Dear All, I wonder if any of you can help me, I am new to XS and want to call a C function from perl. It is actually a pointer to a function which is a member of a structure. A part of the C header file is as follows: 
typedef struct _mystruct1 {
    int val;
    int (* init) (void *object1, void **object2);
} mystruct1;

typedef struct _mystruct2 {
    mystruct *ptr;
} mystruct2;

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In the C code I have: 
mystruct2 *mine;
int res;
struct obj *object1, **object2;
.....

mine->ptr->val = 100;
res = mine->ptr->init(object1, object2);

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and I want to be able to say the same from perl, i.e.: 
print "$mine->{"ptr"}->{"val"}";            # This seems to work
$mine->{"ptr"}->{"init"}(object1, object2); # This does not work

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To do that I used an XS program which has the following: 
rh = (HV *)sv_2mortal((SV *)newHV());
hv_store(rh, "val", 3, newSViv(mine->ptr->val), 0);
hv_store(rh, "init", 4, newRV((SV *)mine->ptr->init), 0); # This line 
+did not work

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I have tried to find out the type of the return pointer as in: 
if (SvROK ((SV *) mine->ptr->init)) {
    type = SvTYPE(SvRV((SV *)mine->ptr->init));
    printf ("Something ..%d", type);
}

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but both statements failed ...

Any help appreciated ...

Maged Messeh

In reply to How to pass a function reference from C to perl by Anonymous Monk

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