Question

If I have structure definitions, for example, like these:

struct Base {
  int foo;
};

struct Derived {
  int foo; // int foo is common for both definitions
  char *bar;
};

Can I do something like this?

void foobar(void *ptr) {
  ((struct Base *)ptr)->foo = 1;
}

struct Derived s;

foobar(&s);

In other words, can I cast the void pointer to Base * to access its foo member when its type is actually Derived *?

Answer 1

Many real-world C programs assume the construct you show is safe, and there is an interpretation of the C standard (specifically, of the “common initial sequence” rule, C99 §6.5.2.3 p5) under which it is conforming. Unfortunately, in the five years since I originally answered this question, all the compilers I can easily get at (viz. GCC and Clang) have converged on a different, narrower interpretation of the common initial sequence rule, under which the construct you show provokes undefined behavior. Concretely, experiment with this program:

#include <stdio.h>
#include <string.h>

typedef struct A { int x; int y; }          A;
typedef struct B { int x; int y; float z; } B;
typedef struct C { A a;          float z; } C;

int testAB(A *a, B *b)
{
  b->x = 1;
  a->x = 2;
  return b->x;
}

int testAC(A *a, C *c)
{
  c->a.x = 1;
  a->x = 2;
  return c->a.x;
}

int main(void)
{
  B bee;
  C cee;
  int r;

  memset(&bee, 0, sizeof bee);
  memset(&cee, 0, sizeof cee);

  r = testAB((A *)&bee, &bee);
  printf("testAB: r=%d bee.x=%d\n", r, bee.x);

  r = testAC(&cee.a, &cee);
  printf("testAC: r=%d cee.x=%d\n", r, cee.a.x);

  return 0;
}

When compiling with optimization enabled (and without -fno-strict-aliasing), both GCC and Clang will assume that the two pointer arguments to testAB cannot point to the same object, so I get output like

testAB: r=1 bee.x=2
testAC: r=2 cee.x=2

They do not make that assumption for testAC, but — having previously been under the impression that testAB was required to be compiled as if its two arguments could point to the same object — I am no longer confident enough in my own understanding of the standard to say whether or not that is guaranteed to keep working.