I have a struct defined as:
typedef struct {
int type;
void* info;
} Data;
and then i have several other structs that i want to assign to the void* using the following function:
Data* insert_data(int t, void* s)
{
Data * d = (Data*)malloc(sizeof(Data));
d->type = t;
d->info = s;
return d;
}
struct {
...
} Struct#;
then i just call
insert_data(1, variable_of_type_Struct#);
When i compile this it gives a warning
warning: assignment from incompatible pointer type
i tried to cast the variable in the insert to (void*) but didn't work
insert_data(1, (void *) variable_of_type_Struct#);
How can i get rid of this warning?
Thanks
Pass in the address of the struct, not a copy of it (i.e. not passed by value):
insert_data(1, &variable_of_type_Struct);
Pass a pointer to the struct object:
struct your_struct_type bla;
insert_data(1, &bla);
Hope this program helps!
#include <stdio.h>
#include <stdlib.h>
typedef struct {
int type;
void* info;
} Data;
typedef struct {
int i;
char a;
float f;
double d;
}info;
Data* insert_data(int t, void* s)
{
Data * d = (Data*)malloc(sizeof(Data));
d->type = t;
d->info = s;
return d;
}
int main()
{
info in;
Data * d;
d = insert_data(10, &in);
return 0;
}
I'm not quite sure what this was:
struct {
...
} Struct#;
So, I cleaned up your program a little bit and got no warnings, after putting the address of the struct into the call, insert_data(1, &variable_of_type_Struct);
#include <stdlib.h>
#include <stdio.h>
typedef struct {
int type;
void* info;
} Data;
Data* insert_data(int t, void* s);
Data variable_of_type_Struct;
Data* insert_data(int t, void* s)
{
Data * d = (Data*)malloc(sizeof(Data));
d->type = t;
d->info = s;
return d;
}
void test()
{
insert_data(1, &variable_of_type_Struct);
}
insert_data
waits for a void*
, you put a Data
.
insert_data(1, &variable_of_type_Struct#);
It miss a level of indirection.