Pointers and Portability

[REH]

Given that the standard guarantees that all pointers to struct are the
same size and representation, and that the same is true for all
function pointers, is this snippet of (contrived) code portable?

typedef struct t1 {int i;} t1;
typedef struct t2 {double d;} t2;

t1* f1() {static t1 x; return &x;}
t2* f2() {static t2 x; return &x;}

struct T;
typedef struct T T;
typedef T* (*F) ();

void foo(T** t, F f) {*t = f();}


int main()
{
t1* p1;
t2* p2;

foo((T**) &p1, (F) f1);
foo((T**) &p2, (F) f2);
return 0;
}



REH

 

[Peter Nilsson]

Given that the standard guarantees that all pointers to
struct are the same size and representation, and that
the same is true for all function pointers, is this
snippet of (contrived) code portable?

typedef struct t1 {int i;} t1;
typedef struct t2 {double d;} t2;

t1* f1() {static t1 x; return &x;}
t2* f2() {static t2 x; return &x;}

struct T;
typedef struct T T;
typedef T* (*F) ();

void foo(T** t, F f) {*t = f();}

int main()
{
    t1* p1;
    t2* p2;

    foo((T**) &p1, (F) f1);
    foo((T**) &p2, (F) f2);
    return 0;
}

No, neither t1*(*)() nor t2*(*)() are compatible with T*(*)().
That the size and representation of the function and struct
pointer types are the same does not negate the fact that you
end up calling a function through an incompatible type, thus
invoking undefined behaviour.

That said, you can convert them back to the original type
and then call them.

 

[REH]

No, neither t1*(*)() nor t2*(*)() are compatible with T*(*)().
That the size and representation of the function and struct
pointer types are the same does not negate the fact that you
end up calling a function through an incompatible type, thus
invoking undefined behaviour.

That said, you can convert them back to the original type
and then call them.

Thanks.

REH