Reputation: 9238
I'm writing a C library which I would like to be usable from both C and C++. At one moment, it should take a callback from the user with 0-3 arguments, which will be called at some pointer later. Like this (a copy of the code is available as GitHub Gist too):
// app_c.c
#include <stdio.h>
#include "lib.h"
double f0(void) {
return 123;
}
double f2(double a, double b) {
return a + b;
}
int main() {
cb_arity = 0;
cb_func = f0;
printf("%f\n", cb_call());
cb_arity = 2;
cb_func = f2;
printf("%f\n", cb_call());
}
I was able to create a pointer to a C function which takes unknown (but still fixed) number of arguments, note it's void (*cb_func)()
, not void (*cb_func)(void)
:
// lib.h
#ifndef LIB_H_
#define LIB_H_
#ifdef __cplusplus
extern "C" {
#endif
extern int cb_arity;
extern double (*cb_func)();
double cb_call(void);
#ifdef __cplusplus
}
#endif
#endif // LIB_H_
// lib.c
#include "lib.h"
#include <stdlib.h>
int cb_arity;
double (*cb_func)();
double cb_call(void) {
switch (cb_arity) {
case 0:
return cb_func();
case 1:
return cb_func(10.0);
case 2:
return cb_func(10.0, 20.0);
case 3:
return cb_func(10.0, 20.0, 30.0);
default:
abort();
}
}
It compiles and runs successfully both on my machine and Wandbox. As far as I understand, no UB is invoked.
Now I would like to make it work in C++ as well. Unfortunately, it looks like I now need reinterpret_cast
because ()
means "no arguments" in C++, not "unknown number of arguments":
// app_cpp.cpp
#include <stdio.h>
#include "lib.h"
int main() {
cb_arity = 0;
cb_func = []() { return 123.0; };
printf("%f\n", cb_call());
cb_arity = 2;
cb_func = reinterpret_cast<double(*)()>(static_cast<double(*)(double, double)>(
[](double a, double b) { return a + b; }
));
printf("%f\n", cb_call());
}
As far as I understand, no UB is invoked here as well: although I convert function pointer double(*)(double, double)
to double(*)(void)
in C++, it's converted back to double(*)(double, double)
in C code right before calling.
Is there any way to get rid of these ugly casts in C++ code? I've tried specifying type of cb_func
as void(*)(...)
, but C++ still won't implicitly convert double(*)(double, double)
to it.
Upvotes: 4
Views: 244
Reputation: 63019
Rather than erase the number of arguments from the callback, you could retain it.
// lib.h
#ifndef LIB_H_
#define LIB_H_
#ifdef __cplusplus
extern "C" {
#endif
typedef struct {
int arity;
union {
void(*zero)(void);
void(*one)(double);
void(*two)(double, double);
void(*three)(double, double, double);
}
} cb_type;
extern cb_type cb;
double cb_call(void);
#ifdef __cplusplus
}
#endif
#endif // LIB_H_
// lib.c
#include "lib.h"
#include <stdlib.h>
cb_type cb;
double cb_call(void) {
switch (cb.arity) {
case 0:
return cb.zero();
case 1:
return cb.one(10.0);
case 2:
return cb.two(10.0, 20.0);
case 3:
return cb.three(10.0, 20.0, 30.0);
default:
abort();
}
}
If you don't expose cb
, you can't mismatch the arity and union member:
// lib.h
#ifndef LIB_H_
#define LIB_H_
#ifdef __cplusplus
extern "C" {
#endif
void register_zero(void(*)(void));
void register_one(void(*)(double));
void register_two(void(*)(double, double));
void register_three(void(*)(double, double, double));
double cb_call(void);
#ifdef __cplusplus
}
#endif
#endif // LIB_H_
Upvotes: 4
Reputation: 9238
Parameterless function declarations is an obsolescent C feature. I would suggest avoiding it.
Essentially, you're trying to circumvent type checking: you want to store an almost arbitrary function in a cb_func
variable, and then be able to call it freely without any signs of danger like explicit casts. However, this code is inherently dangerous: if you mess up cb_arity
, behavior is undefined. Moreover, you can even store double(*)(int, char*)
to cb_func
without any warnings, which is never ok in your example.
On a deeper level, your cb_arity
/cb_func
looks a lot like a tagged union. The canonical way to implement it in C++ is std::variant
(or something function-specific like unique_pseudofunction
). The canonical way to implement it in C is a struct with a "tag" field and an anonymous union
, like this:
struct cb_s {
int arity;
union {
double (*func0)(void);
double (*func1)(double);
double (*func2)(double, double);
double (*func3)(double, double, double);
};
};
extern struct cb_s cb;
Now instead of cb_func = f0
you write cb.func0 = f0
, and instead of cb_func = f2
you write cb.func2 = f2
. Similary in C++ and all casts from lambdas are now gone. The only remaining sign of danger is the underlying union
.
You will have to change your code in two places:
func0
/func1
/... to call, not a big deal.cb_func
. Now it has to know which union member to write into. Presumably, the same code would also write a compile-time constant cb_arity
, so not a big deal agian. If cb_arity
is received from someone else, that someone should then also use tagged union instead of passing cb_arity
and cb_func
separately for better type safety.Sidenote: C++ (as opposed to C) prohibits access to a non-active member of union. This should not affect correctness of the code, because calling a func0
through a pointer to func1
is UB by itself.
Upvotes: 0