2009/9/22 Luke Benstead <kazade(a)gmail.com>:
2009/9/22 Ben Klein <shacklein(a)gmail.com>:
2009/9/22 Vitaliy Margolen <wine-devel(a)kievinfo.com>:
Mike Kaplinskiy wrote:
It actually does not dereference anything. Try passing null into the function - it will work just fine. This is a special case because the array isn't dynamically allocated but is part of the struct, which means that dmW->dmFormName == (dmW+__offset of dmFormName) and not *(dmW+__offset of dmFormName). You can try writing a test program yourself - it will run just fine. It does dereference the pointer. Here is your simple test. Compile it and run it. See what happens.
#include <stdio.h>
typedef struct _s_test { void *pointer;
No. Array, not pointer. E.g.: int array[1];
} s_test;
If it IS the case that this doesn't cause a crash and is perfectly valid, can someone explain to me how/why this works? Or point me (no pun intended) to the bit in the C spec that explains it? Coz the way I read it, it has to dereference dmW, otherwise how would the compiler find the address of the array? ... so confused :)
Luke.
Luke, Wine's current code is basically equivalent to the one above, where there's no dereference : #include <stdio.h> typedef struct _s_test { char pointer[5]; } s_test; int main() { s_test *s = NULL; long diff = (const char*)(&s->pointer[0]) - (const char*)s; printf("diff=%ld\n", diff); return 0; } -- Nicolas Le Cam