Why have a "warning" when initializing an array of function pointers?

Question

i try to initializing an array of function pointers and i have "warning":

ring-buffer.c:57:19: warning: assignment from incompatible pointer type [enabled by default]
RBufP->rbfunc[0] = &RBufPush;
                 ^

but the neighborhood's ok:

/*typedef for func pointer*/
typedef RBRetCod_t (*RBFunc)();

/*RBufP*/
typedef struct {
    RBufSiz_t size; /*size and mask*/
    RBufDat_t rbufdat;
    RBufPoint_t head, tail;
    RBFunc rbfunc[3]; /*announce of function pointers array*/
} RBuf_t;
RBuf_t *RBufP;
...

/*init for func pointers array*/
RBufP->rbfunc[2] = &RBufDel; /*it is ok*/
RBufP->rbfunc[1] = &RBufPull; /*it is ok*/
RBufP->rbfunc[0] = &RBufPush; /*it is bad, why???*/

...

/*body of the functions*/
RBRetCod_t RBufPull(unsigned char *dat)
{
        return RBSUCC;
}
RBRetCod_t RBufDel(void)
{
        return RBSUCC;
}
RBRetCod_t RBufPush(unsigned char dat)
{
        return RBSUCC;
}

please explain to me why the warning occurs in this line: RBufP->rbfunc[0] = &RBufPush;, but in adjacent rows are not there?

could you show how `RBufP` is defined? and if not presented already, whats the way the type is structured. — dhein, May 08 '15 at 10:53
Function types without parameters is an old feature in C, but with some strange behaviours. See http://stackoverflow.com/questions/13950642 — Mats, May 08 '15 at 11:03
Do all of your 3 function have a prototype before the code that's using them ? — nos, May 08 '15 at 11:23
Do you really have to use an array? You could use `struct` instead, and all function pointers could have a correct type. — user694733, May 08 '15 at 11:33

score 7 · Answer 1 · edited May 08 '15 at 13:03

See C11 section 6.7.6.3 §14, specifying when 2 function types shall be considered compatible:

[...] If one type has a parameter type list and the other type is specified by a function declarator that is not part of a function definition and that contains an empty identifier list, the parameter list shall not have an ellipsis terminator and the type of each parameter shall be compatible with the type that results from the application of the default argument promotions. [...]

This is the case for RBufPull and RBufDel, but not RBufPush as unsigned char gets promoted to int.

If you called RBuPush through a pointer of type RBFunc, an int argument would get pushed to the stack, whereas RBufPush would expect an unsigned char. Depending on calling convention and endianness, you'll get incorrect results.

One solution is to change RBufPush to take an int argument. Another one is to use a cast, ie

RBufP->rbfunc[0] = (RBFunc)&RBufPush;

You'll need to cast back to the correct type RBRetCod_t (*)(unsigned char) before calling rbfunc[0].

dhein · Answer 2 · 2015-05-08T11:29:09.587

From ISO/IEC:9899:

J.5.7 Function pointer casts

1 A pointer to an object or to void may be cast to a pointer to a function, allowing data to be invoked as a function (6.5.4).

2 A pointer to a function may be cast to a pointer to an object or to void, allowing a function to be inspected or modified (for example, by a debugger) (6.5.4).

But this rule is limited through:

6.3.2.3 Pointers

[...]

8 A pointer to a function of one type may be converted to a pointer to a function of another type and back again; the result shall compare equal to the original pointer. If a converted pointer is used to call a function whose type is not compatible with the pointed-to type, the behavior is undefined.

So you are trying to acces different functions with the same Pointerobject type. This is undefined behavior and the warning therefor is correct.

Note:

What it seems to me, what you try to achieve would be something like an interface in C#.

But this kind of functionallity is not available in C

@IvanIvanovich: What exactly are you talking about? Sorry don't get it what you are saying. — dhein, May 08 '15 at 11:28
note that it's only UB in case of `RBufPush`, but not `RBufPull` and `RBufDel` due to how empty argument lists work. — Christoph, May 08 '15 at 11:30
@Christoph: You are right and your answer is anyway better than mine. +1 to yours. — dhein, May 08 '15 at 11:31

Why have a "warning" when initializing an array of function pointers?

2 Answers2