If the ID field you test is part of a common initial sequence of fields shared by all the structs, then using a union guarantees that the access will work:
#include <stdio.h>
typedef struct
{
int id;
int junk1;
} Foo;
typedef struct
{
int id;
long junk2;
} Bar;
typedef union
{
struct
{
int id;
} common;
Foo foo;
Bar bar;
} U;
int matches(const U *candidate, int wanted)
{
return candidate->common.id == wanted;
}
int main(void)
{
Foo f = { 23, 0 };
Bar b = { 42, 0 };
U fu;
U bu;
fu.foo = f;
bu.bar = b;
puts(matches(&fu, 23) ? "true" : "false");
puts(matches(&bu, 42) ? "true" : "false");
return 0;
}
If you're unlucky, and the field appears at different offsets in the various structs, you can add an offset parameter to your function. Then, offsetof and a wrapper macro simulate what the OP asked for - passing the type of struct at the call site:
#include <stddef.h>
#include <stdio.h>
typedef struct
{
int id;
int junk1;
} Foo;
typedef struct
{
int junk2;
int id;
} Bar;
int matches(const void* candidate, size_t idOffset, int wanted)
{
return *(int*)((const unsigned char*)candidate + idOffset) == wanted;
}
#define MATCHES(type, candidate, wanted) matches(candidate, offsetof(type, id), wanted)
int main(void)
{
Foo f = { 23, 0 };
Bar b = { 0, 42 };
puts(MATCHES(Foo, &f, 23) ? "true" : "false");
puts(MATCHES(Bar, &b, 42) ? "true" : "false");
return 0;
}