Function pointer doubt

Question

Please tell me what will the call to given function return and how? The code:

typedef struct {
    int size;
    ptrdiff_t index;
    void (*inlet) ();
    int argsize;
    ptrdiff_t argindex;
} CilkProcInfo;


/*
 * Returns a pointer to the slow version for a procedure
 * whose signature is p.
 */

/* the function definition is - */
static void (*get_proc_slow(CilkProcInfo *p)) () {
     return p[0].inlet;
}

/*The function gets called as -*/
   (get_proc_slow(f->sig)) (ws, f);
/*where f->sig is a pointer to CilkProcInfo struct*/

A fine example of how awful the inside-out C type syntax is. — starblue
– starblue, Commented Oct 28, 2009 at 8:21
This is the reason function pointers are often declared with a typedef. — Josh Lee
– Josh Lee, Commented Oct 28, 2009 at 8:24

Wernsey · Accepted Answer · 2009-10-28 11:42:04Z

5

In your CilkProcInfo structure, inlet is a pointer to a function that takes an unspecified number of arguments and does not return a value, like void foo();.

In the line

(get_proc_slow(f->sig)) (ws, f);

the get_proc_slow(f->sig) call returns this function pointer, so it is equivalent to

(f->sig[0].inlet) (ws, f);

So if your f->sig[0].inlet points to the function foo(), it is equivalent to the call

foo (ws, f);

I should admit that the static void (*get_proc_slow(CilkProcInfo *p)) () {... syntax is a bit unfamiliar to me.

edited Oct 28, 2009 at 11:42

answered Oct 28, 2009 at 7:59

Wernsey

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7 Comments

Adrien Plisson Over a year ago

static void (*get_proc_slow(CilkProcInfo *p)) () is the same as: typedef void (proc_sig)(); static proc_sig * get_proc_slow(CilkProcInfo *p);

Naveen Over a year ago

I am a bit confused.. if foo() doesn't take any parameters how you could pass ws and f as params? Isn't that a compiler error?

Lara Bailey Over a year ago

No. void (*inlet)() is a pointer to a function taking unspecified arguments and returning no value, not no arguments.

Lara Bailey Over a year ago

@iWerner: You can always edit your post to include corrections.

Swapnil Over a year ago

Isn't there a type mismatch between what is returned by the function and what is in declaration? I mean that the declaration says - void, but it actually returns a pointer to a function.

|

reko_t · Accepted Answer · 2009-10-28 07:59:51Z

1

get_proc_slow() returns a function pointer of type void(*)() which the code then calls. So when you do:

(get_proc_slow(f->sig)) (ws, f);

It's basically same as doing:

void (*fptr)() = get_proc_slow(f->sig);
fptr(ws, f);

answered Oct 28, 2009 at 7:59

reko_t

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Comments

Zoli · Accepted Answer · 2009-10-28 08:11:38Z

0

It looks like it's a function that returns a pointer to a function whose return value is void that has no parameters (void(*)()) and that accepts a pointer to a CilkProcInfo struct as a parameter. I'm not sure why you'd need the p[0].inlet construct though. Couldn't you just return it as p->inlet?

Oh yeah, and get_proc_slow is the name of the function that returns said function pointer.

edited Oct 28, 2009 at 8:11

answered Oct 28, 2009 at 8:03

Zoli

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1 Comment

AnT stands with Russia Over a year ago

As it has already been noted in other comments, () parameter list in C does not mean "has no parameters`. It means "unspecified number of parameters".

Lara Bailey · Accepted Answer · 2009-10-28 08:21:23Z

0

static void (*get_proc_slow(CilkProcInfo *p)) () {
     return p[0].inlet;
}

Reading from the name out, taking care with the grammar rules: get_proc_slow is a function (with internal linkage) that takes a pointer to a CilkProcInfo struct and returns a pointer to a function taking unspecified arguments and returning no value (void).

(get_proc_slow(f->sig)) (ws, f);

This statement calls the get_proc_slow with an appropriate parameter (f->sig is a pointer to a CilkProcInfo) and then uses the return value (a pointer to a function) to call that function with ws and f as arguments.

answered Oct 28, 2009 at 8:21

Lara Bailey

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