27 Apr
2020
27 Apr
'20
11:45 p.m.
From: Derek Fawcus
I think he means something like:
(*((*((*((*f)()->g))()->h))()->i))()
So I've been confused by this thread, and I'm hoping someone can deconfuse me
- but I think I may have figured it out.
What's confusing me is that in C, the -> operator is followed by "an
identifier [which] designates a member of a structure or union object" (I
checked the spec to make sure my memory hadn't dropped any bits) - but g, h
above are arguments; so I couldn't figure out what was going on.
I think what may have happened is that initially the discussion was about C
("Pretty sure it was not in v7 C"), but then it switched to C++ - with which
I'm not familiar, hence my confusion - without explicitly indicating that
change (although the reference to Bjarne Stroustrup should been a clue). (And
that's why I thought "f()->g()->h()->i()" was ad hoc notation for
"calls f(),
then calls g()".)
Am I tracking now?
Noel
27 Apr
27 Apr
11:56 p.m.
New subject: v7 K&R C
/* struct declarations. */
struct fval;
struct gval;
struct hval;
/* function declarations */
struct fval *f();
struct gval *g();
struct hval *h();
struct fval { struct gval* (*g)(); int dummy; };
struct gval { struct hval* (*h)(); };
struct hval { void (*i)(); };
extern void test();
void test()
{
f()->g()->h()->i();
}
28 Apr
28 Apr
12:02 a.m.
New subject: v7 K&R C
g, h and i are members of structures, the pointer of which is returned by the preceding
function call. They have to be defined as pointers to functions returning a pointer to the
following structure.
A simple example is:
typedef struct Node Node;
struct Node
{
Node *(*f)(void);
};
void
main(void)
{
Node *p;
p->f()->f()->f();
call();
(*((*((*p->f)()->f))())->f)();
}
// (*((*((*((*f)()->g))()->h))()->i))()
On Apr 27, 2020, at 1:45 PM, Noel Chiappa
<jnc(a)mercury.lcs.mit.edu> wrote:
From: Derek Fawcus
I think he means something like:
(*((*((*((*f)()->g))()->h))()->i))()
So I've been confused by this thread, and I'm hoping someone can deconfuse me
- but I think I may have figured it out.
What's confusing me is that in C, the -> operator is followed by "an
identifier [which] designates a member of a structure or union object" (I
checked the spec to make sure my memory hadn't dropped any bits) - but g, h
above are arguments; so I couldn't figure out what was going on.
I think what may have happened is that initially the discussion was about C
("Pretty sure it was not in v7 C"), but then it switched to C++ - with which
I'm not familiar, hence my confusion - without explicitly indicating that
change (although the reference to Bjarne Stroustrup should been a clue). (And
that's why I thought "f()->g()->h()->i()" was ad hoc notation for
"calls f(),
then calls g()".)
Am I tracking now?
Noel
12:47 a.m.
New subject: v7 K&R C
On Mon, Apr 27, 2020 at 01:45:53PM -0400, Noel Chiappa wrote:
So I've been confused by this thread, and
I'm hoping someone can deconfuse me
- but I think I may have figured it out.
What's confusing me is that in C, the -> operator is followed by "an
identifier [which] designates a member of a structure or union object" (I
checked the spec to make sure my memory hadn't dropped any bits) - but g, h
above are arguments; so I couldn't figure out what was going on.
See below.
DF
#include <stdio.h>
struct h_ret { void (*i)(); };
struct g_ret { struct h_ret *(*h)(); };
struct f_ret { struct g_ret *(*g)(); };
void i_fn() { printf("I\n"); }
struct h_ret h_val = { i_fn };
struct h_ret *h_fn() { printf("H\n"); return &h_val; }
struct g_ret g_val = { h_fn };
struct g_ret *g_fn() { printf("G\n"); return &g_val; }
struct f_ret f_val = { g_fn };
struct f_ret *f_fn() { printf("F\n"); return &f_val; }
void fred(struct f_ret *(*f)())
{
#if 1
(*(*(*(*f)()->g)()->h)()->i)();
#else
f()->g()->h()->i();
#endif
}
int main() { fred(f_fn); return 0; }
1700
days inactive
1700
days old
3 comments
4 participants
participants (4)
-
Brantley Coile -
Derek Fawcus -
jnc＠mercury.lcs.mit.edu -
Richard Salz