfunc(&myds);

Your function can tell which is which by checking the value of the type variable, then using either the ds or cs member of the union. All the members of the union overlap.
This User Gave Thanks to Corona688 For This Post:
rupeshkp728
# 5  
Old 05-23-2014
Modifying the myfunc function changing
Code:
data->type

inplace of
Code:
 s->type

Code:
void myfunc(S *data)
{
        switch(data->type) {
        case TYPE_DS:
                data->ds.dc=0; 
                data->ds.dd=0;
                break;
        case TYPE_CS:
                data->cs.cc=0;
                break;
        }
}

This User Gave Thanks to techmonk For This Post:
rupeshkp728
# 6  
Old 05-23-2014
Thanks Coronna and Techmonk for the replies.
I guess(may be I am wrong) the func call is incorrect.

Code:
myfunc(&mycs);
myfunc(&myds);

We need to pass object of type S.
Code:
S sobj = {.type = 1};
myfunc(&sobj);

if(sobj.type == 0)
{
    print sobj.cs.cc
}
else
{
    print sobj.ds.dc; 
    print sobj.ds.dd;
}

# 7  
Old 05-23-2014
Quote:
Originally Posted by rupeshkp728
I guess(may be I am wrong) the func call is incorrect.
Did you try the code or just consider it to look wrong...?
Quote:
We need to pass object of type S.
Why?

The whole point of using a union, and the whole point of adding type to the beginning of your CS and DS structures, was that they'd all have the same memory layout.

The compiler warns you since it doesn't know you built with these assumptions in mind. You can avoid the warning with a typecast if you want:

Code:
myfunc((S *)&obj);

This User Gave Thanks to Corona688 For This Post:
rupeshkp728
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